# Prove that $F(x)=(x^2-17)(x^2-19)(x^2-323)\equiv 0 \pmod{p^j}$ is solvable for all $p$ prime and $j\in\mathbb{N}$

Problem: Prove that $$F(x)=(x^2-17)(x^2-19)(x^2-323)\equiv 0 \pmod{p^j}$$ is solvable for all $$p$$ prime and $$j\in\mathbb{N}$$.

My attempt: With the help of Euler's criterion, I was able to prove that $$F(x)\equiv0 \pmod{p}$$ for any prime $$p$$. Then the hint given for this problem was to use Hensel's lemma to prove the existence of solutions to $$F(x)\equiv0\pmod{p^j}$$ for $$j\geq2$$. But I had trouble applying Hensel's lemma because, for example, $$2\mid F'(x)=2x(x^2-19)(x^2-323)+2x(x^2-17)(x^2-323)+2x(x^2-17)(x^2-19),\forall x\in\mathbb{Z}.$$ so I can't use Hensel's lemma to guarantee a solution for $$F(x)\equiv0\pmod{2^j}$$, $$j\geq2$$.

Could anyone suggest a solution?

You don't apply Hensel's lemma to the whole $$F$$.

Instead, try to apply it to each factor.

More precisely, for any odd prime number $$p$$, one of the three factors $$x^2 - 17$$, $$x^2 - 19$$, $$x^2 - 323$$ admits a root modulo $$p$$ (this is what you've got).

You then ask Hensel to give you a root modulo $$p^j$$ for all $$j > 0$$.

The case $$p = 2$$ is of course special, but obvious.

• Could you still explain why the case $p=2$ is obvious? – WLOG Oct 21 '19 at 4:51
• It's similar to Hensel. You can prove by induction that $x^2-17$ has a root modulo any $2^r$. For $r=3$ it's clear. If $t$ is a root modulo $2^r$, then one of $t$ and $t+2^{r-1}$ is a root modulo $2^{r+1}$. – WhatsUp Oct 21 '19 at 5:11
• I finally got it. Thank you! – WLOG Oct 21 '19 at 5:22