TWO THEOREMS ON THE STRUCTURE OF PYTHAGOREAN TRIPLES AND SOME DIOPHANTINE CONSEQUENCES By Konstantine ‘Hermes’ Zelator 1. INTRODUCTION The major theorem of this paper (Theorem 1) deals with the structure of primitive pythagorean triangles. In Theorem 1, it is proven that there are no primitive 2 2 pythagorean triangles of the form ( s1 x,, s2 y z ), i.e., triangles with one leg equal to s1 times a perfect square and the other equal to s2 times an integer square, if the integers s1 and s2 satisfy the following conditions: a) s1 and s2 are both positive odd squarefree integers. b) s1 s 2= p 1... pn , n≥ 2, p1 ≡ 5, p2 ≡ ... ≡ pn ≡ 1 mod8 , where p1 ,..., pn are primes. c) If s1≡1 mod 8, (and then of course s2≡5mod8) , then s1 is a quadratic nonresidue of every divisor d of s2 , with d > 1. (If s2 ≡1mod8 , then the same condition is assumed on s2 with respect to the divisors d of s1 , d > 1.) One can easily find many examples of integers s1 , s2 satisfying the above conditions. Below, we offer a few. 1. s1 , s2 primes with s1≡ 1, s 2 ≡ 5mod8; s1 a quadratic nonresidue ofs2 . 2. s1= p 1 p 2, p 1≡ p 2 ≡ 1mod8,p1 , p 2 primes, s2 also a prime, p1 a quadratic nonresidue of s2 , while p1 a quadratic residue of s2 . 3. s1 ≡1mod8,s1 a prime, s2= p 1 p 2,, p 1 p 2 prime with p1 ≡1, p2 ≡ 5mod8;s1 a quadratic nonresidue of both p1 and p2 . 4. s1= p 1... pn ; n≥ 3, n an odd integer, s2 a prime, s2 ≡5mod8; p1 ,..., pn primes with p1 ≡... ≡ pn ≡ 1mod8; an even number of integers among the primes p1 ,..., pn being quadratic residues of s2 , while the remaining primes (odd in number) being quadratic nonresudues of s2 . Theorem 2 is somewhat more technical in nature. It is shown that if n is a positive odd squarefree integer such that whenever s1 s 2 = n, then 2 2 4 2 s1≡1, s 2 ≡ 5mod8(or vice versa), the diophantine equation (nx) + y = z , with (nx, y)= 1 is solvable in Z+, if and only if the diophantine equation 2 2 4 2 4 + d1 z= d2 x + d3 y is solvable in Z , for some positive divisors d1,, d 2 d 3 of n with d1 d 2 d 3 = n, d2 < n and d3 < n. It is fairly easy to see that n, satisfying the hypothesis of Theorem 2, is either the product of an odd number of primes p ≡5mod 8, or otherwise the product of an odd of primes p ≡5mod 8, and any number of primes q ≡1mod 8. Theorem 3 is simple in its statement and proof. It states that if p≡ q ≡1mod 4 and q primes, p a quadratic nonresidue of q, then the diophantine equation 2 2 ()()px2 + qy2 = z 2 with (px, qy) = 1 , has no solution in Z+. ~ 2 ~ By combining Theorems 2 and 3, Theorem 4 is established. It states that if p≡1, q ≡ 5mod 8, p a quadratic nonresidue of q , the diophantine equation 2 ()pqx2 + y4 = z 2 , with ()pqx, y = 1, has a solution in Z+, if and only if the equation pqz2= x 4 + y 4 , with ()x, y = 1, has a solution in Z+. 2. THEOREM 1 Let s1, s 2 be two odd squarefree positive integers greater than 1, with (s1, s 2 )= 1and such that s1 s 2 is the product of a prime congruent to 5 mod 8 and primes congruent to 1 mod 8,i.e., s1 s 2= p 1... pn , n≥ 2, p1 ≡ 5mod8,p2 ≡ ...≡ pn ≡ 1mod8.In addition, assume that if s1 ≡1mod8 (same condition on s2 , if s2 ≡1mod8 ), then s1 is a quadratic nonresidue of every divisor d of s2 , including s2 itself, with d > 1. 2 4 2 4 2 Under the above assumptions, the diophantine equation s1 x+ s2 y = z has no + solution in the set of positive integers Z , with (s1 x, s 2 y)= 1. Remark One can easily find examples of s1, s 2 satisfying the hypothesis of the theorem. For instance, s1, s 2 being primes with s1≡ 1, s 2 ≡ 5mod 8 and s1 a quadratic nonresidue of s2 . Also, s1= p 1 p 2, p 1≡ p 2 ≡ 1mod8 , where p1, p 2 are primes, s2 also a prime, s2 ≡5mod 8such that p1 being a quadratic nonresidue of s2 , while p1 a quadratic residue of s2; s 1 ≡ 1mod8 , s1 a prime, while s2= p 1 p 2,, p 1 p 2 primes, p1 ≡1, p2 ≡ 5mod8, and with s1 a quadratic nonresidue of both p1 and p2 . Proof. Let us assume that (x,, y z) is a solution with (s1 x, s 2 y) = 1 in positive integers x , y, z to the equation. 