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WHATIS... a Diophantine 푚-tuple? Andrej Dujella Communicated by Florian Luca and Cesar E. Silva

A Diophantine 푚-tuple is a set of 푚 distinct positive of an element with itself plus 1 is a square. It is obvious integers with the property that the product of any two of that for integers such a condition cannot be satisfied. its distinct elements plus 1 is a square. If a set of nonzero But for rationals there is no obvious reason why the sets rationals has the same property, it is called a rational (called strong Diophantine 푚-tuples) that satisfy these Diophantine 푚-tuple. Fermat found the first Diophantine stronger conditions would not exist. For each element 푎 quadruple in integers {1, 3, 8, 120}. Indeed, we have of such a set we have that 푎2 + 1 is a square; therefore 1 ⋅ 3 + 1 = 22, 1 ⋅ 8 + 1 = 32, 1 ⋅ 120 + 1 = 112, 푎 = 푋/푌, where (푋, 푌, 푍) is a Pythagorean triple, i.e., 푋2 + 푌2 = 푍2. It is known that there exist infinitely many 3 ⋅ 8 + 1 = 52, 3 ⋅ 120 + 1 = 192, 8 ⋅ 120 + 1 = 312. strong Diophantine triples, while no example of a strong Euler was able to extend Fermat’s quadruple to the rational Diophantine quadruple is known. quintuple {1, 3, 8, 120, 777480/8288641}. In the integer case, it is easy to prove that there exist The ancient Greek mathematician Diophantus found infinitely many integer Diophantine quadruples (there the first example of a rational Diophantine quadruple are parametric families for Diophantine quadruples in- 1 33 17 105 volving polynomials and Fibonacci numbers, such as { , , , }. 3 2 16 16 4 16 {푘, 푘 + 2, 4푘 + 4, 16푘 + 48푘 + 44푘 + 12} and {퐹푘, 퐹푘+2, Some of the famous mathematicians of the past, such as 퐹푘+4, 4퐹2푘+1퐹2푘+2퐹2푘+3} for 푘 ≥ 1), while the folklore con- Diophantus, Fermat, and Euler, as well as some modern jecture is that there does not exist a Diophantine quintuple. ones such as Fields Medalist Alan Baker, have made im- The first important result concerning this conjecture was portant contributions to problems related to Diophantine proved in 1969 by Baker and Davenport. Using Baker’s 푚-tuples, but many problems still remain open. theory on linear forms in logarithms of algebraic numbers It is natural to ask how large sets of Diophantine 푚- and a reduction method based on continued fractions, tuples can be. This question is almost completely solved they proved that if 푑 is a positive integer such that in the integer case. On the other hand, it seems that in {1, 3, 8, 푑} forms a Diophantine quadruple, then 푑 has to the rational case we do not have even a widely accepted be 120. This implies that Fermat’s set {1, 3, 8, 120} cannot conjecture. In particular, no absolute upper bound for be extended to a Diophantine quintuple. It was proved in the size of rational Diophantine 푚-tuples is known. The 2004 that a Diophantine sextuple does not exist and that study of this question leads to surprising connections there are only finitely many Diophantine quintuples. Since with elliptic curves. then, the bound on the number of possible Diophantine Note that in the definition of (rational) Diophantine quintuples has been improved by several authors (at the 푚-tuples we excluded the requirement that the product moment the best bound seems to be 5.441 ⋅ 1026 due to Cipu and Trudgian), but the question of the existence of Andrej Dujella is professor at the University of Zagreb and Fellow Diophantine quintuples is still open. of the Croatian Academy of Sciences and Arts. His email address It is known that any Diophantine triple {푎, 푏, 푐} can be is [email protected]. extended to a Diophantine quadruple {푎, 푏, 푐, 푑}. Indeed, For permission to reprint this article, please contact: with 푎푏 + 1 = 푟2, 푎푐 + 1 = 푠2, 푏푐 + 1 = 푡2, we may take [email protected]. DOI: http://dx.doi.org/10.1090/noti1404 푑 = 푎 + 푏 + 푐 + 2푎푏푐 + 2푟푠푡,

