Let p be an odd prime, where a,b are integers such that pł a and płb. Show that the congruence (x^2-a)(x^2-b)(x^2-ab) ≡ 0(modp) is always solvable.

Question

Let p be an odd prime, where a,b are integers such that $p\nmid a$ and $p\nmid b$. Show that the congruence $(x^2-a)(x^2-b)(x^2-ab) \equiv 0\bmod p$ is always solvable.

I know that (a/p) and (b/p) is equivalent to $(-1)^{(p^2-1)/8}$

Answer 1

The Legendre symbol $(\frac{\cdot}{p})$ is the multiplicative function defined on the integers so that $(\frac{a}{p})=0$ when $p|a$. Otherwise, $(\frac{a}{p})=1$ when the congruence $x^2-a\equiv 0\;\text{mod}\;p$ is solvable but $(\frac{a}{p})\equiv-1$ when said congruence is not solvable. To see the originally posted congruence has a solution it suffices to show that at least one of $(\frac{a}{p}),(\frac{b}{p}),(\frac{ab}{p})$ equals $1$. If $(\frac{a}{p})=(\frac{b}{p})=-1$ then $(\frac{ab}{p})=(\frac{a}{p})\cdot(\frac{b}{p})=1$.

Answer 2

Hint:

If $(a/p)=-1$ and $(b/p)=-1,$ then $(ab/p)=(a/p)(b/p)=1$.