# Applying Hensel's lemma

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

• Wait, what ring are you trying to solve this in? Hensel's lemma only works on local rings with some other mild hypotheses. Commented Feb 1, 2015 at 21:30
• I'm trying to prove there is a solution $\mod {p^k}$ for $k = 1,2, \dots$ Commented Feb 1, 2015 at 21:32
• @AdamHughes, problem 121 on page 79 of Gouvea, $p$-adic Numbers. Presumably, several other books. Commented Feb 1, 2015 at 21:33
• @WillJagy I always use the 3,4,5 example myself, but mostly I was trying to make sure we were on the same page--the op and I--since that's a critical bit of info to solve the problem. Thanks for clarifying! Commented Feb 1, 2015 at 21:51
• @AdamHughes, this may be peripheral, but $34$ is the smallest number where $x^2 - 34 y^2 = -1$ is impossible in $\mathbb Z,$ despite $x^2 - 2 y^2 = -1$ and $x^2 - 17 y^2 = -1$ being easy in integers. The first odd (not square) number that does that is $205,$ as $x^2 - 205 y^2 \neq -1$ in $\mathbb Z.$ Next odd is $221.$ Commented Feb 1, 2015 at 21:56

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 $\pm 1$.
• This is good, so +1, but it's also incomplete IMO, you should address $p=17$ (easy) and $p=2$ (almost as easy). Commented Feb 1, 2015 at 22:16