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The measure H of “size” of a Diophantine equation (1) suggested in [37] is the following one: substitute 2 in the polynomial P instead of all variables, absolute values instead of all coefficients, and evaluate. In other words, if P has k monomials of degrees d1,..., dk with coefficients a1,..., ak, respectively, then k di H(P)= ∑ |ai|2 . (2) i=1 For example, for the equation x3 + y3 + z3 − 33 = 0, (3) we have H(x3 + y3 + z3 − 33)= 23 + 23 + 23 + 33 = 57. Problem 1.1 for the equation (3) was open at the time the question [37] was asked, but was later solved by Andrew R. Booker [8]. The answer turned out to be “Yes”, and the solution Booker found is
8, 866, 128, 975, 287, 5283 + (−8, 778, 405, 442, 862, 239)3 + (−2, 736, 111, 468, 807, 040)3 = 33 (4) To the best of our knowledge, Problem 1.1 remains open for the equation x3 + y3 + z3 − 114 = 0 with H = 138. At the time of writing, we are unaware of any equation with smaller H for which the Problem 1.1 is formulated and discussed in the literature but remains open. In this paper, we study the solvability of Diophantine equations systematically, starting from equations with H = 0, 1, 2, 3, . . . , and so on. In the process, we review some techniques for solving Diophantine equations, and list certain families of equations which are solvable by these techniques. We then present the “smallest” equations outside of these families, solve some of these equations, and suggest many other equations to the readers as exercises and open questions. We conclude the introduction by discussing why this particular choice of measure of equation “size” has been chosen in this paper. The first obvious property of H is that, for any integer B > 0, there are only finitely many of Diophantine equations with H ≤ B. This property, which we call the finiteness property, fails for many standard “measures of simplicity” of polynomials. For example, if we define the height of a polynomial as the maximum absolute value of its coefficients, then there are infinitely many polynomials with height 1. Similarly, there are infinitely many equations with given degree or given number of variables. Of course, one may define many other “measures of equations size” satisfying the finiteness property, for example, substitute any other constant (such as 3) instead of 2 in (2). We next give some justification of this particular formula and constant 2 in it. For simplicity, let us consider monomials, and discuss how many symbols we need to write down a given monomial. If we do not use power symbol, and write, for example, x3y2 as xxxyy, we need exactly d symbols to write a monomial of degree d and coefficient a = 1. If a 6= 1, we also need about log2(|a|) symbols to write down a in binary (ignoring sign). So, let us define the length of monomial M of degree d with coefficient a as
l(M) := log2(|a|)+ d. Length l(M) has disadvantage of being not always integer, but ordering monomials by l is equivalent to order- ing them by l(M) (|a|)+d d 2 = 2log2 = |a|2 . This is exactly the size H of M defined in (2). So, at least for monomials, H has the meaning of being a monotone transformation of the (approximation of) the number of symbols needed to write M, and constant 2 corresponds to the fact that coefficients are written in binary, which is a reasonable and standard assumption. The discussion above explains why H is a natural choice, but of course does not imply that H is the only possible choice. For example, a natural alternative would be to accept the fact that length can be irrational, not transform it to integers, and just define the length of a polynomial P consisting of monomials of degrees d1,..., dk with coefficients a1,..., ak as k k k k ∑i=1 di l(P)= ∑ log2(|ai|)+ ∑ di = log2 ∏ |ai| · 2 . (5) i=1 i=1 i=1 !
2 In Section 3.4, we will investigate what happens if we order the polynomials by l instead of H, and conclude that we will end up of studying a similar set of equations, just arriving in a different order. The reason is that there are some equations that are hard to solve but at the same time are amazingly simple to write down, and these equations come up reasonably soon in any natural ordering of equations. In this sense, the particular choice how to order equations (by H, by l, or in some other natural way) does not matter too much. The contribution and organization of this work are as follows. Section 2 reviews some known methods and algorithms for determining whether an equation has integer solutions, such as Hasse principle, prime factors analysis using the law of quadratic reciprocity, Vieta jumping technique, etc. We then use these methods to solve, with computer assistance, the Hilbert’s tenth problem for all1 equations of size H ≤ 30, and also for all equations of size H = 31 with a single exception, see equation (59). In Section 3, we consider special classes of equations, such as equations in two variables, symmetric equations, and equations with three monomials, and, in each category, solve all equations up to a certain size. In addition, Section 3.4 solves all equations of length l < 10, and also all equations of length l = 10 with three exceptions. Section 4 concludes the work and lists some directions for further research.
2 General diophantine equations
2.1 H ≤ 16: trivial and well-known equations. This section investigates Problem 1.1 for Diophantine equations (1) for polynomials P with given H = 0, 1, 2, 3, . . . . We used a simple computer program for enumerating all such equations. The program returns equation 0 = 0 for H = 0, equations ±1 = 0 for H = 1, equations ±2 = 0 and ±x = 0 for H = 2, and so on. For values of H ≤ 14, all the equations are uninteresting and belong to at least one of the following families of trivially solvable equations.
• Equations with no variables like 0 = 0, ±1 = 0, ±2 = 0, etc.
• Equations with small solutions, e.g. equation xy + y + 1 = 0 has solution x = 0, y = −1. In this case, the answer to Problem 1.1 is trivially “Yes”, and such equations can be safely excluded from the analysis. Specifically, we have excluded all the equations that have a solution with maxi |xi| ≤ 100. In particular, this excludes equations with no free term, for which P(0,...,0)= 0. In fact, after multiplying by −1 if necessary, we may assume that P(0,...,0) > 0. After this modification, the program returns no equations with H ≤ 4 but returns equations x2 + 1 = 0 and ±2x + 1 = 0 with H = 5.
• Equations in one variable m amx + · · · + a1 x + a0 = 0 (6)
with a0 6= 0, in which any integer solution must be a divisor of a0. In fact, all integer solutions to (6) can be listed in time polynomial in the size of input [12]. We have added the condition to the code to exclude all such equations, which immediately allowed us to exclude all the equations up to H ≤ 8. More generally, this is the way we proceed further in this paper: identify the class of equations for which Problem 1.1 can be solved, modify the code to exclude all equations from this class from further consideration, find the equations with smallest H not excluded so far, and then repeat the process.
• Equivalent equations. The program next returns many equations with H = 9, including, for example, four equations ±2x ± 2y + 1 = 0. As a next step, we call equations equivalent if they can be transformed to each other by substitutions xi → −xi and/or permutations of variables, and exclude all equations except of one from every equivalence class. For example, the program then returns only one equation 2x + 2y + 1 = 0 from the aforementioned four.
1Here, by “all” we mean all equations that has not been solved by other people. For equations solved by others, a reference is given.
3 • We next exclude all linear equations a1x1 + a2x2 + · · · + an xn + b = 0, because they all can be easily solved, in fact in polynomial time [10]. More generally, we may exclude all equations in the form
ax1 + Q(x2,..., xn)= 0 (7)
where a 6= 0 is an integer and Q is a polynomial with integer coefficients. For such equations, Problem 1.1 has a “Yes” answer if and only if Q(x2,..., xn) is a multiple of a for some integers x2,..., xn. Note that (7) covers all linear equations and also some non-linear ones like 3y = x2 + 1.
• Equations that have no real solutions. Obviously, such equations have no integer solutions as well. A simple example is x2 + y2 + 1 = 0 with H = 9. All equations with no real solutions are recognizable in finite time [34]. Moreover, we have also excluded equations for which inequality P(x1,..., xn) ≤ 0 (or P(x1,..., xn) ≥ 0) has a finite number of integer solutions, which also allows to solve Problem 1.1 in finite time. A simple example is 3 − x2 − y2 = 0 with H = 11.
• Equations with no solutions modulo a. We next exclude equations P(x1,..., xn) = 0 for which there exists an integer a ≥ 2 such that P(x1,..., xn) is not a multiple of a for any integers x1,..., xn. Obviously, then the equation has no integer solutions. A simple example is the equation 2xy + 1 = 0 with H = 9, where the left-hand side is always odd. The divisibility condition can be easily checked for any fixed a by enumerating all possible reminders x1,..., xn can give after division by a. Moreover there is an algorithm developed by Ax [3] in 1967 that checks this condition for all values of a in finite time2. This allows us to exclude all equations with H ≤ 12.
• Equation (x2 + 2)y = 1 with H = 13 is the smallest one which has real solutions and also solutions modulo every integer, but still has no integer solutions, because 1 is never an integer. We next exclude x2+2 all equations in the form P1 · P2 · ... · Pm − a = 0,
where Pi are polynomials and a is a constant, for which there is no factorisation a = u1 · ... · um of a such that equations Pi = ui are solvable in integers for all i. For all such equations, the answer to Problem 1.1 is “No”. This allows us to exclude all equations with H ≤ 14.
For all the equations listed above Problem 1.1 is trivial and we have excluded all such equations from the consideration. This automatically excludes all equations with H ≤ 14, and the smallest equation the program classified as non-trivial is the equation y2 = x3 − 3 (8) with H = 15. As a next step, we have also excluded some classes of equations for which Problem 1.1 can be highly non-trivial but is known to be solvable in finite time. Specifically, we have excluded:
• Quadratic equations. A deep theorem of Grunewald and Segal [20] proves the existence of a finite algorithm for solving Problem 1.1 for quadratic equations in any number of variables.
Theorem 2.1. There is an algorithm, which, given any quadratic polynomial
n n n P(x1,..., xn)= ∑ ∑ aij xi xj + ∑ bixi + c i=1 j=1 i=1
with integer coefficients, determines whether equation P = 0 has an integer solution.
2The main theorem in [3] gives an algorithm for solving a diophantine equation modulo every prime p. However, on page 11 of [3], it is remarked that “This gives an algorithm for determining whether a given system of diophantine equations has, for all primes p, a solution over Zp”, where Zp is the set of p-adic integers. This is equivalent to the solvability of the equation modulo every prime power, which in turn is equivalent to its solvability modulo every integer a by the Chinese remainder theorem.
4 For n = 2 variables, there is an implementation available online [1]. Theorem 2.1 allows us to exclude all quadratic equations. • Cubic equations in two variables, the smallest non-trivial example is (8). This is the smallest equation whose solvability problem requires at least some thinking and/or at least minimal background in number theory. However, this equation belongs to a family for which Problem 1.1 is known to be solvable. Indeed, we have the following well-known results. Theorem 2.2. Let P(x, y) be a polynomial with integer coefficients which is irreducible over Q but not absolutely irreducible3. Then there is an algorithm for determining all integer solutions (in fact, all rational solutions) to the equation P(x, y)= 0.
Proof. For every polynomial P(x, y) there exists a unique homogeneous polynomial Q(x, y, z) of the same degree such that P(x, y)= Q(x, y, 1). Solutions to the system of equations ∂Q ∂Q ∂Q = = = 0 ∂x ∂y ∂z are called singular points. If P(x, y) is irreducible over Q but not absolutely irreducible, all rational solutions to P(x, y) = 0 corresponds to singular points. There is a finite number of singular points and an algorithm [32] that can list them all. Then we can check which of these singular points lead to a rational (or integer) solutions to P(x, y)= 0.
Theorem 2.3. There is an algorithm that, given a polynomial P(x, y) of degree at most 3 with integer coefficients, determines4 all integer solutions to the equation
P(x, y)= 0. (9) In particular, Problem 1.1 is solvable for this class of equations.
