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.

  • $\begingroup$ Could you still explain why the case $p=2$ is obvious? $\endgroup$ – WLOG Oct 21 '19 at 4:51
  • 1
    $\begingroup$ 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 $2^{r+1}$. $\endgroup$ – WhatsUp Oct 21 '19 at 5:11
  • $\begingroup$ I finally got it. Thank you! $\endgroup$ – WLOG Oct 21 '19 at 5:22

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