1
$\begingroup$

For $m>1$ $$ (m-2)^3\cdot(m+2)^4\equiv 4 \mod m $$ How many possible integers $m$ can be? My modular arithmetic intuition is a bit scarce, so hints and explanations are more appreciated.

$\endgroup$
  • 4
    $\begingroup$ Well, $m \equiv 0 \pmod m$. So your equation says $(-2)^3 \cdot 2^4 \equiv 4 \pmod m$. $\endgroup$ – Fredrik Meyer May 23 '17 at 8:53
1
$\begingroup$

As hinted in the comments by Fredrik Meyer:

$${(m - 2)^3}\cdot{(m + 2)^4} \equiv {(0-2)^3}\cdot{(0+2)^4} \equiv 4 \pmod m,$$

so that $m \mid (-128 - 4) = -132$. In other words, if you restrict $m > 1$, then $m \mid 132 = {2^2}\cdot{3}\cdot{11}$. Thus, there are $d(m) - 1 = (2+1)\cdot(1+1)\cdot(1+1) - 1 = 3\cdot{2}\cdot{2} - 1= 12 - 1 = 11$ possible choices for $m$ (because of the restriction $m > 1$).

$\endgroup$
  • $\begingroup$ I get that you've calculated how many integers that divides $132$, little question: What $d$ stands for at function? $\endgroup$ – user373239 May 23 '17 at 9:10
  • $\begingroup$ $d(m)=\sigma_{0}(m)$, which is just the number-of-divisors function (see OEIS sequence A000005 for some references). $\endgroup$ – Jose Arnaldo Bebita-Dris May 23 '17 at 9:14
  • 1
    $\begingroup$ So if $$m = \prod_{i=1}^{r}{{p_i}^{\alpha_i}}$$ is the prime factorization of $m$, then $$d(m) = \prod_{i=1}^{r}{\bigg(\alpha_i + 1\bigg)}.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris May 23 '17 at 9:18
  • 1
    $\begingroup$ I knew the strategy but notations are new to me :D Thank you $\endgroup$ – user373239 May 23 '17 at 9:21
  • $\begingroup$ You're welcome, @AbdullahUYU! =) $\endgroup$ – Jose Arnaldo Bebita-Dris May 23 '17 at 9:21

Your Answer

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