Proof. A combination of main theorems in [4], [28], and [29] imply the result in the case when P(x, y) is absolutely irreducible. If P(x, y) is irreducible over Q but not absolutely irreducible, then the result follows from Theorem 2.2. Finally, if P(x, y) is reducible over Q, it can be written as a product of a linear polynomial P1(x, y) and a quadratic polynomial P2(x, y) with rational coefficients. Then the equations Pi(x, y)= 0, i = 1, 2 can be solved by known algorithms [10, 1].
While the original algorithm in [4] takes a finite but long time to run, subsequent authors [27, 33] made it practical. An algorithm for finding integer solutions for a lagre class of cubic equations P(x, y)= 0 is implemented in SageMath, which is an open-source and free to use mathematical software system [38]. In particular, if one enters equation (8) in SageMath, the program outputs that it has no integer solutions. We will give a direct proof of this fact in Proposition 2.9 below, but all further equations covered by Theorems 2.2 and 2.3 will be excluded from further analysis.
• In 1969, Baker developed a general method for solving many equations in the form ym = P(x). Theorem 2.4. There is an algorithm for determining: (a) all integer solutions of the equation
2 n y = P(x)= an x + · · · + a1x + a0,
provided that all ai are integers, an 6= 0, and P(x) has at least three simple (possibly complex) zeros;
3 A polynomial P with integer coefficients is called absolutely irreducible if it cannot be written as a product P = P1 · P2 of non-constant polynomials, even if we allow complex coefficients. 4See [29] for the meaning of word determines in case if (9) has infinitely many solutions. For our purposes, it suffices that the algorithm checks whether (9) has a finite number of integer solutions, and if so, lists all the solutions.
5 (b) all integer solutions of the equation
m n y = P(x)= an x + · · · + a1x + a0, (10)
provided that m ≥ 3, n ≥ 3, all ai are integers, an 6= 0, and P(x) has at least two simple zeros.
Theorem 2.4 (a) can be used to solve essentially all two-variable equations that are quadratic in one of the variables (say y). For example, the equation
2y2 + y − x4 − x − 2 = 0
reduces to the question whether 12 − 4 · 2 · (−x4 − x − 2) = 8x4 + 8x + 17 can be a perfect square, which is covered by Theorem 2.4 (a) because this polynomial has 4 simple zeros. The next exercise suggests the reader to try to solve this problem directly.
Exercise 2.5. Does there exist an integer x such that 8x4 + 8x + 17 is a perfect square?
In the same way, if the equation f (x)y2 + g(x)y + h(x)= 0 (11) has any integer solution with f (x) 6= 0, then D(x) = g2(x) − 4 f (x)h(x) must be a perfect square. If D(x) has at least three simple zeros, there are finitely many such x and they can all be found by Theorem 2.4 (a). We can then check if any of these x leads to a solution of (11). The values of x such that f (x)= 0 can also be checked directly, unless f (x) ≡ 0. Finally, if f (x) ≡ 0 then (11) reduces to the question h(x) whether g(x) can be an integer (or g(x)= h(x)= 0), which can also be solved in finite time. The simplest equation not covered by the previous results but covered by Theorem 2.4(b) is
y3 = x4 + 2, (12)
because P(x) = x4 + 2 has 4 simple zeros. We also remark that Theorem 2.4(b) covers slightly more general equations aym = P(x) (13) under the same conditions on m and P(x) and for any integer a. Indeed, we can consider a cases, x = az, x = az + 1,..., x = az +(a − 1), where z is a new integer variable, and in each case, equation (13) is either not solvable because the right-hand side is not divisible by a, or, after cancelling a, reduces to an equation in the form (10). All such equations will be also excluded from further analysis.
The types of equations listed above are rigorously defined, and we were able to make our program to exclude them automatically. However, there are other types of equations which program returns but we will consider them uninteresting and ignore. These are:
• Equations which do not have integer solutions because of trivial inequality-based obstructions, such as < < a (a) If 0 |a| |b|, then b cannot be an integer, (b) If a2 < b < (a + 1)2 for some integer a, then b cannot be a perfect square.
An example of (a) is the equation
y(2 + y + x2 + z2)= x2 + 1.
Because x2 + 1 > 0, either the factors on the left-hand side are both positive or both negative. If both positive, 2 + y + x2 + z2 > x2 + 1, if both negative, y < −(x2 + z2 + 2) and |y| > x2 + 1. In both cases a contradiction by (a).
6 An example of (b) is the equation5
y2 − (x2 + z2)y − 3x + 2 = 0.
If considered as a quadratic equation in y, its determinant is D = (x2 + z2)2 + 12x − 8. It is easy to check that for |x|≥ 7 we have
(x2 + z2 − 1)2 < D < (x2 + z2 + 1)2,
and the values with |x| ≤ 6 can be checked directly. Also, D 6=(x2 + z2)2 because 12x − 8 is never 0. Hence, D cannot be a perfect square. It is difficult to rigorously define an exact family F of the equations “with no integer solutions because of inequality-based obstructions” and exclude them automatically. Instead, we allow the program to return such equations but will ignore them in the rest of the paper.
• Equations that can be reduced to the equations with smaller H. If polynomial P in the equation P = 0 is representable as a product of polynomials with smaller H, the resulting equation m ∏ Pi(x1,..., xn)= 0 (14) i=1
has an integer solution if and only if at least one of the equations Pi(x1,..., xn)= 0, i = 1,..., m has an integer solution. Obviously, there are many other ways to reduce an equation to a smaller one. Initially, the program returned not only equation (8) with H = 15, but, for example, its variation (y + 1)2 = x3 − 3 with H = 20. After we excluded all 2-variable cubic equations, the program did not give up, and, when reached H = 27, returned 3-variable equations (y + z)2 = x3 − 3 and (yz)2 = x3 − 3. All these equations can be reduced to the smaller equation (8) by a trivial change of variables. In general, if an equation A can be reduced to a smaller equation B, it does not mean that A is trivial, because B may be far from trivial itself. However, because we study all equations systematically in order, any such reduction implies that equation B has been already considered, hence A can be ignored.
After excluding (or ignoring) all the types of equations listed above, we get no equations with H ≤ 16, which implies that all such equations are either trivial or belong to some family for which a finite algorithm for Problem 1.1 is well-known.
2.2 H ≥ 17: prime factors of quadratic forms The smallest equation which is not excluded by the criteria described in Section 2.1 is −x2y + y2 + z2 + 1 = 0 with H = 17. This equation can be written as
y(x2 − y)= z2 + 1. (15)
In fact, the only equations the program returns for H ≤ 20 are (15) and a similar equation
y(x2 + 3)= z2 + 1, (16) with H = 19. These equations are not completely trivial, but they are also not difficult. The key idea is to understand what prime numbers can be divisors of z2 + 1. For example, it is easy to see that z2 + 1 is never divisible by 3, because if z ≡ 0, 1, or 2(mod 3), then z2 + 1 ≡ 1, 2, or 2(mod 3), respectively6. More generally, we have the following well-known fact.
5There are smaller equations which can be solved by this method, but they are covered by Theorem 2.4. 6For integer m, we will write a ≡ b(mod m) if a − b is a multiple of m.
7 Proposition 2.6. For any integer z, all odd prime factors p of z2 + 1 are in the form p = 4k + 1.
2 2 p−1 2 p−1 Proof. Let p be any odd prime divisor of z + 1. Then z ≡−1(mod p), which implies that z =(z ) 2 ≡ p−1 p−1 p−1 (−1) 2 (mod p). But Fermat’s little theorem implies that z ≡ 1(mod p). Hence 1 ≡ (−1) 2 (mod p). p−1 Because 1 6≡ −1(mod p), this implies that 2 is even, hence p = 4k + 1 for some integer k. We now apply Proposition 2.6 to solve the equations (15) and (16). We remark that equation (15) has been first solved by Victor Ostrik in the comments section to the mathoverflow question [37].
Proposition 2.7. Equations (15) and (16) have no integer solutions.
Proof. We start with the easier equation (16). If x ≡ 0, 1, 2 or 3(mod 4), then x2 ≡ 0, 1, 0 or 1(mod 4), respectively. In particular, we never have x2 ≡ 2 or 3(mod 4). Hence, if x is odd, then x2 + 3 is divisible by 4. But then (16) implies that z2 + 1 is divisible by 4, or z2 ≡ 3(mod 4), a contradiction. If x is even, then 2 2 x + 3 is odd and x + 3 = ∏i pi for some odd prime factors pi. Then (16) implies that each pi is also a prime 2 2 factor of z + 1, hence pi ≡ 1(mod 4) by Proposition 2.6. But then x + 3 = ∏i pi ≡ ∏i 1 = 1(mod 4), and x2 ≡ 2(mod 4), again a contradiction. We next consider equation (15). First note that if y ≤ 0 then x2 − y ≥ x2 ≥ 0, hence y(x2 − y) ≤ 0, or z2 + 1 ≤ 0, a contradiction. Hence y > 0 which implies that x2 − y > 0. Now, Proposition 2.6 implies that all odd prime factors of z2 + 1 are in the form p = 4k + 1 for integer k. If z2 + 1 is odd, then all prime factors of y and x2 − y must be in this form as well. The product of any number of such primes is 1 modulo 4, hence, y ≡ 1(mod 4) and x2 − y ≡ 1(mod 4). Thus, x2 ≡ 2(mod 4), a contradiction. If, conversely, z2 + 1 is even, then its prime factorization has exactly one prime factor of 2, which goes to either y or x2 − y. In the first case, y ≡ 2(mod 4) and x2 − y ≡ 1(mod 4). In the second case, y ≡ 1(mod 4) and x2 − y ≡ 2(mod 4). In both cases, x2 ≡ 3(mod 4), again a contradiction. Hence, (15) has no integer solutions.
