If $ a+1=n^{2}$, then $ (2n)^{2}+(2a)^{2}=(2a+1)^{2}+3$, since there are infinitely many $ a$ for which $ a+1$ is a perfect square, this solves the problem.

$ a=2k, b=2k^2-2,c=2k^2-1$

Johan Gunardi wrote: $ a = 2k, b = 2k^2 - 2,c = 2k^2 - 1$ I think that's not correct: look at the coefficient of $ k$...

$ (2k)^2+(2k^2-2)^2 - (2k^2-1)^2 = 4k^2+4k^4-8k^2+4-4k^4+4k^2+1 = 3$, so it works.

I'm sorry, must have missed the squares. Never mind.

Can we find all solutions ?