<s>[INST] Congruence modulo ($2^{10}$) Good evening, i’d like to discuss the following congruence which i’m stuck with, with you, hoping to find answers :
Find the number of solution of
$$x^5-16x\equiv 0 \mod 2^{10}$$
I think i have to imply a congruence (mod $2^5$) to find conditions on x and think i’d like to say that x has to be even.
Then i’d like to return mod $2^{10}$ or at least being helped by Group Theory finding condition on the moltiplicative order of x.
In this way i could find an isomorphism between Z/2^5z and Z/2^3z x Z/2z and easily conclude after knowing the order of x.
I think i should procede this way but i don’t really have any idea to find the solution.
Any tip or advice would be amazing, 
Thanks!
Ps. I think i know the number of solution, should be eight.
 [/INST] First, we note that 
$$
x^5 - 16x \equiv (x-2)x(x+2)(x^2+4) \equiv 0 \mod 2^{10}
$$
Hence $x$ is even, $x=2t$. The equation becomes:
\begin{gather}
(2t-2)(2t)(2t+2)(4t^2+4) \equiv 2^5(t-1)t(t+1)(t^2+1)\equiv 0 \mod 2^{10}\\
(t-1)t(t+1)(t^2+1)\equiv 0 \mod 2^{5}
\end{gather}
Now we have again two cases:
Case 1: $t$ is even, $t=2w$. Hence we have
\begin{gather}
(2w-1)2w(2w+1)(4w^2+1) \equiv 0 \mod 2^5\\
w\equiv 0 \mod 2^4
\end{gather}
So we have solutions $x$ of the form $x=4w$ with $0\leq x < 2^{10}$ and $w\equiv 0 \mod 2^4$. Then we have $0\leq w<2^8$ with $w\equiv 0 \mod 2^4$. We obtain $\frac{2^8}{2^4} = 2^4$ solutions.
Case 2: $t$ is odd, $t=2w+1$. Hence we have
\begin{gather}
(2w)(2w+1)(2w+2)(4w^2+4w+1+1) \equiv 0 \mod 2^5\\
w(w+1)\equiv 0 \mod 2^2
\end{gather}
So we have again to split:


*

*$w\equiv 0\mod 2^2$: then we have solutions $x$ of the form $x=2t=4w+2$ with $0\leq x < 2^{10}$ and $w\equiv 0 \mod 2^2$. Then we have $0\leq w < \frac{2^9-1}{2}$ with $w\equiv 0 \mod 2^2$. We obtain $\left\lceil\frac{2^9-1}{2\cdot 2^2}\right\rceil = 2^6$ solutions.

*$w\equiv 3\mod 2^2$: then we have solutions $x$ of the form $x=2t=4w+2$ with $0\leq x < 2^{10}$ and $w\equiv 3 \mod 2^2$. Then we have $0\leq w < \frac{2^9-1}{2}$ with $w\equiv 3 \mod 2^2$. We obtain $\left\lceil\frac{2^9-1}{2\cdot 2^2}\right\rceil = 2^6$ solutions.


Finally we get $2^4+2^6+2^6 = 16+64+64 = 144$ solutions.
</s>