Proposition 2.7 finishes the analysis of all equations with H ≤ 20. For H = 21, the only equation the program returns is y(x2 + 2)= 2z2 − 1. (17) For H = 22, it returns several equations, including
y(x2 − 1 − y)= z2 + 4 (18) and y(x2 + 4)= z2 + 2. (19) To solve equations (18) and (19) by the method as in the proof of Proposition 2.7, we need to understand the possible odd prime factors of z2 + 2 and z2 + 4. Let us consider, more generally, odd prime factors of z2 − a for any integer a. Given an odd prime p and integer a with gcd(a, p) = 1, if there exists an integer z 2 a such that z ≡ a(mod p), we say that a is a quadratic residue modulo p and write p = 1, otherwise we say a that a is a quadratic non-residue modulo p and write p = −1. In this notation, Proposition 2.6 states that if −1 p = 1 then p ≡ 1(mod 4). In fact, the converse direction is also true, and this can be written in a compact form as −1 p−1 =(−1) 2 . (20) p Moreover, we have the formula 2 p2−1 =(−1) 8 , (21) p the multiplicative law ab a b = , (22) p p p
8 and, most importantly, the law of quadratic reciprocity, which states that if p and q are distinct odd prime numbers, then p q p−1 q−1 =(−1) 2 2 . (23) q p a These formulas suffice to easily compute p in general. For example,
2 −4 −1 2 −1 p−1 = = =(−1) 2 (24) p p p p and −2 −1 2 p−1 p2−1 (p−1)(p+5) = =(−1) 2 (−1) 8 =(−1) 8 . (25) p p p p p−1 3−1 p−1 3 2 2 2 As another example, for odd p 6= 3, we have 3 p =(−1) =(−1) , which easily implies that 3 = 1 ⇔ p ≡ 1 or 11(mod 12) (26) p and −3 = 1 ⇔ p ≡ 1(mod 3). (27) p a Exercise 2.8. Use (20)-(23) to compute p for all integers |a| ≤ 12 which are not perfect squares. We next use formulas (24) and (25) to solve equations (8), (18) and (19). Proposition 2.9. Equations (8), (18) and (19) have no integer solutions. Proof. We start with equation (8). We first remark that x cannot be divisible by 4, because this would imply that y2 + 3 ≡ 0(mod 16), or y2 ≡ 13(mod 16). One may check directly that this is impossible. We next rewrite the equation as y2 + 4 = x3 + 1 =(x + 1)(x2 − x + 1). Formula (24) implies that all odd prime factors of y2 + 4 are in the form p = 4k + 1 for integer k. Hence, the same is true for all prime factors of odd positive integer x2 − x + 1. The product of any number of such primes is 1 modulo 4, hence, x2 − x + 1 ≡ 1(mod 4), and x(x − 1) is divisible by 4. Because x is not divisible by 4, we must have x ≡ 1(mod 4). But then x3 + 1 ≡ 2(mod 4), which implies that y2 + 4 ≡ 2(mod 4), or y2 ≡ 2(mod 4), a contradiction. We next consider equation (18). Because z2 + 4 > 0, factors y and x2 − 1 − y have the same sign. Because their sum x2 − 1 ≥ −1, they cannot be both negative hence they are both positive. By (24), z2 + 4 has no 2 2 m prime factors in the form p = 4k + 3. Hence, y(x − 1 − y)= z + 4 = 2 ∏ pi, where all pi are in the form 4k + 1, and m can be 0, 2 or 3. In particular, y 6≡ 3(mod 4) and x2 − 1 − y 6≡ 3(mod 4). (28) If x is even, (28) is possible only if y ≡ 1(mod 4) (then x2 − 1 − y ≡ 2(mod 4)) or y ≡ 2(mod 4) (then x2 − 1 − y ≡ 1(mod 4)). But then in both cases we have m = 1, a contradiction. If x is odd, then (28) is possible only if y ≡ 0(mod 4) (then x2 − 1 − y ≡ 0(mod 4)) or y ≡ 2(mod 4). In the first case, we have m ≥ 4, a contradiction. In the second case, y = 8r + 2 or 8r + 6. If y = 8r + 6 = 2(4r + 3), then it must have a prime factor in the form 4k + 3, a contradiction. Hence, y = 8r + 2. Because x is odd, x2 ≡ 1(mod 8). Then x2 − 1 − y ≡ 6(mod 8), and must therefore have a prime factor in the form 4k + 3, again a contradiction. Finally, consider equation (19). Let p be an arbitrary prime factor of x2 + 4. If p = 2, then x2 + 4 is even, hence x is even as well. But then x2 + 4 ≡ 0(mod 4), hence z2 + 2 ≡ 0(mod 4) and z2 ≡ 2(mod 4), a contradiction. Thus, p must be odd. Then (24) implies that p ≡ 1(mod 4), hence p ≡ 1(mod 8) or p ≡ 5(mod 8). Next, (19) implies that p is also a prime factor of z2 + 2, hence (25) excludes the possibility that p ≡ 5(mod 8). Hence, all the prime factors of x2 + 4 are in the form p = 8k + 1. But then x2 + 4 ≡ 1(mod 8), and x2 ≡ 5(mod 8), which is a contradiction.
9 We next consider equation (17). To solve it by the same method, we need to understand possible prime factors of 2z2 − 1. The following Proposition describes, more generally, the prime factors of an arbitrary quadratic form ax2 + bxy + cy2.
Proposition 2.10. Let a, b, c be integers such that gcd(a, b, c) = 1, and let D = b2 − 4ac. Let p be a prime 2 2 D factor of ax + bxy + cy for some integers x, y. Then either p is a divisor of 2D, or p = 1, or p is a common divisor of x and y. Proof. Assume that an odd prime p is a divisor of ax2 + bxy + cy2, but not a divisor of D and not a common divisor of x and y. By symmetry, we may assume that p is not a divisor of y, so that gcd(y, p)= 1. Then there exists an integer z such that yz ≡ 1(mod p). As p is a divisor of ax2 + bxy + cy2, we have
0 ≡ 4az2(ax2 + bxy + cy2)= z2(2ax + by)2 − D(yz)2 ≡ (2axz + byz)2 − D(mod p).
D Hence, p = 1. Let us apply Proposition 2.10 to some specific quadratic forms.
Corollary 2.11. Let x, y be integers and let p be an odd prime that is not a common divisor of x and y. Then:
if p is a prime factor of then x2 + y2 p ≡ 1(mod 4) x2 + 2y2 p ≡ 1 or 3(mod 8) x2 − 2y2 p ≡ 1 or 7(mod 8) x2 + 3y2 p = 3 or p ≡ 1(mod 3) x2 − 3y2 p = 3 or p ≡ 1 or 11(mod 12)
Proof. Let us prove for example the statement about prime divisors of x2 − 3y2. The proofs of all other statements are similar. We will apply Proposition 2.10 with a = 1, b = 0, c = −3, hence D = b2 − 4ac = 12. If p 6= 3 is an odd prime, it is not a divisor of 2D = 24. Hence, if p is a divisor of x2 − 3y2 and gcd(p, x, y)= 2 12 12 3 2 3 3 1, Proposition 2.10 implies that p = 1. Because p = p p = p , this implies that p = 1, and (26) implies that p ≡ 1 or 11(mod 12). We will now use Corollary 2.11 to solve equation (17).
Proposition 2.12. Equation (17) has no integer solutions.
Proof. Let p be any prime factor of x2 + 2. Then p is also a prime factor of 1 − 2z2. Because 1 − 2z2 is odd, p 6= 2. Because x2 + 2 = x2 + 2 · 12, Corollary 2.11 implies that p ≡ 1(mod 8) or p ≡ 3(mod 8). On the other hand, because 1 − 2z2 = 12 − 2z2, Corollary 2.11 also implies that p ≡ 1(mod 8) or p ≡ 7(mod 8). Hence, all the prime factors of x2 + 2 are in the form p = 8k + 1. But then x2 + 2 ≡ 1(mod 8), and x2 ≡ 7(mod 8), which is a contradiction.
Proposition 2.12 finishes the analysis of the equations with H ≤ 21. With H = 22, the program returns sign variations of equation (19), y(x2 + 4) = z2 − 2 and y(x2 − 4) = z2 − 2, which can be treated in a way similar to (19). There is also one more equation with H = 22 of different type, which we will discuss in the next subsection. For higher H, the number of equations of this type increases, but they are all similar to (15)-(19), and can be solved by exactly the same methods. The following exercise suggests the reader to use Corollary 2.11 to solve the equations similar to (15)-(19) up to H ≤ 26. Exercise 2.13. Prove that equations y(x2 ± 4)= 2z2 − 1, y(x2 + 4)= 2z2 + 1,
10 y(x2 + 4 − y)= z2 + 1, y(x2 + 2)= z2 +(z + 1)2, x(x2 + y2 + 2)= z2 + 1, y(x2 + z2 − y)= x2 + 1, y(z2 − x2 + y)= x2 − 2, y(2 − x2 + 2y)= z2 − 2 and y(2 + x2 − 2y)= z2 + 2 have no integer solutions.
Starting with H ≥ 27, equations which can be solved by a straightforward application of Corollary 2.11 will not be even listed.
2.3 H ≥ 22: Markoff-type equations and Vieta jumping The only equation with H ≤ 22 that is not excluded by the criteria of Section 2.1 and is not of the type discussed in Section 2.2 is x2 + y2 − z2 = xyz − 2, (29) with H = 22. This is an equation of the type
ax2 + by2 + cz2 = dxyz + e (30) which generalizes famous Markoff equation
x2 + y2 + z2 = 3xyz.
The standard approach in solving such equations is the techniques of Vieta jumping and infinite descent. One assumes the existence of a solution, and then uses this solution to produce another solution, which is strictly “smaller” in a certain sense. The process is then repeated to get a contradiction. We next illustrate this technique by presenting the solution of Problem 1.1 for equation (29), developed by Will Sawin and Fedor Petrov [6].
Proposition 2.14. Equation (29) has no integer solutions.
Proof. By contradiction, assume that an integer solution to (29) exists, and choose some specific solution (x, y, z). Let us replace x by a new variable t, to get a quadratic equation in t
t2 − yz · t + y2 − z2 + 2 = 0.
Because (x, y, z) is a solution to (29), one root of this quadratic equation is t1 = x. By the properties of quadratic equations, the second root t2 satisfies t1 + t2 = yz, or t2 = yz − x1 = yz − x. In other words, if (x, y, z) is a solution to (29), then (yz − x, y, z) is also a solution. This can be also checked directly:
(yz − x)2 − (yz − x)yz =(yz − x)(yz − x − yz)= x2 − xyz = −y2 + z2 − 2, where the last equality follows from the fact that (x, y, z) solves (29). Now, fix any solution (x, y, z) with |x| + |y| + |z| minimal. First, assume that |x| = max(|x|, |y|, |z|). Because |x| + |y| + |z| is minimal, we must have |x| ≤ |yz − x|. But then
x2 ≤ |yz − x||x| = |y2 + 2 − z2|,
11 where the equality follows from (29). Because |x| = max(|x|, |y|, |z|), the last inequality is possible only if z2 ≤ 2. However, (29) with z = 0 and z = ±1 reduces to equations x2 + y2 + 2 = 0 and x2 ∓ xy + y2 + 1 = 0, and none of them has integer solutions. The case |y| = max(|x|, |y|, |z|) can be excluded similarly, hence it must be |z| = max(|x|, |y|, |z|). If |z| = |x|, then z2 = x2, and (29) implies that y is a divisor of 2, but trying all divisors we find no solutions, a contradiction. The case |z| = |y| is excluded similarly, hence |z| > max(|x|, |y|). By changing signs, we may assume that x, y > 0. If z ≥ 0, then z > max(x, y), and
z2 + xyz ≥ (x + 1)2 + y2 > x2 + y2 + 2, which is a contradiction with (29). Hence, we must have z < 0. In this case, note that (x, y, −xy − z) is another solution to (29). Indeed,
(−xy − z)2 + xy(−xy − z)=(−xy − z + xy)(−xy − z)= z2 + xyz = x2 + y2 + 2.
Note that z(−xy − z)= −x2 − y2 − 2 < 0, hence z < 0 implies that −xy − z > 0. Because |x| + |y| + |z| is minimal, we must have |− xy − z| ≥ |z|. But |− xy − z| = −xy − z, and |z| = −z, and inequality −xy − z ≥−z is impossible if x, y > 0. Exercise 2.15. By adapting the proof of Proposition 2.14, prove that equation
x2 + y2 − z2 = xyz − 6, (31) has no integer solutions.
