Primitive Pythagorean Triples: An Alternative Formula

Primitive Pythagorean Triples: An Alternative Formula Benson Schaeffer∗ Portland, OR, USA Abstract In this paper I prove that the formula below generates all the possible primitive Pythagorean triples (2x qr + 22x−1 q 2 )2 + (2x qr + r2 )2 = (2x qr + 22x−1 q 2 + r2 )2 , x is a positive integer, q and r are relatively prime odd integers. 1 Introduction The standard formula for the production of all possible primitive Pythagorean triples [3, pp. 237] is well known. There is an alternative, however, (2x qr + 22x−1 q 2 )2 + (2x qr + r2 )2 = (2x qr + 22x−1 q 2 + r2 )2 , (1) x a positive integer, q and r relatively prime positive odd integers. After developing equation (1), I discovered that it had been previously worked out by Bottari [2, pp. 169]. Below I present a proof independent of his that equation 1 generates all possible primitive Pythagorean triples. AMS 2020 subject classification: 11A99, 11D41 Diophantine equations, Fermat’s equation ∗ The corresponding author. E-mail: bookie@hevanet.com 1 First I present and demonstrate the validity of an alternative to the standard binomial expansion formula [3, pp. 9]. Then, in the proof proper, I use the alternative binomial formula to show that equation 1 generates all possible primitive Pythagorean triples. 2 Lemma: Alternative Binomial Formula An alternative to the standard binomial expansion formula [2, p. 9] for a and b nonzero integers, n a positive integer, is n n (a + b) = a + b n X an−i (a + b)i−1 . (2) i=1 The validity of the alternative binomial formula can be demonstrated using the formula d(1 − rn ) , with d = 1 and for the sum of a finite geometric series [1, p. 10], Sn = 1−r   a r= , as follows, a+b n n X X an−1 (a + b)i−1 n n−i i−1 n n−i a +b a (a + b) = a + b(a + b) (a + b)n−1 i=1 i=1 n−i n  X a n n−i = a + b(a + b) a+b "i=1
# a n 1 − a+b
= an + b(a + b)n−i a 1 − a+b   n n n n−i [(a + b) − a ](a + b) = a + b(a + b) b(a + b)n = an + (a + b)n − an = (a + b)n . 3 Proof: To start, let A, B, C be a Pythagorean triple such that A2 + B 2 = C 2 , (3) A, B, C pairwise relatively prime positive integers. Since either A or B must be even and the other odd [3, pp 236], equation (3) can be rewritten as A2 + (A + d)2 = (A + d + e)2 , d, e positive integers, A, (A + d) = B and (A + d + e) = C pairwise relatively prime. 2 (4) I now use the alternative binomial formula to show for A2 and rewrite equation (4), A2 = (A + d + e)2 − (A + d)2 , 2 2 A = (A + d) + e p X (A + d)2−i (A + d + e)i−1 − (A + d)2 i=1 A2 = e 2 X (A + d)2−i (A + d + e)i−1 . i=1 Since either of the terms in the sum above, e(A + d) or e(A + d + e) would be greater than A2 where e ≥ A, A must be greater than e, A = f + e, f a positive integer. Equation (3) can be rewritten, first, as (f + e)2 + (f + e + d)2 = (f + e + d + e)2 , and letting e + d = g, then as (f + e)2 + (f + g)2 = (f + e + g)2 . (5) Inspection of (5) makes it clear that f must be even. I therefore rewrite it as (2x h + e)2 + (2x h + g)2 = (2x h + e + g)2 , (6) x a positive integer, h an odd positive integer, and apply the alternative binomial formula determine the value of (2x h)2 = 22x h2 . (2x h + e)2 + (2x h + g)2 = (2x h + e + g)2 , (2x h)2 + e +(2x h)2 + g 2 X i=1 2 X (2x h)2−i (2x h + e)i−1 (2x h)2−i (2x h + g)i−1 = (2x h)2 + e i=1 2 X (2x h)2−i (2x h + e + g)i−1 i=1 +g 2 X (2x h)2−i (2x h + e + g)i−1 i=1 x 2 2x 2 (2 h) = 2 h =e 2 X i=2 +g i starting at 2 because 2 X i=2   (2x h)2−i (2x h + e + g)i−1 − (2x h + e)i−1   (2x h)2−i (2x h + e + g)i−1 − (2x h + g)i−1 , 2x h + e + g)1−1 − (2x h + e)1−1 = 2x h + e + g)1−1 − (2x h + g)1−1 = 1 − 1 = 0. 3 Therefore, 22x h2 = eg + ge 22x h2 = 2eg 22x−1 h2 = eg. (7) Equation (7) tells us that 22x−1 h2 contains all of the prime factors of e and g are relatively prime, because otherwise (2x h + e), (2x h + g) and (2x h + e + g) would not be pairwise relatively prime,; and that the odd prime factors of e and g are squares [3, pp.236]. The above allows me to rewrite e, g and 2x h as e = 22x−1 k 2 g = m2 22x h2 = 2eg = 2.22x−1 k 2 m2 , 22x h2 = 22x k 2 m2 , 2x h = 2x km, k and m relatively prime odd positive integers, or k = 1 and m 6= 1, k 6= 1 and m = 1, or k = m = 1. Equation (6) thus becomes (2x km + 22x−1 k 2 )2 + (2x km + m2 )2 = (2x km + 22x−1 k 2 + m2 )2 , directly analogous to equation (1), (2x km + 22x−1 k 2 ), (2x km + m2 ) and (2x km + 22x−1 k 2 + m2 ) relatively prime. Conversely (2x qr + 22x−1 q 2 )2 + (2x qr + r2 )2 = (2x qr + 22x−1 q 2 + r2 )2 22x q 2 r2 + 23x q 3 r + 22x q 2 r2 + 2x+1 qr3 + r4 = 22x q 2 r2 + 23x q 3 r + 2x+1 qr3 + r4 + 22x q 2 r2 , q and r relatively prime by assumption, or q = 1 and r 6= 1, q 6= 1 and r = 1, or q = r = 1, and (2x qr + 22x−1 q 2 ), (2x qr + r2 ) and (2x qr + 22x−1 q 2 + r2 ) relatively prime. This completes the proof that equation (1) generates all possible primitive Pythagorean triples. References [1] M. Abramowitz and I. A. Stegun (Eds.), Handbook of Mathematical Functions, pp. 10, Dover Publications, New York, 1965. 4 [2] A. Bottari, Periodico di Matematica, 23 (1908), pp. 104-110. Cited in L.E. Dickson, History of the Theory of Numbers: Diophantine Analysis, pp. 169, AMS Chelsea Pub- lishing, 1966. [3] D.M. Burton, Elementary Number Theory, Fifth Edition, pp. 9, 236, 237, McGraw-Hill, Boston, 2002. 5