Are there any positive integers $a, b, c, d$ such that both $(a, b, c)$ and $(b, c, d)$ are Pythagorean triples? 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:


*

*$a,b,c$ is primitive (otherwise reduce by the gcd)

*likewise $b,c,d$ is primitive

*Using the traditional solution for a triple:
$2mn$ ; $m^2-n^2$ ; $m^2+n^2$, then $c$ is odd

*if $c$ is odd in the $(b,c,d)$ triple, then $b$ must be even = $2mn$

*Then we have ($2mn$, $m^2+n^2$, $d$ ) for the second triple

*Then $(2mn)^2$ + $(m^2+n^2)^2$ = $m^4 + 6m^2n^2 + n^4 = d^2$ which is a $\square$


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.
