Theorem (Legendre): Let $a,b,c$ coprime positive integers, then $ax^2 + by^2 = cz^2$ has a nontrivial solution in rationals $x,y,z$ iff $\left(\frac{-bc}{a}\right)=\left(\frac{-ac}{b}\right)=\left(\frac{ab}{c}\right)=1$.
I'm trying to prove this theorem using
Lemma Modified global square theorem: The rational number $cz$ is a $c$ times a square iff it is a $c$ times a square in $\mathbb{Q}_p$ for every prime $p$.
So far, it is possible to show that the equation can be solved locally for all primes powers except $2^r$:
To solve this equation $\pmod {abc}$ just put $x = bc$, $y = ac$, $z = ab$. Reducing this gives a solution for every odd prime $p | abc$. Hensel lifts it to $\mathbb{Q}_p$.
If an odd prime $p \not| abc$ then put $z = 1$ so that we have $ax^2 \equiv c – by^2 \pmod p$, $x$ and $y$ can take on $(p+1)/2$ values each so they must have an intersection which solves this congruence. Hensel lifts it to $\mathbb{Q}_p$ again.
It's easy to solve when $p=2$ but I think it $p=4$ needs to be done for Hensel to apply.
My question is how to get a rational number out of this, so that I can apply the global square theorem and conclude? It seems like the $p$-adic numbers have to have a finite expansion but that seems just as hard to prove as anything. Also if anyone has a hint for the $2^r$ case that would be great too! Thanks a lot.
Best Answer
Best link about solvability $ax^2+by^2=cz^2$ is a chapter 2 of book
K. Kato, N. Kurokawa, T. Sato Number Theory 1