Find three distinct integers, X, Y and Z, such that X + Y, X + Z, Y + Z, X - Y, X - Z, and Y - Z are all squares of integers.
Apparently, there are many solutions.
Find the set [X, Y, Z] with the smallest X + Y + Z.
With the restriction that Z = -Y, the necessary squares are reduced to three in number; namely X + Y, X � Y and 2Y.
If we writeX + Y = a2,X � Y = b22Y = c2
thenX = (a2 + b2)/2,Y = (a2 � b2)/2
and we require that a2 = b2 + c2. We can use standard formulae to generate all coprime Pythagorean triples (a, b, c) using positive integers p, q, k, with p and q coprime and of opposite parity, and with p > q. Case 1:a = k(p2 + q2),b = k(p2 � q2),c = 2kpq
giving:X = k2(p4 +q4)andY = 2k2p2q2
For small values of p, q, k, these generate [X,Y] pairs like: [17,8], [68,32], [97,72], [153,72], [257,32], [272,128], [337,288]...
Since p and q have opposite parity, in case 2, X and Y will be integers iff k is even. So we could put k = 2K and for K = 1, 2, .. use the modified formula:
X = 2K2(p4 + q4 + 6p2q2)andY = 2K2(p2 � q2)2
For small values of p, q, k, these generates [X,Y] pairs like: [82,18], [328,72], [626,50], [706,450], [738,162], [1312,288]...
If k = 1, only coprime triples (a, b, c) will be generated by the coprime pairs (p, q). So the parameter k is needed to produce all multiples of these triples.
Some thoughts (spoiler)
