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.

  • 1
    $\begingroup$ There are two very obvious solutions to begin with. $\endgroup$ – Arnaud Mortier May 17 '18 at 20:35
  • $\begingroup$ @ArnaudMortier yes obviously x congruous to 0 and 2 mod (2^10) $\endgroup$ – jacopoburelli May 17 '18 at 20:37
  • 2
    $\begingroup$ There are actually $144$ solutions in the range $0 \le x < 2^{10}$. $\endgroup$ – 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 $32$.

  • $\begingroup$ 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. $\endgroup$ – Mark Bennet May 17 '18 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.