<s>[INST] Pythagorean triples of a specific form I would like to construct infinitely many Pythagorean triples where c = b+3. Proving there are infinitely many should not be difficult but finding the right form for a b and c is proving to be difficult. I have tried modifying the typical a=n^2 + m^2   b=2nm   c=n^2-m^2  but that hasn't worked out for me yet. Any suggestions on what to start with? Thanks!
 [/INST] First, note that we have $a^2=c^2-b^2=6b+9$. We want this to be a perfect square; it's easy to see that we must have $3|b$ and modular considerations show we have to have $2|n$; a little computation suggests to us that $b=6n(n+1)$ is always a perfect square, with $a^2=36n^2+36n+9=9(2n+1)^2$. Now $a^2+b^2=36n^2+36n+9+36n^4+72n^3+36n^2=9(4n^4+8n^3+8n^2+4n+1) = (6n^2+6n+3)^2=(b+3)^2$.
So the requested triple is $(6n+3,6n^2+6n,6n^2+6n+3)$. I doubt there are others, but I'm not going to prove that.
</s>