Pythagorean triple is a triple of integers $(a, b, c)$ such that $a^2+b^2=c^2$. Is there any Pythagorean triple such that, not only $a^2+b^2$, but also $b^2+c^2$ is a square number? If not, how to prove it?
I tried to prove non-existence the following way: If true, it would mean that there is a pair of integers such that both sum and difference of their squares is a square number. Let's call these integers $a$ and $b$ and $a<b$. Then, there are integers $c$ and $d$ such that: \begin{align} &a^2+b^2=c^2 \\ &a^2-b^2=d^2 \end{align}
Multiplying those equations gives: \begin{equation} a^4=(cd)^2+b^4 \end{equation}
This is similar to Fermat's Last Theorem for $n=4$, but using it only shows that $cd$ can't be square number, not that there are no integer solutions.
For the $(a,b,c)$ and $(b,c,d)$ Pythagorean triples to work, we note:
But H.C. Pocklington of St John's College in 1913 proved that 6 for the coefficient of $x^2y^2$ in the general equation $x^4 + dx^2y^2+y^4 = z^2$ is impossible by prime moduli, in this case 6 != 7 mod 8 See H. C. Pocklington. "Some diophantine impossibilities" Proc. Camb. Phil. Soc, 17: pf 110 – 118, 1914.