772 Notices of the AMS Volume 63, Number 7 THEGRADUATESTUDENTSECTION

and then 푎푑+1 = (푎푡+푟푠)2, 푏푑+1 = (푏푠+푟푡)2, 푐푑+1 = P + Q (푐푟 + 푠푡)2. Quadruples of this form are called regular, and the stronger version of the Diophantine quintuple conjecture is that all Diophantine quadruples are regular. Fujita proved that any Diophantine quintuple contains a P * P regular Diophantine quadruple. P P Here we sketch the ideas used in the proof of finite- ness of Diophantine quintuples and other similar results. Extending the Diophantine triple {푎, 푏, 푐}, 푎 < 푏 < 푐, to a Diophantine quadruple {푎, 푏, 푐, 푑} leads to the system Q 2 2 2 푎푑 + 1 = 푥 , 푏푑 + 1 = 푦 , 푐푑 + 1 = 푧 , and by eliminating P + P = 2 P 푑, we get the system of simultaneous Pellian equations: 푐푥2 − 푎푧2 = 푐 − 푎, 푐푦2 − 푏푧2 = 푐 − 푏. Solutions of Pellian equations are contained in finitely P * Q many binary recursive sequences. Thus, the problem leads to finding intersections of binary recursive sequences, i.e., secant line tangent line finitely many equations of the form 푣푚 = 푤푛. These se- 푚 푛 quences satisfy 푣푚 ≈ 훼훽 , 푤푛 ≈ 훾훿 for certain algebraic Figure 1. If 푃 and 푄 have different 푥-coordinates, numbers 훼, 훽, 훾, 훿 (e.g. in the case of the Diophantine triple then the straight line through 푃 and 푄 intersects the {1, 3, 8} treated by Baker and Davenport, 훼 = (3 + √3)/3, curve in exactly one more point, denoted by 푃 ∗ 푄, 훽 = 2 + √3, 훾 = (4 ± √2)/4, 훿 = 3 + 2√2), which im- and we define 푃 + 푄 as −(푃 ∗ 푄) (where −(푃 ∗ 푄) is 훼 the point with the same 푥-coordinate but negative plies 푚 log 훽 − 푛 log 훿 + log 훾 ≈ 0. But a consequence of Baker’s theory is that a linear combination, with integer 푦-coordinate as 푃 ∗ 푄). If 푃 = 푄, then we replace the coefficients, of logarithms of algebraic numbers which is secant line by the tangent line at the point 푃. nonzero cannot be very close to 0. We therefore obtain upper bounds for 푚, 푛. To obtain lower bounds, we can use the congruence method, introduced in joint work with to a quadruple, we have to find a rational 푥 such that 2 Pethő in 1998, and consider 푣푚 ≡ 푤푛 (mod 푐 ). If 푚, 푛 푎푥 + 1, 푏푥 + 1, and 푐푥 + 1 are all squares of rationals. are small (compared with 푐), then ≡ can be replaced by =, By multiplying these three conditions, we obtain a single and this (hopefully) leads to a contradiction (if 푚, 푛 > 2, condition i.e., if 푑 does not correspond to a regular quadruple). 푦2 = (푎푥 + 1)(푏푥 + 1)(푐푥 + 1), Therefore, we obtain lower bounds for 푚, 푛 (small powers of 푐). Comparing the upper and lower bounds we get a which is in fact the equation of an elliptic curve (nonsin- contradiction for large 푐. gular cubic curve with a rational point). We will explain It is likely that we cannot surpass Fermat by construct- below which points on the curve satisfy the original ing Diophantine quintuples. However, in the rational case, system of equations and give extensions to Diophantine there exist larger sets with the same property. Euler quadruples. found infinitely many rational Diophantine quintuples. The set 퐸(ℚ) of rational points on an elliptic curve 퐸 The question of the existence of rational Diophantine over ℚ (affine points [푥, 푦] satisfying the equation along sextuples remained open for more than two centuries. In with the point at infinity) forms an abelian group with the 1999 Gibbs found the first rational Diophantine sextuple law of addition defined by the secant and tangent method as described in the figure. 11 35 155 512 1235 180873 { , , , , , }, Moreover, by the Mordell-Weil theorem, the abelian 192 192 27 27 48 16 group 퐸(ℚ) is finitely generated, and hence it is the while in 2016 Dujella, Kazalicki, Mikić, and Szikszai proved product of the torsion group and 푟 ≥ 0 copies of the 푟 that there exist infinitely many rational Diophantine infinite cyclic group: 퐸(ℚ) ≅ 퐸(ℚ)tors × ℤ . sextuples. For example, there are infinitely many such Let us denote the curve 푦2 = (푎푥 + 1)(푏푥 + 1)(푐푥 + 1) sextuples containing the triple {15/14, −16/21, 7/6}, with by ℰ. We say that ℰ is induced by the Diophantine triple the simplest example being {푎, 푏, 푐}. There are three rational points on ℰ of order 2, 15 16 7 1680 910 624 namely 퐴 = [−1/푎, 0], 퐵 = [−1/푏, 0], 퐶 = [−1/푐, 0], and { , − , , − , − , }. 14 21 6 3481 1083 847 also two other obvious rational points: No example of a rational Diophantine septuple is known. 푃 = [0, 1], Moreover, we do not know any rational Diophantine 푆 = [1/푎푏푐, (푎푏 + 1)(푎푐 + 1)(푏푐 + 1)/푎푏푐]. quintuple (or even quadruple) that can be extended to √ two different rational Diophantine sextuples. Note that the 푥-coordinate of the point 푃 − 푆 is exactly We now describe connections between rational Dio- the number 푑 from the definition of regular Diophantine phantine 푚-tuples and elliptic curves. Let {푎, 푏, 푐} be a quadruples. In general, 푃 and 푆 will be independent rational Diophantine triple. In order to extend this triple points of infinite order. But an important question, with

