Let $r,s,t$ be positive integers. Do there exist integers $a,b,c$, not all 0, satisfying $$r a^2+s b^2=t c^2\enspace ?$$
Let's call $(r,s,t)$ "valid" if such a solution exists, and "invalid" otherwise. The goal is to characterize the valid $(r,s,t)$ triples.
Note that we can always divide out by factors common to $(r,s,t)$, without affecting validity. We can also divide any of the terms by any square factor. Therefore, it suffices to characterize $r,s,t$ that are square free and share no common factors. If some prime factor $p$ divides two of $(r,s,t)$, we can first multiply by $p$, and then divide the two terms by $p^2$. Therefore, we can even assume that any prime factor divides at most one of $r$, $s$, or $t$.
Not all $(r,s,t)$ triples are valid: by looking at things mod 4, we can conclude that for valid triples, there must exist $A,B,C\in\{0,1\}$, not all 0, such that $r A+s B\equiv t C\bmod 4$. To see this, simply eliminate all common factors from $(a,b,c)$, and then set $(A,B,C)=(a^2,b^2,c^2)\bmod 4$. So for example, $(r,s,t)=(1,1,3)$ has no solution.
On the other hand, if $t=1$ and either of $r,s$ are equal to 1, then there exists a solution. Concretely, $(1,s,1)$ has a solution $$(a,b,c)=\left(s-1\;,\;2\;,\;s+1\right)$$ An analogous solution shows that the triple $(r,1,1)$ is also valid.
Do any other valid $(r,s,t)$ triples exist? If not, how do we prove that the above captures all valid $(r,s,t)$ triples?
