Postby **pex** » Mon Mar 28, 2016 4:09 am UTC

I believe the general statement is "nonlinear Diophantine equations are hard".

Let k be an arbitrary integer. The solution family that you found is (a,b,c,d) = (k, k^{3}, 0, k^{2}); a couple of others are (0, k, -k, k^{2}) and (k^{3}, k, k^{5}-k, k^{2}), as well as (obviously) the same things with b and c swapped. I suppose there must be even more families, but haven't really looked into it yet.

Interestingly, the only solutions that I have encountered where d is not a square all have d=-5. Some solutions are

(-1, 2, 3, -5)

(-2, 1, 9, -5)

(-3, 1, 14, -5)

(-9, 2, 43, -5)

(-14, 3, 67, -5)

(-43, 9, 206, -5)

(-67, 14, 321, -5)

(-206, 43, 987, -5)

(-321, 67, 1538, -5)

(-987, 206, 4729, -5)

(-1538, 321, 7369, -5)

and of course, for every (a,b,c,d), there are also (a,c,b,d), (-a,-b,-c,d), and (-a,-c,-b,d). There seems to be a pattern among the solutions for d=-5 (if a number shows up as b or c, it also appears as a in another solution), but I haven't bothered proving this yet. Such chains of solutions are not uncommon in Diophantine equations, though.

In general, you want to find a and d such that M^{2} = (ad)^{2} - 4a^{2} + 4d is a perfect square; the corresponding b and c are (ad ± M) / 2. I do not know of a generic way to find such solutions, but then again, there is much that I don't know about Diophantine equations.