The next equation of type (30) the program returns is
x2 + y2 + z2 = xyz − 4, (32) which has H = 24. This is a special case of well-known equation
x2 + y2 + z2 = dxyz + e. (33)
In 1953, Mordell [24] showed how to use Vieta jumping method to check the solvability of (33) for any d and e. In particular, it is well-known that (32) has no integer solutions. You will be asked to provide a direct proof of this fact in Exercise 2.19 below. The only equation with H ≤ 25 we have not discussed so far is the equation
y2 − x2yz + z2 + 1 = 0. (34)
The proposition below uses Vieta jumping technique to investigate a more general equation
y2 − ayz + z2 + C = 0, (35) where a > 0 and C > 0 are integer parameters.
C Proposition 2.16. If equation (35) has any integer solutions, then either a−2 is a perfect square or there is a < C−1 solution to (35) such that 0 z ≤ 2 . Proof. Because ayz = y2 + z2 + C > 0, y and z are non-zero and of the same sign. Hence, by changing their signs if needed, we may assume that y > 0 and z > 0. If there is a solution with y = z, then (35) implies that 2 C 2 y (a − 2) = C, and a−2 = y is a perfect square. Otherwise y 6= z in any solution. Let us choose a positive solution with max(y, z) minimal. By symmetry, we may assume that y > z. Then, by considering (35) as quadratic equation in y, we conclude that it has another solution y′, such that yy′ = z2 + C. Because y > 0 2 > ′ z2+C > and z + C 0, we have y = y 0. Because we have selected a solution with max(y, z) minimal, we ′ 2 ′ 2 2 2 C−1 must have y ≥ y. Hence z + C = yy ≥ y ≥ (z + 1) = z + 2z + 1, and z ≤ 2 .
12 C < C−1 For example, with C = 1, a−2 may be a perfect square only if a = 3, while inequality 0 z ≤ 2 = 0 is impossible. Hence, equation y2 − ayz + z2 + 1 = 0 may have a solution only if a = 3. Because 3 is not a perfect square, this implies that equation (34) has no integer solutions.
Exercise 2.17. Use Proposition 2.16 to prove that equations
y2 − x2yz + z2 + 4 = 0 and y2 − x2yz + z2 + 5 = 0 have no integer solutions.
Equation (34) has been the last one with H ≤ 25. One of the equations with H = 26 the program returns is − yx2 − xyz + y2 + z2 + 2 = 0. (36) This equation is not of the form (30) or (35). However, we next show that it can also be solved using Vieta jumping technique. Let us replace the variable z with w − x for a new variable w, and variable y by t − 2 for a new variable t. Then we get
−(t − 2)x2 − (t − 2)x(w − x)+(t − 2)2 +(w − x)2 + 2 = 0, which simplifies to x2 + w2 + t2 − 4t = xwt − 6. This equation is almost of type (33), except that is has extra linear term −4t. We next extend Mordell’s method [24] for solving (33) to equations with such linear terms.
Proposition 2.18. There is an algorithm that, given integers a, b, c, d, e, determines in finite time (in fact, in time polynomial in a, b, c, d, e) whether equation
x2 + y2 + z2 + ax + by + cz = dxyz + e. (37) has an integer solution.
Proof. We may assume that d ≥ 0, otherwise replacing d with −d and (say) a with −a leads to an equivalent equation. In fact, we may assume that d > 0, because the equation is easy for d = 0. Let us rewrite (37) as
(x + a/2)2 +(y + b/2)2 +(z + c/2)2 = dxyz + f , (38) where f = e +(a2 + b2 + c2)/4. Note that for every fixed x equation (37) is a 2-variable quadratic equation, and its solvability can be checked easily. Hence, we can easily check if there exist a solution with x (and similarly with y or z) belonging to any bounded range. In particular, we can check whether there is any solution with |xyz| ≤ | f /d|, and if so, stop, and otherwise assume that |xyz| > | f /d|, or d|xyz| > | f |. Because dxyz + f ≥ 0 as a sum of squares, this implies that xyz ≥ 0. It suffices to develop an algorithm A that, given a, b, c, d, e, determines whether there exist solutions to (37) in non-negative integers. Indeed, given such an algorithm, we can apply it four times with inputs (i) a, b, c, d, e, (ii) −a, −b, c, d, e, (iii), −a, b, −c, d, e and (iv) a, −b, −c, d, e, and conclude that (37) has an integer solution if and only if the algorithm outputs “Yes” at least once in these four runs. We next check for solutions such that (y + b/2)2 +(z + c/2)2 ≤ f , or (x + a/2)2 +(z + c/2)2 ≤ f , 2 2 < |a| < |b| < |c| or (x + a/2) +(y + b/2) ≤ f , or x 5 2 , or y 5 2 , or z 5 2 , output “Yes” if find a solution, and |a| otherwise assume that the opposite inequalities hold for any solution. In particular, x ≥ 5 2 implies that 6 a 4 x ≥ x + ≥ x. (39) 5 2 5
13 Now fix any solution with x + y + z minimal. We may assume that x ≥ y ≥ z ≥ 0, because other ordering can be treated similarly. Since we have excluded solutions with (y + b/2)2 +(z + c/2)2 ≤ f , equation (38), considered as a quadratic in x, has two positive roots, whose sum is dyz − a. If one solution is x, another one is dyz − a − x. Because we have selected a solution with x + y + z minimal, we must have x ≤ dyz − a − x, or 2(x + a/2) ≤ dyz. Together with (39), this implies that 2(4x/5) ≤ dyz, or
5 x ≤ dyz. 8
36 2 2 36 2 2 36 2 2 On the other hand, (39) implies that 25 x ≥ (x + a/2) . Similarly, 25 y ≥ (y + a/2) , 25 z ≥ (z + a/2) , and, by (38), 36 (x2 + y2 + z2) ≥ dxyz + f . (40) 25 36 2 Because x ≥ y ≥ z, this implies that 25 (3x ) ≥ dxyz + f , or 108 x dyz − x ≤− f . 25 108 > If dyz − 25 x 0, then we must have − f ≥ 0 and x ≤− f . Existence of such solutions can be checked by a 108 direct search, so we may assume that dyz − 25 x ≤ 0, or 25 x ≥ dyz. 108 Next observe that (40) implies that
25 25 x2 + y2 + z2 − dxyz ≥ f , 36 36 or 25 2 252 25 x − dyz + y2 + z2 − d2y2z2 ≥ f . 72 722 36 5 25 25 5 25 5 Because 8 dyz ≥ x ≥ 108 dyz, we have x − 72 dyz ≤ max{ 18 dyz, 216 dxyz} = 18 dyz, hence
5 2 252 25 dyz + y2 + z2 − d2y2z2 ≥ f , 18 722 36 or 25 25 y2 + z2 − d2y2z2 ≥ f 576 36 Because y ≥ z, we have 2y2 ≥ y2 + z2, and we obtain
25 2y2Q(z) ≥ f , 36
25d2 2 2 25 where Q(z) = 1 − 1152 z . This inequality is possible if either Q(z) ≥ 0 or |2y Q(z)| ≤ 36 | f |. Both possibilities can be checked directly in finite time.
Exercise 2.19. Use the algorithm from the proof of Proposition 2.18 to check whether equations (32), (36) and
x2 + y2 + z2 = xyz − 8. (41) have any integer solutions.
14 So far, we have only considered equations in the form (30) with |a| = |b| = |c| = 1. The smallest equation of type (30) returned by the program which does not satisfy this condition is
y2 + z2 − 2x2 = xyz − 5. (42) with H = 29. The following Proposition demonstrates that Vieta jumping technique applies with no problems to such equations as well.
Proposition 2.20. Equation (42) has no integer solutions.
Proof. By contradiction, assume that an integer solution to (42) exists, and choose a solution (x, y, z) with |x| + |y| + |z| minimal. By changing signs and exchanging y and z if necessary, we may assume that y ≥ z ≥ 0. In fact, with z = 0 (42) reduces to y2 − 2x2 = −5 that is not solvable in x, y, hence we may assume that y ≥ z > 0. First assume that x ≤ 0. If y and z would be both odd, then x would also be odd, but then modulo 8 (42) yz reduces to 0 = −4, a contradiction. Hence, 2 is an integer. Considering (42) as a quadratic equation in x, it ′ yz ′ y2+z2+5 < has two integer solutions, x and x = − 2 − x. Note that xx = − 2 0. Because x ≤ 0, we must have x′ > 0. Because we have selected a solution with |x| + |y| + |z| minimal, we must have |x′| ≥ |x|, or ′ yz > x ≥−x, or − 2 − x ≥−x, hence yz ≤ 0, a contradiction with y ≥ z 0. The remaining case is x > 0. Then considering (42) as a quadratic equation in y, we conclude that it has two integer solutions, y and y′ = xz − y. Because we have selected a solution with |x| + |y| + |z| minimal, we must have |y′| ≥ |y| = y, hence either −y′ ≥ y or y′ ≥ y. In the first case, we have −y′ = y − xz ≥ y, or xz ≤ 0, a contradiction. In the second case, z2 − 2x2 + 5 = yy′ ≥ y2. Because y ≥ z, this may be possible only if −2x2 + 5 ≥ 0. Because x > 0, this implies that x = 1, but in this case (42) reduces to y2 − yz + z2 = −3, which has no solutions in y, z.
2.4 H ≥ 26: Sum of squares values of a polynomial In the previous sections we have solved all equations with H ≤ 25 and some equations with H ≥ 26. The only equations with H = 26 we have not discussed so far are
y(x2 − y)+ z(x2 − z)= 2, (43)
y(x3 − y)= z2 + 2 (44) and y2 + y(x2 + 1)+ z2 + x2 + 4 = 0. (45) The direct application of Corollary 2.11 seems not to work for these equations. For example, in the equation (44), if z is odd, Corollary 2.11 implies that all prime factors p of z2 + 2 are of the form p = 8k + 1 or p = 8k + 3. Hence, if y and x3 − y are positive, then prime factorizations of y and x3 − y must also consist only of primes of this form, hence y and x3 − y are 1 or 3 modulo 8. But then x3 = y +(x3 − y) must be 2, 4 or 6 modulo 8, which is a contradiction. However, it is unclear how to use Corollary 2.11 to get a contradiction 2 if z is even. In this case, z + 2 = 2 ∏ pi, where all pi are at the form pi = 8k + 1 or pi = 8k + 3. However, a simple example x = 3, y = 9 demonstrates that the prime factorization of y(x3 − y) may also have this form. Proposition 2.22 below, proved by Will Sawin, solves equations (43) and (44) by first reducing them to the instances of following problem: given a polynomial P in one variable, does there exist an integer x such that P(x) can be represented as a sum of two squares? This problem can then be solved using the following Corollary from Proposition 2.10.
Corollary 2.21. Let a, b, c be integers such that gcd(a, b, c)= 1, let D = b2 − 4ac, and let p be an odd prime D not dividing D such that p = −1. Then, for any integers x and y, p enters the prime factorization of m = ax2 + bxy + cy2 with even multiplicity.
15 Proof. Let be the maximal integer that such k is a common divisor of and , and let x and y . k p x y u = pk v = pk Then p is not a common divisor of u and v, hence, by Proposition 2.10, p cannot be a divisor of au2 + buv + 2 m . Hence, enters the prime factorization of with even multiplicity . cv = p2k p m 2k Proposition 2.22. Equations (43) and (44) have no integer solutions.
