# Can the equation $ax^2+by^2=cz^2$ be solved in integers (excluding trivial solutions)?

Suppose $a,b,c\in\mathbb{N}$ and are each squarefree. Is there a general solution for this equation?

I found that for this equation to be soluble in integers there are three necessary and sufficient conditions which are $ab,bc$ and $ac$ should be quadratic residues $\bmod c,\bmod a$ and $\bmod b$ respectively. That is, the equations

$$ab\equiv\alpha^2 \hspace{-0.8em}\pmod{c} \\ bc\equiv\beta^2 \hspace{-0.8em}\pmod{a} \\ ac\equiv\gamma^2 \hspace{-0.8em}\pmod{b}$$ should be solvable. These conditions can be easily derived from the given equation assuming that $a,b$ and $c$ are relatively prime in pairs, but I cannot understand how these are the only sufficient conditions and how the equation is solvable if these conditions are satisfied.

• It would be very helpful to point out what the symbol $\Box$ in the referenced theorem of Legendre means. Is it just any constant? Or some kind of determinant, maybe?