# 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.

• There are two very obvious solutions to begin with. – Arnaud Mortier May 17 '18 at 20:35
• @ArnaudMortier yes obviously x congruous to 0 and 2 mod (2^10) – jacopoburelli May 17 '18 at 20:37
• There are actually $144$ solutions in the range $0 \le x < 2^{10}$. – Derek Holt May 17 '18 at 20:53

This is an extended hint which is a start to one way of approaching the problem. The strategy is to work with a power of $2$ times an odd number to cancel as much as possible.
Write $x=2^ry$ with $y$ odd ($1\lt x\le 2^{10}$). Clearly $x$ is even so $1\le r\le 10$. Use equality to represent equivalence mod $2^{10}$
Then $2^{5r}y^5-2^{r+4}y=0$ and $\left(2^{4(r-1)}y^4-1\right)2^{r+4}y=0$
Now either $r+4\ge 10$ or the first factor in brackets must be even. Since $y$ is odd, this second possibility implies $r=1$ and you need $y^4-1$ to be divisible by $32$.
• Note that if $y$ is odd, one of $y\pm 1$ is divisible by $4$ and $y^2-1$ is therefore divisible by $8$. $y^2+1$ is therefore divisible by $2$ but not $4$ so that $(y^4-1)=(y^2+1)(y^2-1)$ is always divisible by $16$. You need, therefore, to find an additional factor of $2$ and this constrains $y$ to half the odd numbers - I'm sure you can work out which half. – Mark Bennet May 17 '18 at 21:03