Proof. Equation (43) can be rewritten as
(y − x2/2)2 +(z − x2/2)2 = x4/2 − 2, or (2y − x2)2 +(2z − x2)2 = 2x4 − 8 = 2(x2 − 2)(x2 + 2). Corollary 2.21 implies that if a sum of squares has a prime factor p congruent to 3 modulo 4, then the multi- plicity of p must be even. If x is even, the right-hand side is equal to 8 times something congruent to 3 modulo 4, and cannot be a sum of two squares. Hence, we may assume that x is odd. In this case, numbers x2 − 2 and x2 + 2 are odd and relatively prime, because if they had a common odd prime factor p, it would also be a divisor of (x2 + 2) − (x2 − 2)= 4, a contradiction. Because x2 + 2 is odd, positive, and congruent to 3 mod 4, it must have a prime factor p congruent to 3 modulo 4 of odd multiplicity. Because x2 − 2 and x2 + 2 are relatively prime, p is not a factor of x2 − 2. Hence, the product 2(x2 − 2)(x2 + 2) has a prime factor congruent to 3 modulo 4 of odd multiplicity, and therefore cannot be a sum of two squares. Similarly, equation (44) can be written as
(y − x3/2)2 + z2 =(x3/2)2 − 2, or (2y − x3)2 +(2z)2 = x6 − 8 =(x2 − 2)(x4 + 2x2 + 4). Once again, if x is even then the right-hand side is 8 times something congruent to 3 modulo 4 and cannot be a sum of two squares. If x is odd then x2 − 2 and x4 + 2x2 + 4 are odd and relatively prime, because if they had a common (odd) prime factor p, it would divide gcd(x2 − 2, x4 + 2x2 + 4) = 12, hence p must be 3, but x2 − 2 is never divisible by 3. But then x4 + 2x2 + 4 is odd, positive, and congruent to 3 mod 4, hence it must have a prime factor p congruent to 3 mod 4 of odd multiplicity. Then p is a prime factor of odd multiplicity of the product (x2 − 2)(x4 + 2x2 + 4), hence this product cannot be a sum of two squares.
Exercise 2.23. Use the method in the proof of Proposition 2.22 to show that equation (45) has no integer solutions.
Proposition 2.22 and Exercise 2.23 finish the analysis of all equations with H ≤ 26. We next discuss the equations in the range 27 ≤ H ≤ 29. We start by listing the equations in this range which can be solved by exactly the same method as in the proof of Proposition 2.22.
Exercise 2.24. Use the method in the proof of Proposition 2.22 to show that equations
y2 + z2 = x4 − 4,
y2 − 2x2y + z2 = −4 and y2 − x2y + z2 = 2x2 − 5 have no integer solutions.
16 Proposition 2.22 has been proved using the information about prime factors of a2 + b2. In a similay way, Corollary 2.21 may also be used to determine for which polynomials P there exist x such that P(x) is repre- sentable in the forms a2 ± 2b2, a2 ± 3b2, and so on. To illustrate the technique, we will solve the following equation 3y2 − x2y − z2 + 3 = 0 (46) with H = 27. This equation can be rewritten as
(6y − x2)2 − 3(2z)2 = x4 − 36 =(x2 − 6)(x2 + 6).
Hence, the problem is whether there exists x such that P(x)= x4 − 36 can be represented as a2 − 3b2.
Proposition 2.25. Equation (46) has no integer solutions.
Proof. If x is even, then (46) implies that 3y2 − z2 + 3 is divisible by 4. If y even, then z2 is 3 modulo 4, while if y odd, then z2 is 2 modulo 4, in both cases a contradiction. If x is divisible by 3, then (46) implies that z is divisible by 3 as well, hence 3(y2 + 1) = x2y + z2 is divisible by 9, but then y2 + 1 is divisible by 3, a contradiction. Hence, x is coprime with 6, and so is x4 − 36. Then x4 − 36 may have only prime factors such that p ≡ ±1(mod 12) (call ot type one) or such that p ≡ 5(mod 12) or p ≡ 7(mod 12) (call it type 2). If x4 − 36 = (x2 − 6)(x2 + 6) is representable as a2 − 3b2 then all type two prime factors must be of even multiplicity by Corollary 2.21. Also, no prime factor p of type two can be a common prime factor of x2 − 6 and x2 + 6, otherwise p would be a factor of 12, but 12 does not have type two prime factors, a contradiction. Hence, x2 + 6 can have only type one prime factors, as well as type two prime factors of even multiplicity. But this implies that x2 + 6 is ±1 modulo 12, which is a contradiction. Exercise 2.26. Use the method in the proof of Proposition 2.25 to show that equations
y2 − x2y + 2z2 = x2 − 3,
y2 − x2y − 2z2 = −x2 − 3, y2 +(2 − x2)y − 2z2 = −4, 2y2 +(1 − x2)y + 2z2 = −2, y2 − (x2 + 4)y + 2z2 = −1, y2 − (x2 + 4)y − 2z2 = −1 and y2 − x2y + 2z2 = −9 have no integer solutions.
If an equation reduces to the question whether a polynomial P(x) can take values representable as a sum of squares (or other quadratic form), but Corollary 2.21 provides no obstructions, this is an indication that an equation may be solvable. To illustrate this, consider equation
y2 − xy + 2z2 = x3 + x − 1. (47)
It is equivalent to (2y − x)2 + 2(2z)2 = 4x3 + x2 + 4x − 4, hence we need to know if there exist x such that P(x) = 4x3 + x2 + 4x − 4 is representable as a2 + 2b2. Polynomial P(x) is irreducible, and divisibility analysis modulo 8 does not preclude P(x) from having only prime factors allowed by Corollary 2.11. Motivated by this, we performed a computer search and found that x = 30, y = −51, z = 107 is a solution. The program did not discover this automatically, because it searched only for solutions up to 100.
17 Exercise 2.27. Reduce equation y2 − x3y + z2 = −4 to the question whether a polynomial can be a sum of squares. Conclude that there is no obstructions for being so. Then perform a computer search to find a solution.
Propositions and Exercises above cover all equations with H ≤ 28, and some of the equations with H = 29. Other equations with H = 29 require some transformations before we can apply the listed methods. For example, consider equation y(x2 + 2)=(z2 + z)x + 2z + 1 (48) with H = 29. It may not be immediately obvious how to write the right-hand side as a sum of squares. However, one may note that, modulo x2 + 2,
0 ≡ (z2 + z)x + 2z + 1 ≡ (z2 + z)x2 +(2z + 1)x ≡ (z2 + z)(−2)+(2z + 1)x
≡ 4(z2 + z) − 2(2z + 1)x +(x2 + 2)=(2z − x + 1)2 + 1. The rest is easy. Because z2 + z is even, it follows from (48) that x2 + 2 is odd. Because x2 + 2 is a divisor of a sum of squares, all its prime factors are 1 modulo 4, hence x2 + 2 ≡ 1(mod 4), a contradiction. Exercise 2.28. Use the same argument as above to prove that equation
y(x2 + 2)= 2z2 + 2xz + 1 has no integer solutions.
As another example, consider equation
y(x2 − z2 − y)= x2 + z2 + 1, (49) also with H = 29. Here, the right-hand site is a sum of three squares. However, (49) can be rewritten as
(y + 1)(x2 − z2 − y + 1)=(x2 + z2 + 1)+(x2 − z2 − y)+ y + 1 = 2(x2 + 1), and then we may apply Proposition 2.6.
Exercise 2.29. Finish the proof that equation (49) has no integer solutions.
Another nice example is the equation
y(x2 − z2 − y + 1)= x2 + 3. (50)
Corollary 2.11 states that the prime factors of x2 + 3 other than 2 and 3 must be in the form p = 3k + 1, but example x = y = 7, z = 6 shows that the left-hand side can also have only such prime factors. However, (50) still has no integer solutions, as we prove next.
Proposition 2.30. Equation (50) has no integer solutions.
Proof. Denote x2 − z2 − y + 1 by a new variable t. Then ty = x2 + 3 by (50). Now, by definition of t, t = x2 + 3 − 3 − z2 − y + 1 = ty − z2 − y − 2, which can be written as (t − 1)(y − 1) = z2 + 3. Hence, equation (50) reduces to a nice system of equations
(t − 1)(y − 1)= z2 + 3 (ty = x2 + 3
Now we can use Corollary 2.11 to prove that this system has no integer solutions. It is clear that y and t have the same sign, and we may assume that they are positive, otherwise substitution y′ = 1 − y, t′ = 1 − t leads to
18 the same system with positive variables. First assume that both y and t are even. Because x2 + 3 is not divisible by 8, this is possible only if both y and t are equal to 2 modulo 4. But then both t − 1 and y − 1 are equal to 1 modulo 4, and so is their product z2 + 3. But then z2 is 2 modulo 4, a contradiction. Similarly, if both y and t are odd, then both y − 1 and t − 1 are even, which is possible only if they both are equal to 2 modulo 4. But then y and t are both 3 modulo 4, and their product x2 + 3 is 1 modulo 4, a contradiction. Finally, assume that y is even and t is odd (the case when y is odd and t is even is similar). Then x2 + 3 is even, and must therefore be 4 modulo 8, which implies that y = 4a for some odd integer a. For the same reason, t − 1 = 4b for an odd integer b, and we have
4b(4a − 1)= z2 + 3 ((4b + 1)4a = x2 + 3.
By Corollary 2.11, all prime factors of odd positive integers a, b, 4a − 1 and 4b + 1 must be either 3 or in the form p = 3m + 1. This implies that none of these numbers can be equal to 2 modulo 3. But this is possible only if a is 1 modulo 3 and b is 0 modulo 3. But in this case both 4b and 4a − 1 are divisible by 3, hence their product z2 + 3 is divisible by 9, which is a contradiction.
The only equations with H = 29 we did not discussed so far are
y2 − xyz + z2 = x3 − 5 (51) and x(x2 + y2 + 1)= z3 − z + 1. (52) These equations have been left open in the first version of this paper, but were quickly solved since that. Equation (51) turned out to have no integer solutions, as shown by Majumdar and Sury in [22].
Proposition 2.31. Equation (51) has no integer solutions.
Proof. The idea is to do analysis modulo 4 to check that in any solution to (51) x must be 1 modulo 4 and z must be even. Then substitute z = 2v, rewrite the equation as
(y − vx)2 − 3 = x3 − 8 − (4 − x2)v2 =(x − 2)(x2 + 2x + 4 +(x + 2)v2), use Corollary 2.11 to conclude that all odd prime factors of the left-hand side are p = 3 or p ≡±1(mod 12), and then consider possibilities x ≡ 1, 5, and 9 modulo 12 to arrive at a contradiction in each case. See [22] for details.
In contrast, equation (52) turned out to be solvable in integers. We have performed a computer search and found no solutions to (52) with |z|≤ 106. But then Andrew R. Booker performed a check up to |z|≤ 107 and found a solution x = 4280795, y = 4360815, z = 5427173. This finishes the analysis of the equations with H ≤ 29. It turns out that the methods introduced above suffice to solve all equations with H = 30. The most interesting of them are listed in the following exercise.
