### 26th Putnam 1965

Problem B3

Show that there are just three right angled triangles with integral side lengths a < b < c such that ab = 4(a + b + c).

Solution

Answer: 12, 16, 20; 10, 24, 26; 9, 40, 41.

We need the result that for some integral d, m, n we have c = d(m2 + n2) and b = 2dmn, a = d(m2 - n2) or b = d(m2 - n2), a = 2dmn (*).

It follows that 4(a + b + c) =4d(2m2 + 2mn), ab = 2d2mn(m2 - n2). Hence, 4 = dn(m - n). So we must have n = 1, 2 or 4 and (d, m, n) = (1, 5, 1), (2, 3, 1), (4, 2, 1), (1, 4, 2), (2, 3, 2) or (1, 5, 4), giving the three answers above.

It remains to prove (*). Let d be the gcd of a and b. It follows that d also divides c. Put a = dA, b = dB, c = dC, so that A, B, C are relatively prime in pairs. C cannot be even, for then A and B would both be odd and hence A2 + B2 would be congruent to 2 mod 4, which is impossible, since C is a square. So C must be odd and just one of A, B must be even. Assume A is odd. Then A2 = (C - B)(C + B). If an odd prime p divides A, then it must divide C - B or C + B. It cannot divide both, for then it would also divide B and C. So C - B and C + B must both be odd squares. Say C + B = h2, C - B = k2. Then A = hk, B = (h2 - k2)/2, C = (h2 + k2)/2, with h and k odd. Put m = (h + k)/2, n = (h - k)/2, then h = m + n, k = m - n and A = m2 - n2, B = 2mn, C = m2 + n2, so we have put a, b, c, in the form (*). On the other hand, it is obvious that if a, b, c have this form then they satisfy a2 + b2 = c2.

© John Scholes
jscholes@kalva.demon.co.uk
25 Jan 2002