Try as I may, I cant devise a short, self-contained proof for this problem (any ideas..?). However, I can get there by building on a theorem that is quite well known but would, itself, take up too much space here so Ill just quote it

Theorem: A positive integer can be written as the sum of two squares if and only if all its prime factors of the form 4n + 3 occur to even powers. (See any basic text on number theory or Google The Two Squares Problem.).

4xy x y = w^{2}can be written as(x + y)^{2} (x y)^{2} (x + y) = w^{2}

So thatu(u 1) = v^{2} + w^{2}(1)whereu = x + y and v = x y

If we can prove that u(u 1) has a prime factor of the form 4n + 3 that occurs to an odd power, then equation (1) together with the theorem will give us the contradiction we need.

All odd integers are ofthe form 4n + 1 (type 1) or 4n + 3 (type 2), so type 1 x type 1 = (4n + 1)(4m + 1) = 4(4mn + m + n) + 1 = type 1 type 1 x type 2 = (4n + 1)(4m + 3) = 4(4mn + m + 3n) + 3 = type 2 type 2 x type 2 = (4n + 3)(4m + 3) = 4(4mn + 3m + 3n + 2) + 1 = type 1

It follows that if an odd integer of type 2 is written as a product of prime numbers, then an odd number of primes of type 2 must be involved (counting repetitions). So at least one of those type 2 primes must be raised to an odd power.(2)

Sinceu v = 2x, u and v must both be even or both odd. Also, since u(u 1) is even, (1) shows that v and w must both be even or both odd. Hence, u, v and w must all be even or all odd.

Case 1:u, v and w even. Here the RHS of (1) must be divisible by 4, and, since u 1 is odd, it follows that u is divisible by 4, and that u 1 is of the form 4n + 3. Case 2: u, v and w odd. Writing u = 2U + 1, v = 2V + 1, w = 2W + 1, (1) now gives (2U+1)2U = (2V + 1)^{2} + (2W + 1)^{2} 2U = 4(V^{2} + V + W^{2} + W U^{2}) + 2 u = 2U + 1 = 4(V^{2} + V + W^{2} + W U^{2}) + 3 So u is of the form 4n + 3.

In both cases, the odd part of the product u(u 1) is of the form 4n + 3, and therefore, by statement (2), must contain a prime factor of the form 4n + 3 raised to an odd power. Since u and u 1 are consecutive integers they have no common factors so, in both cases, u(u 1) will also have that same prime factor raised to the same odd power, and cannot therefore be expressed as the sum of two squares, which contradicts (1) and proves that the original equation has no integer solutions.