Exercise 2.32. Prove that equations
6 + x2 + x2y + 2y2 + z2 = 0,
2 + x3y + y2 + 2z2 = 0, y(y + z2)= x4 − 2 and 2 + x2 + y2 ± x2yz ∓ z2 = 0 with H = 30 have no integer solutions.
19 2.5 H ≥ 31: Entering the open territory In the previous section we have finished the analysis of equations with H ≤ 30. For H = 31, the number of equations that the program could not automatically solve increased dramatically. First of all, at this size we have got the first two-variable equations not covered by Theorem 2.4. They are
y3 + x3y − 2x − 3 = 0, (53)
y3 + x3y + y + 2x − 1 = 0, (54) xy3 − 2y + x3 + x + 1 = 0 (55) and y3 + xy − x4 − 3 = 0. (56) However, equations (53)-(55) are covered by another deep result, see e.g. [35] for a proof.
Theorem 2.33. Let F be a family of polynomials
m n i j P(x, y)= ∑ ∑ aij x y i=0 j=0
7 with integer coefficients aij of degree m > 0 in x and n > 0 in y which are irreducible over Q[x, y], and satisfy at least one of the following conditions:
(C1) either there exists a coefficient aij 6= 0 of P such that ni + mj > mn, or
i j (C2) the sum of all monomials aij x y of P for which ni + mj = nm can be decomposed into a product of two non-constant relatively prime polynomials in Z[x, y].
Then there is an algorithm that, given any polynomial P ∈ F, determines all integer solutions of equation P = 0.
In 1887, Runge [31] proved that for every P ∈ F equation P = 0 has at most finitely many integer solutions. Theorem 2.33 is an effective version of this result. In particular, it implies that Problem 1.1 is solvable for this class of equations. Now, equations (53) and (54) have degrees m = n = 3 in x and y, and coefficient a31 = 1 6= 0. But then ni + mj = 3 · 3 + 3 · 1 > 9 = mn, (C1) holds, and Theorem 2.33 is applicable. Equation (55) has a13 6= 0, and Theorem 2.33 is applicable by a similar argument. The next exercise suggest the reader to sovle these equations directly, without using Theorem 2.33.
Exercise 2.34. Solve equations (53), (54) and (55).
Equation (56) is not covered by Theorem 2.33. Indeed, for this equation m = 4, n = 3, and the only non-zero coefficients are a03, a11, a40, and a00. This implies that (C1) does not hold. In (C2), the monomials with ni + mj = nm are y3 and −x4, and their sum y3 − x4 is irreducible8 and therefore is not a product of relatively prime polynomials. However, equation (56) can be solved using Corollary 2.11.
Proposition 2.35. Equation (56) has no integer solutions.
Proof. If x is divisible by 3, then (56) implies that y is divisible by 3 as well, but then y3 + xy − x4 is divisible by 9 and cannot be equal to 3. Now rewrite the equation as
y(y2 + x)=(x2)2 + 3.
7That is, P(x, y) is not a product of non-constant polynomials with rational coefficients 8This can be checked directly or using the following easy sufficient condition of Ehrenfeucht [15]: if degrees of polynomials f (x) and g(y) are relatively prime, then f (x) − g(y) is irreducible.
20 Because (x2)2 + 3 > 0, y and y2 + x must have the same sign. If y < 0 and y2 + x < 0, then y2 < −x, hence |y| ≤ y2 < −x and |y2 + x| = −x − y2 ≤ −x, resulting in y(y2 + x) < (−x)2 < x4 + 3, a contradiction. Hence, both y and y2 + x are positive. If x is even, then by Corollary 2.11 all prime factors of x4 + 3 are in the form p = 3k + 1. But then y and y2 + x also have only such prime factors, hence y ≡ y2 + x ≡ 1(mod 3), which is possible only if 4 x ≡ 0(mod 3), a contradiction. If x is odd, then x + 3 = 4 ∏i pi with pi ≡ 1(mod 3) by Corollary 2.11. Because x is odd, y and y2 + x cannot be both even, hence either (i) y = 4v with v and y2 + x odd, or (ii) y2 + x = 4u with u and y odd. In case (i), all prime factors of v and y2 + x are 1 modulo 4, hence v ≡ y2 + x ≡ 1(mod 3), but then y ≡ y2 + x ≡ 1(mod 3), which is impossible as shown above. The case (ii) is similar.
Proposition 2.35 finishes the analysis of two-variable equations with H = 31, so we next consider three- variable equations. For some of these equations, it was at least clear from the beginning what method to apply. Exercise 2.36. Use Proposition 2.10 and Corollary 2.11 to prove that equations
1 + 5y + x2y − 3z2 = 0,
1 + 3y + x2y + 2xz − 2z2 = 0, 1 + 2y + x2y + 3z + 3z2 = 0, 1 + 3y + x2y + y2 − 3z2 = 0, 1 + y + xy + x2y + y2 + 3z2 = 0 and 3 − x2 − xy + x3y − z2 = 0 have no integer solutions. Exercise 2.37. Use the algorithm from the proof of Proposition 2.18 to check whether equations
5 + x + x2 + y + y2 + z + xyz + z2 = 0,
3 + x2 + xy + y2 − 2z + xyz + z2 = 0 and 1 + 2x + x2y + y2 + z + xyz + z2 = 0 have any integer solutions. Exercise 2.38. Use Corollary 2.21 to prove that the equations
3 + 2x2 + x2y − y2 + 2z2 = 0 and 3 + 2x2y + y2 + yz + z2 = 0 have no integer solutions. For some other equations, however, the proof method was not clear from the beginning. Consider, for example, equation 3 − x2 + x2y + y2 + yz − yz2 = 0. (57) One may note that this equation is equivalent to y(x2 + y + z − z2)= x2 − 3 and use Corollary 2.11 to analyse the prime factors of x2 − 3. However, it turns out that this analysis do not lead to a contradiction. Instead, one may note that the cubic terms of this equation are yx2 − yz2 = y(x − z)(x + z), which with new variables x − z = u and x + z = v reduces to yuv. All the other terms are at most quadratic, which is an indication that Vieta jumping technique may work. And it indeed works, but requires some efforts.
21 Proposition 2.39. Equation (57) has no integer solutions.
Proof. After substitution z + x → u, z − x → v and multiplication by 4, the equation reduces to 4y2 − u2 − v2 + 2uv + 2yv + 2yu − 4yuv + 12 = 0. Then we do substitution u → u/2, v → v/2, y → y/2 and get a new equation 48 − u2 + 2uv − v2 + 2uy + 2vy − 2uvy + 4y2 = 0, and we need to prove that it has no solutions in even integers. Next we do substitution u → u + 1, v → v + 1, y → y + 1 to get the equation 54 − u2 − v2 + 10y − 2uvy + 4y2 = 0 that should have no solutions in odd integers. Finally, we do the substitution y → y/2 to get the equation
54 − u2 − v2 + 5y − uvy + y2 = 0, and we will prove that this equation has no solutions such that u and v are odd while y is even. Let us call such solutions “eligible”. Let us choose an eligible solution with the smallest |u| + |v| + |y|. From symmetry, we may assume u ≥ v > 0. First consider the case y > 0. Then equation, as a quadratic in y, has another integer solution y′ = uv − 5 − y. Note that if u and v are odd and y even, then y′ is even, hence the new solution is eligible. Note also that yy′ = 54 − u2 − v2. The case of small u and v can be checked by an easy computer search, so we may assume that 54 − u2 − v2 < 0. Then y′ < 0, and the inequality |y′| ≥ |y| reduces to −y′ ≥ y, or −(uv − y) ≥ y, or uv ≤ 0, a contradiction. Now consider the case y < 0. Then the equation as quadratic in u has another integer solution u′ = −vy − u. If y is even and v and u are odd, then u′ is odd, hence the new solution is eligible. Hence we must have |u′| ≥ u. Because u′ + u = −vy > 0, this implies that u′ > 0. Then u′u ≥ u2, or v2 − y2 − 5y − 54 ≥ u2 ≥ v2, or −y2 − 5y − 54 ≥ 0, a contradiction. Exercise 2.40. Use the method of proof of Proposition 2.39 to show that equations
3 − x2 + x2y + 2y2 − yz2 = 0,
3 + x2 + x2y + 2y2 − yz2 = 0 and 3 + x2 + x2y + y2 + z2 − yz2 = 0 have no integer solutions. We next consider the equation 3 + 2y + y2 + x2yz + z2 = 0, (58) which is similar to the ones discussed in Proposition 2.16 and Exercise 2.17, but “extra” term 2y requires a more careful analysis. Proposition 2.41. Equation (58) has no integer solutions.
Proof. Denote x2 = a ≥ 0 and z →−z to get 3 + 2y + y2 − ayz + z2 = 0. Because ayz = z2 +(y + 1)2 + 2 > 0, y, z are non-zero and have the same sign. Choose a solution with |y| + |z| minimal. Then equation ′ z2+3 ′ 2 2 has another solution y = y , and |y | ≥ |y| reduces to z + 3 ≥ y . Hence, we have |z| ≥ |y| with the ′ y2+2y+3 ′ only possible exception |y| = 2 and |z| = 1. Similarly, we have z = z and |z | ≥ |z| reduces to (y + 1)2 + 2 ≥ z2. Hence we have |y + 1| ≥ |z|, with the only possible exception |z| = 1 and |y + 1| = 0. In the first exceptional case we either have y = 2 and z = 1 (and then a = 6) or y = −2 and z = −1 (and then a = 2). In the second exceptional case we have y = z = −1 (and then a = 3). Otherwise we have |y + 1| ≥ |z| ≥ |y|. Inequality |y + 1| ≥ |y| implies that y ≥ 0, hence z = y or z = y + 1. In the first case the equation reduces to (2 − a)y2 + 2y + 3 = 0. Any integer solution y must be a divisor of 3, so by trying y = ±1 and ±3 we find a = 3 and a = 7. In the second case (z = y + 1) the equation reduces to (2 − a)y2 +(4 − a)y + 4 = 0. By trying all divisors of 4 for y, we find that possible positive integer values for a are a = 2 and a = 5. In conclusion, the equation is solvable in integers y, z only for a = 2, 3, 5, 6,and 7. None of there values is a perfect square.
We leave several equations similar to (58) to the readers as an exercise.
22 Exercise 2.42. Check whether equations 1 + y + y2 − yz + x2yz + z2 = 0, 3 + y + y2 + z + x2yz + z2 = 0, and 1 + 3x + y2 − x2yz + z2 = 0 have any integer solutions. The only equation with H = 31 that we did not discussed so far is the equation y(x3 − y)= z3 + 3. (59) This equation is cubic in both x and z, and it is not clear how to apply the methods discussed above (like the information about prime factors of quadratic forms or Vieta jumping technique) to prove that it has no integer solutions. On the other hand, we have performed a computer search and found no solutions to (59) with |z|≤ 107. We leave the solvability of this equation to the reader as an open question. Open Question 2.43. Do there exist integers x, y, z satisfying (59)? Equation (59) can be rewritten as (y − x3/2)2 − x6/4 + z3 + 3 = 0. We next do some heuristic analysis of how many solutions to expect to a more general equation
d1 d2 dn a1x1 + a2 x2 + · · · + anxn = C, (60) provided that there are no obstructions (such as divisibility, quadratic residues, Vieta jumping, or any other obstructions) to the solution existence. Let us choose some large constant B and look at the solutions such that < di 1/di di B/2 maxi |ai xi | ≤ B. There are O(B ) choices of each variable xi such that |ai xi | ≤ B, resulting in about KBS combinations of variables, where n 1 S = ∑ , (61) i=1 di and K > 0 is a constant depending on the equation. From this, we need to subtract about K(B/2)S combina- di S tions with maxi |ai xi | ≤ B/2, but the result is still K1B for a different constant K1. Now, there are O(B) S−1 possible values of the left-hand side of (60), hence each value, including C, occurs on average ≈ K2B times. Thus, we have the following cases. S−1 • If S > 1, K2B grows reasonably fast with B. Hence, if a solution cannot be easily found by a computer search, we expect that there should be a special reason (obstruction) to a solution existence, and that we can in principle discover this obstruction and use it to prove that there are no solutions.
