Playing with Bezout's identity and Pythagorean triples, it seems that $a^{2}+b^{2}=c^{2}$ for coprime positive integers $a<b<c$ if and only if:
$$bx+cy=a^{2}$$
$$b^{2}x+c^{2}y=1$$
Or which is the same, when
$$bx+cy=a^{2}$$
$$(ab)^2x+(ac)^2y=a^{2}$$
However, I have no clue about how to prove or reject it, any help or hint you could provide would be welcomed.
Thanks!
If you solve the pair of simultaneous equations for $x$ and $y$, you get:
$$y = -\frac{a^2b-1}{c(c-b)}\\ x = \frac{a^2c-1}{b(c-b)}$$
A counterexample to the conjecture would be any Pythagorean triple where the above solution is not a pair of integers. It seemed to me that this was more likely to occur when $c-b>1$, and indeed, the first such triple, $(8,15,17)$ gives $y=\frac{959}{34}$ and is a counterexample.
Note that if you have any triple with $c-b=1$, then the conjecture is true. Such triples are parameterised by $(a,b,c)=(2k+1, 2k^2+2k, 2k^2+2k+1)$, and you can verify that the expressions for $x$ and $y$ simplify to integers.
I don't know if all triples with $c-b>1$ serve as counterexamples, but many do. For example, any primitive triple where $c-b$ is even will be a counterexample ($b,c$ have the same parity, both odd by primitivity, $a$ must then be even, and parity considerations show that $x,y$ are not integers).