# Given any integers $a,b,c$ and any prime $p$ not a divisor of $ab$, prove that $ax^2+by^2\equiv c\pmod{p}$ is always solvable.

The fact that there are $\dfrac{p+1}{2}$ quadratic residues seem to me to help solving the question, but I don't know how to go on from that point. Could you give me any hint?

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Yes, $m$ and $n$ are any integers which satisfy $(m,p)=(n,p)=1$. –  Aran Komatsuzaki Jun 7 at 5:30
I will rewrite the question to make it clear. I'm sorry! –  Aran Komatsuzaki Jun 7 at 5:33
I answered this before, here –  i707107 Jun 7 at 5:37
If $c\not\equiv0,1$ then multiply by $c^{-1}$. Handle $c\equiv0$ separately. –  blue Jun 7 at 5:44
This statement is a private more General. Formula more General there math.stackexchange.com/questions/794510/… –  individ Jun 7 at 12:01