< di • If S = 1, we expect a constant number of solutions in the range B/2 maxi |ai xi |≤ B. By summing up solutions in the intervals [B/2, B],[B/4, B/2], [B/8, B/4], and so on, we conclude that there should di be about K2 log B solutions with maxi |ai xi | ≤ B. If the constant K2 is small, we expect that equation (60) should have infinitely many of solutions, but the smallest one may be astronomical. So, for this type of equations, it makes sense to do a deeper computer search aiming to find large solutions.
S−1 • If S < 1, then K2B decreases fast with B. Summing up over intervals [B, 2B], [2B, 4B], and so on, we di > S−1 ∞ i(S−1) conclude that the total number of solutions with maxi |ai xi | B is expected to be K2B ∑i=1 2 = S−1 K3B for some K3 > 0. Hence, if the equation happen to have no small solutions, it may well have no solutions at all, even if there are no obstructions to the solution existence. The absence of obstructions may make the proof of the solution non-existence especially challenging. 1 1 1 For the equation (59), we have S = 2 + 3 + 6 = 1, hence there are two reasonable ways how one may try to solve this equation: either find a reason (obstruction) why an integer solution cannot exist, or use computer search to try to find a large solution.
23 3 Equations of special type
3.1 Equations in two variables We may classify equations by the number of variables. Equations in one variable are easy to solve, so the first open case are equations in two variables. All equations with H ≤ 31 are discussed in the previous sections. For H = 32, the program returns the following two-variable equations:
y3 + xy + x4 + 4 = 0, (62)
y3 + xy + x4 + x + 2 = 0, (63) y3 + y = x4 + x + 4 (64) and y3 − y = x4 − 2x − 2 (65) Equation (62) looks similar to (56), but it is not clear how to solve it using Corollary 2.11. If both x and y are even, substitution x = 2u, y = 2v reduces the equation to −v(2v2 + u)=(2u2)2 + 1. The right hand side is a sum of squares and by Corollary 2.11 has only the prime factors in the form p = 4k + 1. However, an example v = −1, u = 3 demonstrates that v(2v2 + u) may also have only such prime factors. We leave the solvability of these equations to the reader as an open question.
Open Question 3.1. Determine whether each of the equations (62)-(65) have any integer solution.
Let us do some heuristic analysis for the equations (62)-(65). They have monomials x4, y3, and some 1 1 < monomials of lower degree, which we will ignore. Then S = 4 + 3 1, where S in defined in (61). This means that if such equations have no small solutions, they likely to have no solutions at all, even if there are no clear obstructions for the solution existence. This is what makes such equations difficult. On the other hand, these equations are in 2 variables, and correspond to the study of integer points on the curves. Comparing to the study of integer points on surfaces, this question is much better understood. For example, we have the following general result.
Theorem 3.2. There is an algorithm that, given a polynomial P(x, y) with integer coefficients, decides whether the equation P(x, y)= 0 has finite or infinite number of integer solutions.
See the introduction of [5] for an explanation how to deduce Theorem 3.2 from the classical 1929 theorem of Siegel. However, all the known proofs of Theorem 3.2 are ineffective. For a given P, the algorithm can output that the equation P(x, y) = 0 has a finite number of integer solutions, and even give an explicit upper bound for the number of such solutions [7], but not for their size. Hence, it cannot be used to find the solutions and even to decide if any integer solution exists.
3.2 Symmetric equations Symmetric equations are those which does not change after permutations of variables. They usually look particularly nice. We start with the case of two variables and ignore Theorem 2.33 for a moment. Then the smallest symmetric equation program returned was
x3 + x2y2 + y3 = 7 (66) with H = 39. We remark that this equation is covered by Theorem 2.33, because it has m = n = 3 and condition (C1) holds with i = j = 2. However, it is instructive to present the direct proof that (66) has no integer solutions due to Rouse [30].
Proposition 3.3. Equation (66) has no integer solutions.
24 Proof. A standard tool for solving symmetric equations in 2 variables is change of variables u = −(x + y), v = xy. Then −u3 =(x + y)3 = x3 + y3 + 3xy(x + y)= x3 + y3 − 3uv, hence x3 + y3 = −u3 + 3uv, and (66) reduces to
v2 + 3uv = u3 + 7.
This is an example of elliptic curve, for which there are known algorithms (one of them implemented in Sage- Math [38]) to compute integer solutions. In this example, the solutions are (−3, 4), (−3, 5), (186, −2831) and (186, 2273). If, for example, u = −3 and v = 4, then x + y = 3 and xy = 4. Then x(3 − x) = 4, or x2 − 3x + 4 = 0. But this quadratic equation has no integer solutions. The other three cases can be considered similarly.
We next invite the reader to use the same method as in the proof of Proposition 3.3 to solve the following similar equations.
Exercise 3.4. Check whether equations x3 + x2y2 + y3 = 9, x3 + x2y2 + y3 − xy = 4, x3 + x2y2 + y3 − xy = 5 and x3 + x2y2 + y3 + x2 + y2 + 1 = 0 have any integer solutions.
Note that all the listed equations are covered by Theorem 2.33. After we excluded all equations covered by this Theorem, the program returned no symmetric equations in 2 variables with H ≤ 50. For 3 variables, the smallest symmetric equations the program returns are equations (32) and (41), see Exercise 2.19. The next one is
xyz + x2 + y2 + z2 + x + y + z + 5 = 0 (67) with H = 31. This equation is of the type (37) and is covered by Proposition 2.18.
Exercise 3.5. Use the algorithm from the proof of Proposition 2.18 to check whether equation (67) has any integer solutions.
We next asked the program to exclude all equations of the type (37). Then the smallest equation it returns is x2 + y2 + z2 − xy − yz − xz = xyz − 4 with H = 36. However, the substitution x → x − 1, y → y − 1, z → z − 1 reduces this equation to exactly (67). The smallest symmetric 3-variable equation which is truly not of the type (37) is
x3 + y3 + z3 + xyz = 5 (68) with H = 37. We leave the solvability of this equation to the reader as an open question.
Open Question 3.6. Do there exist integers x, y, z satisfying (68)?
1 1 1 For the equation (68), we have S = 3 + 3 + 3 = 1, where S in defined in (61). Hence, it is quite likely that this equation has integer solutions but the smallest one may happen to be quite large.
25 3.3 Equations with 3 monomials Equation (67), while symmetric and has reasonably low H, has 8 monomials and therefore does not look particularly simple. Another possible measure of simplicity of the equation is the number of monomials. Note than one monomial must be a non-zero free term, otherwise 0 is a solution. An equation with 2 monomials has the form k1 kn ax1 ... xn = b, (69) b b < where a, b are non-zero integers and k1,..., kn are positive integers. If either a is not an integer, or a 0 and m1 ml b all ki are even, then (69) has no integer solution. Otherwise let p1 ... pl be the prime factorization of a , and let S(k ,..., k ) be the set of non-negative integers N representable as N = ∑n c k for some non-negative 1 n i=1 i i integers c1,..., cn. Then (69) is solvable in integers if and only if
mj ∈ S(k1,..., kn), j = 1,..., l. (70)
l cij Indeed, if x1,..., xn is any solution to (69), then each xi can be written as xi = ∏j=1 pj for some integers n cij ≥ 0, hence mj = ∑i=1 cijki, j = 1,..., l, and (70) follows. The proof of the converse direction is similar. We next consider equations with 3 monomials. We have already met several such equations, see (8) and (12), but such equations, while may be non-trivial to solve, are covered by the general algorithm in Theorem 2.4. In fact, Theorem 2.4 implies the existence of an algorithm for solving an arbitrary 3-monomial equation in 2 variables. Proposition 3.7. There exists an algorithm which, given any polynomial P(x, y) with 3 monomials and integer coefficients, determines whether equation P(x, y)= 0 has an integer solution. Proof. We may assume that P(0, 0) 6= 0, otherwise x = y = 0 is a solution. Then P(x, y) has the form
k1 k2 m1 m2 P(x, y)= a1 x y + a2x y + a3, where a1, a2, a3 are non-zero integers and k1, k2, m1, m2 are non-negative integers. If both k1 and m1 are posi- tive, then P(x, y)= 0 only if x is a divisor of a3, and we may solve the equation by trying all the divisors and solving the resulting equations in one variable. The case when both k2 and m2 are positive is similar. Otherwise equation P(x, y)= 0 can be written as axk = bym + c, where we may assume that a > 0 and c 6= 0. By interchanging the roles of x and y if necessary, we may assume that k ≤ m. This equation is easy if k ≤ 1 or k = m = 2, so we may also assume that k ≥ 2 and m ≥ 3. Because c 6= 0, bym + c has no repeated roots, and the equation is solvable in finite time by Theorem 2.4.
With Proposition 3.7, we may restrict the attention on 3-monomial equations with at least 3 variables. The first such equations we get from out program are
x3y2 = z2 + 6 (71) and x3y2 = z2 − 6 (72) with H = 42. Recall that the program automatically excludes equations which have a solution such that max{|x|, |y|, |z|} ≤ 100, and most of the equations it returned so far turned out to have no solutions. However, it is easy to see that equation (71) is one of the few exceptions: by doing search up to 500 instead of 100, we can easily find that it has a solution x = 7, y = 25, z = 463. Does equation (72) have integer solutions as well? When the direct search up to z ≤ 107 returned no results, we then tried a slightly different strategy. For each fixed x, equation (72) is a quadratic equation in 2 variables. There is a general algorithm for solving all such equations, which is implemented online at [1]. We have tried a few values of x, and, for x = 19, found that the equation 193y2 = z2 − 6 is solvable, with the smallest solution being y = 755, 031, 379, z = 62, 531, 004, 125.
26 Exercise 3.8. By using website [1] or writing an own computer program for solution search, check that each of the following equations x3y2 = 2z2 + 3 x3y2 = z2 + 10 x3y2 = 3z2 − 5 x3y2 = z2 + 14 x3y2 = z2 − 14 has an integer solution.
Exercise 3.8 covers all equations of the form x3y2 = az2 + b the program retuned up to H ≤ 50. After excluding these equations, the smallest 3-monomial equation it returns is
x3y2 = z3 + 6 (73) with H = 46. The famous and well-believed abc conjecture of Masser and Oesterl´e[26] states that if a, b, c are coprime positive integers such that a + b = c, then the product of district prime factors of abc is not much smaller than c. More precisely, if rad(n) denotes the product of district prime factors of n, then the abc conjecture predicts that for every real number ǫ > 0, there exists only a finite number of triples (a, b, c) of coprime positive integers such that a + b = c and ǫ c > rad(abc)1+ . It is easy to check that this conjecture implies that equation (73) may have at most finitely many integer solu- tions. However, the abc conjecture remains open, and even its truth would not exclude a possibility of some solutions to (73). We leave the solvability of this equation as an open question.