2 2 2 2 2 ()()s1 x+ s2 y = z ( ) 1 In virtue of (s1 x, s 2 y)= 1, we see that equation (1) describes a primitive Pythagorean triangle. Assume first x to be odd any y even. From (1), it follows that 2 2 2 2 2 2 s1 x= m − n , s2 y= 2 mn , z= m + n ( ) 2 For positive integers m, n with ()m, n = 1 and m+ n ≡ 1mod2 The first equation in (2) implies, since (m , n )= 1, ~ 3 ~ 2 2 2 2 d1 m 1 − d2 n 1 d1 m 1 + d2 n 1 x= km n , n= k , m= k , (3) 1 1 2 2 for positive integers m1,,,, n 1 d 1 d 2 k with (m1 , n 1 )= 1, d1 d 2= s 1 and k = 1or 2. (Refer to References[1]or [2].) Note that s1 is squarefree, so must be d1 and d 2 .Now, x is odd and so, from the first equation in (3), we see that k = 1 and m1≡ n 1 ≡ 1mod2. On the other hand, the second equation in (2) implies the following two possibilities or sub cases: 2 2 m= 2δ1 M , n= δ 2 N ( ) 4 2 2 m=δ1 M, n = 2δ 2 N ( ) 5 Where δ1,δ 2 are positive integers with δ1δ 2= s 2 , (they are squarefree since s2 is), and MN, are positive integers with (MN,= 1), since (m , n )= 1. Also, since, as we have shown above, k=1, the last two equations in (3) take the form 2 2 2 2 2n= d1 m 1 − d2 n 1 , 2 m= d1 m 1 + d2 n 1 ( ) 6 First assume (4) to be the case.Then of course, since m and n have different parities, m must be even and n odd, and so N must also be odd. Combining equations (4) and (6), we obtain 2 2 2 2 2 2 2δ 2 N= d1 m 1 − d2 m 1 , 4δ1M= d1 m 1 + d2 n 1 . (7) By virtue of N≡ m1 ≡ n 1 ≡ 1 mod2 (7) yields 2 2δ 2≡d 1 − d 2 mod8 , 4δ1M≡ d1 + d 2 mod8. (8) According to the hypothesis of Theorem 1, either s1 is a prime congruent to 5mod 8 or a product of a prime congruent to5mod 8 and primes congruent to 1mod8 and s2 is a product of primes congruent to 1mod8, or vice versa. Suppose, under the hypothesis of Theorem 1, that s2 ≡5mod8, while s1 ≡ 1mod8; then from d1 d 2= s 1 and δ1δ 2= s 2 , we conclude d1≡ d 2 ≡ 1mod8and δ1 ≡ 1,δ 2 ≡ 5 or δ1 ≡ 5,δ1 ≡ 1mod8 . Then, however, the first congruence in (8) implies 2δ 2 ≡ 0mod8, a contradiction since δ 2 is odd. Now, if s1 ≡ 5and s2 ≡ 1mod8 , we obtain, from s1= d 1 d 2 and s2= δ 1δ 2 , under the hypothesis of the theorem, ~ 4 ~ d1 ≡1, d2 ≡ 5 or d1 ≡5, d 2 ≡ 1and δ1≡ δ 2 ≡1mod8 The first equation in (8) implies 2δ 2 ≡ 4mod8, a contradiction since δ 2 is odd. Next, assume equation (5) to be the case; then M is odd. By combining equations (5) and (6), we arrive at 2 2 2 2 2 2 4δ 2 N= ±( d1 m 1 − d2 n 1 ) , 2δ1 M= d1 m 1 + d2 n 1 . (9) Repeating the reasoning we applied when we considered (8), we see that if s1 ≡ 5 and s2 ≡ 1mod8, then d1 ≡1, d2 ≡ 5 or d1 ≡ 5, d 2 ≡ 1mod8 while δ1≡ δ 2 ≡ 1mod8. 2 The second of (9) yields 2δ1M ≡ 2 δ1 ≡d 1 + d 2 mod8; but 2δ1 ≡ 2, while d1+ d 2 ≡6mod8, whence a contradiction. Finally, assume s1 ≡ 1and s2 ≡ 5mod8. According to Legendre’s theorem, since Nm1 n 1 ≠ 0, the first equation in (9) implies that d1⋅ d 2 is a quadratic residue of δ 2 ; but d1 d 2= s 1 ≡ 1mod8 and δ 2 divides s2 .Therefore, if δ 2 is a divisor of s2 , with δ 2 > 1, by the hypothesis of the theorem, it would follow that s1 is a quadratic nonresidue of δ 2 , whence a contradiction. Now, if δ 2 = 1 then δ1δ 2=s 2 ⇒ δ 1= s 2 . Then, however, the second equation of (9) and Legendre’s theorem imply that −d1 d 2 = − s 1 is a quadratic residue of δ1= s 2 ; but -1 is a quadratic residue of s2 (since s2 is a product of primes congruent to 1 mod 4). Hence s1 is a quadratic residue of s2 , contrary to the hypothesis of the theorem.