August 2016 Notices of the AMS 773 THEGRADUATESTUDENTSECTION significant consequences, is whether they can have finite over rings of integers of certain quadratic fields which orders and which orders are possible. show that there is a close connection between existence Now we can answer the question which points on ℰ of a 퐷(푛)-quadruple and representability of 푛 as a give extensions to Diophantine quadruples. Namely, the difference of two squares in the ring. Note that integers 푥-coordinate of a point 푇 ∈ ℰ(ℚ) satisfies the original ≡ 2 (mod 4) are exactly those that cannot be represented three conditions if and only if 푇 − 푃 ∈ 2ℰ(ℚ). It can be as a difference of two squares of integers. verified that 푆 ∈ 2ℰ(ℚ). This implies that if 푥(푇) satisfies More details on Diophantine 푚-tuples and the com- the original conditions, then also the numbers 푥(푇 ± 푆) plete list of references can be found at the webpage satisfy them. It can be shown that 푥(푇)푥(푇 ± 푆) + 1 is web.math.hr/˜duje/dtuples.html. always a perfect square. Thus, {푎, 푏, 푐, 푥(푇 − 푆), 푥(푇), 푥(푇 + 푆)} is “almost” a rational Diophantine sextuple. The only missing condition is that 푥(푇 − 푆)푥(푇 + 푆) + 1 is a square, and this last condition is satisfied if the ABOUT THE AUTHOR point 푆 is of order 3. In that way, the problem of construction of rational Diophantine sextuples becomes Andrej Dujella received a PhD in closely connected with elliptic curves with torsion group mathematics from the University of ℤ/2ℤ×ℤ/6ℤ. Elliptic curves induced by Diophantine triples Zagreb in 1996. were used by Dujella and Peral in 2014 in constructing His research interests include Di- elliptic curves with given torsion and high rank (details ophantine equations, elliptic curves, of the current rank records can be found at the webpage polynomial root separation, and web.math.hr/˜duje/tors/tors.html). It is interesting applications of Diophantine approx- that any elliptic curve over ℚ with torsion group ℤ/2ℤ × imation to cryptography. ℤ/8ℤ can be induced by a Diophantine triple. There are several natural generalizations of the notion Credits of Diophantine 푚-tuples. We can replace squares by 푘-th Figure 1 and p. 774 photo, courtesy of Andrej Dujella. powers for fixed 푘 ≥ 3 (in a joint work with Bugeaud we showed that there are no such quadruples for 푘 ≥ 177) or by perfect powers (Luca showed in 2005 that the cardinality of such a set is uniformly bounded assuming the 푎푏푐-conjecture). L’Hospital We can replace the number 1 in the conditions “푎푏 + 1 is a square” by a fixed integer 푛. Such sets are called 퐷(푛)-푚-tuples. It is easy to show that there are no 퐷(푛)- quadruples if 푛 ≡ 2 (mod 4). Indeed, assume that {푎1, 푎2, 푎3, 푎4} is a 퐷(푛)-quadruple. Since the square of an integer is ≡ 0 or 1 (mod 4), we have that 푎푖푎푗 ≡ 2 or 3 (mod 4). This implies that none of the 푎푖’s is divisible by 4. Therefore, we may assume that 푎1 ≡ 푎2 (mod 4). But now we have that 푎1푎2 ≡ 0 or 1 (mod 4), a contradiction. On the other hand, it can be shown that if 푛 ≢ 2 by Nam Nguyen (mod 4) and 푛 ∉ 푆 = {−4, −3, −1, 3, 5, 8, 12, 20}, then there exists at least one 퐷(푛)-quadruple. For 푛 ∈ 푆, the question of the existence of 퐷(푛)-quadruples is still open. No 퐷(−1)-quintuple exists, and there are only finitely many such quadruples (and all of them must contain the element 1); among the main contributors here are Filipin and Fuchs. These results solve an old problem investigated by Diophantus and Euler by showing that there does not exist a set of four positive integers with the property that the product of any two of its distinct elements plus their sum is a perfect square. Indeed, since 푥푦 + 푥 + 푦 = (푥 + 1)(푦 + 1) − 1, the existence of such a set would imply the existence of 퐷(−1)-quadruples with elements ≥ 2. Instead of over the integers and rationals, the problem can be considered over any commutative ring with unity. There are interesting results, due to Franušić and Soldo,

774 Notices of the AMS Volume 63, Number 7