Open Question 3.9. Do there exist integers x, y, z satisfying (73)? The following heuristic analysis shows that this equation may have no solutions but proving this may be quite difficult. For an integer B > 0, there are about O(B1/2+1/3) = O(B5/6) triples (x, y, z) such that B/2 < max{|x3y2|, |z|3} ≤ B. Hence, if we search for a solution to (73) in this range, there will be about O(B−1/6) expected solutions. This makes the existence of large solutions quite unlikely, even if there are no clear obstructions for the solution existence. In case of absence of such obstructions, proving that a surface has no integer points may be very difficult.
3.4 The shortest open equations In this section we discuss what happens if, instead of ordering the equations by H, we order them by length l defined in (5), or, equivalently, by an integer
k k l(P) ∑ di L(P) := 2 = ∏ |ai| · 2 i=1 . i=1 It turns out that the shortest equation the program cannot automatically exclude is the equation (15) of length l = 7, which has H = 17, and is also the smallest non-excluded equation with respect to H. Then the program returns many other equations that have H ≤ 31 and was therefore considered earlier in this paper. The shortest equations with H > 31 the program returns are the equations
1 + x2 − x3y2 + z2 = 0
1 + y2 + x3yz + z2 1 + x2 + x2y2z + z2
27 of length l = 9. The insolubility of the first equation follows from the insolubility of (15) after substitution t = xy, the second one has no integer solutions due to Proposition 2.16, while the last one, considered as quadratic in z, has two roots z and z′, with |z + z′| = x2y2 and zz′ = x2 + 1. Unless min(|x|, |y|) ≤ 1, |z + z′| > |zz′|, which is possible only if min(|z|, |z′ |) ≤ 1. The listed cases of small |x|, |y|, |z|, |z′ | can be easily checked. By similar methods, we were able to solve all equations of length l ≤ 10, with three exceptions. The following Proposition and Exersize lists the most interesting equations out of the solved ones. Proposition 3.10. Equations x2 − 5xyz + y2 + 1 = 0, (74) 1 + x4 + 4z + y2z = 0, (75) 1 + 2x2y + 4z + y2z = 0, (76) 1 + x3y + 2y2 + 2z2 = 0 (77) and 1 + 2x2y + xy2z + z2 = 0 (78) have no integer solutions, while the equation
2 + y2 − x3y2 + z2 = 0 (79) has integer solutions.
Proof. For the equation (74), it is easy to check that there are no solutions with x = 0 or y = 0 or x = y. Choose a solution with |x| + |y| + |z| minimal. By changing signs and symmetry, we may assume x > y > 0. ′ y2+1 As a quadratic in x, (74) has another root x = x . Because |x| + |y| + |z| was minimal, we must have x′ ≥ x, or y2 + 1 ≥ x2 ≥ (y + 1)2, which is a contradiction. Equation (75) can be written as −z(y2 + 4)= x4 + 1. Because x4 + 1 =(x2)2 + 12 =(x2 − 1)2 + 2x2, Corollary 2.11 implies that all its odd prime factors are (a) 1 modulo 4 and (b) 1 or 3 modulo 8. Hence, all odd prime factors must in fact be 1 modulo 8. Prime factorization of x4 + 1 may also contain at most one factor of 2, hence all positive divisors of this number must be 1 or 2 modulo 8. But this is impossible for y2 + 4. For the equation (76), if y2 + 4 is a divisor of 2x2y + 1, then it must also be a divisor of 2x2(y2 + 4) − y(2x2y + 1) = 8x2 − y and 2x2(8x2 − y)+(2x2y + 1)=(2x)4 + 1. But this is a contradiction with the insolubility of (75). After multiplication by 8, equation (77) can be written as (4y + x3)2 +(4z)2 = x6 − 8 = (x2 − 2)(x4 + 2x2 + 4). Let us call primes of the form p = 4k + 3 “special”. By corollary 2.21, any special prime enters the prime factorization of a sum of square with even multiplicity. If x is odd, then x2 − 2 is 3 modulo 4, and its prime factorization must therefore contain some special prime p with odd multiplicity. This is possible only if p is a common factor of x2 − 2 and x4 + 2x2 + 4, but then p divides x4 − 2x2, 4x2 + 4, x2 + 1, and 3, hence p = 3. But x2 − 2 is not divisible by 3, a contradiction. If x is even, substitution x → 2t transforms (x2 − 2)(x4 + 2x2 + 4) into 8(2t2 − 1)(4t4 + 2t2 + 1). If t is even, then 2t2 − 1 is 3 mod 4, while if t is odd, then 4t4 + 2t2 + 1 is 3 mod 4, in both cases we get a contradiction in the same way as above. After multiplication by 4, equation (78) can be rewritten as (2z + xy2)2 + 22 = x2y(y3 − 8) = x2y(y − 2)(y2 + 2y + 4). Corollary 2.11 implies that no of the divisors of the left-hand side is 3 modulo 4. If y is odd, then either y or y − 2 is 3 modulo 4, a contradiction. If y is even, substitution y → 2t reduces the right hand side to 2t(2t − 2)(4t2 + 4t + 4)= 16t(t − 1)(t2 + t + 1). Whatever t is modulo 4, one of the factors t, t − 1, or t2 + t + 1 is 3 modulo 4, a contradiction. For the equation (79), we note that for each fixed x it is quadratic in y and z and can be solved by the available software. Checking x = 1,2,... , we get that for x = 103 there is a solution
y = 56781721110114762679275339, z = 5935594047429827052541057073.
28 Exercise 3.11. Use the methods from the proof of Proposition 3.10 to check whether equations
2 + x2 − y2 + 3xyz = 0,
1 + y + y2 − x3yz + z2 = 0, 1 − x2y + 5y2 + z2 = 0, 1 + 5x2y + y2 + z2 = 0, 1 + 2y + x2y − 6z2 = 0, 1 + 2y + x4y + z − z2 = 0, 1 − 3x2y + y2 + 2z2 = 0, 1 + x3y + x2y2 + z2 = 0, 2y2 − x2yz + 2z2 − 1 = 0, 2y2 − x2yz + 2z2 + 1 = 0, 1 + 2y2 + x3yz + z2 = 0 and 1 + 2y2 − x3y2 + z2 = 0 have any integer solutions.
The only equations of length l ≤ 10 that seem to require a new idea are the equations
y(x3 − y)= z4 + 1, (80)
2y3 + xy + x4 + 1 = 0 (81) and x3y2 = z4 + 2 (82) that have length l = 10 and sizes H equal to 37, 37, and 50, respectively. Equation (80) resembles the current smallest open equation (59), but with z4 instead of z3. This difference seems to be crucial, because S defined in 1 1 1 11 < (61) is now S = 2 + 4 + 6 = 12 1, hence the equation is unlikely to have large solutions. The same analysis is true for the equations (81) and (82) as well. On the other hand, we have checked that all these equations have no solutions in the range up to a million. We leave their solubility as open questions.
Open Question 3.12. Do there exist integers x, y, z satisfying (80)?
Open Question 3.13. Do there exist integers x, y satisfying (81)?
Open Question 3.14. Do there exist integers x, y, z satisfying (82)?
As mentioned above, equation (80) is similar to (59). Further, equation (81) is similar to the smallest open 2-variable equations (62)-(65), while equation (82) resembles the smallest open 3-monomial equation (73). It is likely that if we resolve equations with small l, we may use similar methods to resolve the equations with the smallest H, and vice versa.
29 4 Conclusions
We have considered, in a systematic way, all “small” polynomial Diophantine equations and identify those ones for which the solvability problem (Problem 1.1) is interesting. As expected, a vast majority of small equations either have small easy-to-find solutions or have no solutions for trivial reasons. However, we also have found some simple-to-state equations which are not solvable in integers for deeper reasons, as well as equations whose smallest solutions are quite large and are difficult to find by a direct search. Some of the equations are solved, some others are left to the reader as exercises and open questions. An obvious direction for future research is to solve open questions listed here, proceed to the equations with higher values of H or l, and see how far can we go. The following table summarizes all smallest open equations listed in this paper, just to have them all in one place.
Equation Size Comment y(x3 − y)= z3 + 3 H = 31 The smallest (in H) open equation y3 + xy + x4 + 4 = 0 H = 32 The smallest open 2-variable equations y3 + xy + x4 + x + 2 = 0 H = 32 y3 + y = x4 + x + 4 H = 32 y3 − y = x4 − 2x − 2 H = 32 x3 + y3 + z3 + xyz = 5 H = 37 The smallest open symmetric equation x3y2 = z3 + 6 H = 46 The smallest open 3-monomial equation y(x3 − y)= z4 + 1 l = 10 The shortest (in l) open equations 2y3 + xy + x4 + 1 = 0 l = 10 x3y2 = z4 + 2 l = 10
This project is very active, and some of the equations listed as open may be solved by the time you read this paper. Please check the mathoverflow question [19] for the up-to-date list of the current smallest open equations. Another possible research direction is to investigate, in addition to Problem 1.1, the following problems, in the increasing level of difficulty. Let P be the set of multivariate polynomials P(x1,..., xn) with integer coefficients.
Problem 4.1. Given P ∈P with constant term 0, does equation P = 0 has a non-zero integer solution?
Problem 4.2. Given P ∈ P, determine whether equation P = 0 has a finite number of integer solutions, and if so, list them all.
Problem 4.3. Given P ∈ P, find all integer solutions to equation P = 0. If there are infinitely many of them, find a convenient way to describe them all (one can use expressions with parameters, recurrence relations, etc.)
Problem 4.4. Investigate the same questions as above for rational solutions.
Because these problems look more difficult than Problem 1.1, one is expected to find even smaller equations for which the listed problems are highly non-trivial. Such equations may serve as nice test cases for any known or new technique for determining integer and/or rational points on curves and surfaces.
Acknowledgements
I thank anonymous mathoverflow user Zidane whose question [37] inspired this work. I also thank other mathoverflow users for very helpful discussions on this topic, especially Will Sawin for solving equations (29), (43) and (44), Fedor Petrov for the alternative solution of (29), Andrew R. Booker for finding a solution to (52), Majumdar and Sury for solving (51) in a recent preprint [22], Victor Ostrik for outlining the solution of equation (15), and Jeremy Rouse for solving equation (66) (see Proposition 3.3) and for pointing me to references [28] and [29]. I thank University of Leicester students Daniel Bishop-Jennings, Sander Bharaj and Will Moreland
30 for doing the projects on this topic under my supervision and independently verifying some of the results presented here. I thank William Gasarch for carefully reading an earlier draft of this paper and providing me with the detailed and helpful list of comments and suggestions. Last but not least, I thank Aubrey de Grey for many useful suggestions (including the suggestion to include the analysis with an alternative measure of the equations sizes), and for writing his own version of the computer program for enumerating equations, which provides an independent validation that none of the equations of small size has been accidentally missing.
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