Given any integers $a,b,c$ and any prime $p$ not a divisor of $ab$, prove that $ax^2 + by^2
\equiv с \pmod{p} $ is solvable.
$p$ does not divide $ab$ implies $p$ does not divide both $a$ and $b$. But from here how can I show that the aforesaid equation is solvable?
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abcdmath
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pigeonhole principle: there are, including $0,$ exactly $\frac{p+1}{2}$ distinct values of $t^2 \pmod p ; $ ( this is for odd prime $p$ ). There are $\frac{p+1}{2}$ distinct values of $a x^2 \pmod p ; $ and $\frac{p+1}{2}$ distinct values of $c-b y^2 \pmod p , ; $ so there is an overlap. – Will Jagy Nov 24 '18 at 01:28
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2You can apply the method in one of my answer, and obtain a formula for the number of solutions to your equation: https://math.stackexchange.com/questions/398200/the-number-of-solutions-of-ax2by2-equiv-1-pmodp-is-p-frac-abp/398245#398245 – Sungjin Kim Nov 24 '18 at 02:00