Let a prime $p>2$ such that $p\nmid a$.Prove that the equation:
$$ax^2+bx+c \equiv 0 \pmod p$$
has a solution iff $p \mid b^2-4ac$ or $b^2-4ac$ is a quadratic residue $\mod p$.
Solve : $x^2+7x+10 \equiv 0 \pmod {11}$
Prove that $6x^2+5x+1 \equiv 0 \pmod p$ has a solution for each prime $p$.
Can I just take the discriminant $D=b^2-4ac$??
So the equation $ax^2+bx+c \equiv 0 \pmod p$ would have a solution,if $D \geq 0$.
If $D \equiv 0 \pmod p \Rightarrow p \mid D \Rightarrow p \mid b^2-ac$
Is it right so far? And how can I continue?
Best Answer
Consider $ax^2+bx+c \equiv 0 \pmod p$ and multiply by $4a$, this wont change things since $p\not \mid a$, $$4a^2x^2+4abx+4ac=(2ax+b)^2 -(b^2-4ac)$$ So we see that the original congruence has a solution iff the congruence
$$(2ax+b)^2 \equiv b^2-4ac\pmod p$$ holds in which case $b^2-4ac$ is a quadratic residue modulo $p$.