Applying Hensel's lemma

Question

I'm trying to prove that the following equation:

$$(x^2 - 2) (x^2 - 17) (x^2 - 2\cdot 17) = 0$$

has solutions $ \pmod{p^k}$ for all $p,k$. It's easy to find nonzero solutions $ \pmod{2,17} $ - and for all other primes it follows from the fact that either $ 2 $ or $ 17 $ is a square residue $ \pmod{p} $, or their product is. Now I'm trying to use Hensel's lemma to lift the $\mod p$ solutions to solutions $\mod p^k$ for $k=2,3\ldots$ - but how do I prove that

$$ \frac{d}{dx} (x^2 - 2) (x^2 - 17) (x^2 - 2\cdot 17) \neq 0 \pmod{p} \text{ for some } x \text{ satisfying the equation?} $$

Thanks in advance - I would appreciate some help

Answer 1

For every prime $p>3$ we have that $G=\mathbb{Z}/(p^k\mathbb{Z})^*$ is a cyclic group and: $$ |G| = (p-1)\,p^{k-1}. $$

If $p\neq 17$, then both $2,17$ and $2\cdot 17$ belong to $G$.

For every $g\in G$, we have that the Legendre symbol $ g^{\frac{|G|}{2}} $ can be only $\pm 1$, and it equals one iff $g$ is a square in $G$. Since the Legendre symbol is multiplicative, we have that at least one element among $2,17,34$ is a square in $G$.

^{Footnote: The square of the Legendre symbol $g^{\frac{|G|}{2}}$ is one, and the equation $x^2-1=0$ has the same number of solutions in $\mathbb{Z}/(p\mathbb{Z})^*$ and in $\mathbb{Z}/(p^k\mathbb{Z})^*$ by Hensel's lemma. The first group is a field, hence the only square roots of one in $\mathbb{Z}/(p^k\mathbb{Z})^*$ are $\pm 1$.}