1

Find all solutions $x^{12} \equiv 1 \pmod{13}$. Hint: the computation of high powers is better accomplished by using binary expansion of the exponent.

Approach: This time I don't have a clear approach. 1 is clearly a solution, so maybe I should tru using the factor theorem by saying there exists a polynomial $g(x)$ such that $x^{12}-1=(x-1)g(x) \equiv 0 \pmod{12}$ and see how it goes from there. How do I start ?. The topics related to that problem are factor theorem and lagrange's theorem.

Michael Hardy
  • 1
TheMathNoob
  • 2,011
  • Do you know Fermat's theorem? – lhf Jul 05 '16 at 01:57
  • I know it's quite famous but not yet. Probably in the next chapters. – TheMathNoob Jul 05 '16 at 01:58
  • Do you know Lagrange's theorem in group theory? – lhf Jul 05 '16 at 01:58
  • Just Lagrange's theorem in number theory – TheMathNoob Jul 05 '16 at 02:00
  • 2
    Then calculate! There are in principle $13$ things to check, $x\equiv 0\pmod{13}$ (bad), $x\equiv 1\pmod{13}$ (good) and so on. You can save time by noting that $(13-x)^{12}\equiv (-x)^{12}\equiv x^{12}$. And there are lots of tricks to keep numbers small during calculations (reduce mod $13$). – André Nicolas Jul 05 '16 at 02:01
  • 2
    Calculating is a good thing. You will find that $x^{12}\equiv 1\pmod{13}$ for all $x$ from $1$ to $12$. So $x^{12}\equiv 1\pmod{13}$ unless $x\equiv 0\pmod{13}$. A little later, you will find that the calculation could have been avoided. Yet doing it, and looking for tricks to cut down on the work, is valuable experience. Do not use a calculator. – André Nicolas Jul 05 '16 at 02:23
  • Or note $,2^6\equiv -1,$ so $,2,$ has order $12,,$ so if $,x\not\equiv 0,$ then $,x\equiv 2^k$ for some $,k,,$ so $,x^{12} \equiv (2^k)^{12}\equiv (2^{12})^k\equiv 1^k\equiv 1\ \ $ – Bill Dubuque Jul 05 '16 at 02:33
  • See the Order Test for a general way to make such order inferences. Namely, we have the following Order Test $\ ,a,$ has order $,n\iff a^n \equiv 1,$ but $,a^{n/p} \not\equiv 1,$ for every prime $,p\mid n\ \ $ – Bill Dubuque Jul 05 '16 at 02:53
  • So, Noob, any thoughts about the comments/answers? – Gerry Myerson Jul 07 '16 at 09:16
  • I will work on it as soon as I get a good handle on fermat's little theorem. – TheMathNoob Jul 07 '16 at 10:07
  • OK. But my answer doesn't rely on Fermat. – Gerry Myerson Jul 08 '16 at 12:51

2 Answers2

1

Hint: show that if $a$ is a solution, then so are $a^2,a^3,\dots,a^{12}$. Then compute $2^r$ for all $r$, $1\le r\le12$.

Gerry Myerson
  • 179,216
0

This problem has little things to do with Lagrange's theorem and factor theorem.

First, one can paraphrase the Lagrange's theorem:

If p is a prime, then either all coefficient of $f(x)$ is a multiplier of p or the number of in-congruent solution to $f(x)$ is at most deg $f(x)$.

From this theorem, one can draw a conclusion that:

The number of in-congruent solution to $x^{12}-1$ is at most 12.

But it tells one literally nothing because the low bound of numbers of in-congruent solution in this case is unknown. Even if one has already known the low bound, it may give an inequality which can be confined to find the solutions unless the low bound equals to the upper bound. (i.e. The number of in-congruent solution to $x^{12}-1$ is at least 12.)

Second, one can state the factor theorem:

$f(x)$ has a polynomial factor $(x-x_1)$ if and only if $x_1$ is a root of $f(x)$.

Using this theorem, one can factorize the $f(x)$. However, it doesn't give any apparent clue for finding roots.

So, how to solve the problem?

Notice that $12 = 13-1$ and $13$ is a prime.

One can recall Fermat's little theorem :

For all $a$ cannot divide $13$, then $a^{12} \pmod {13}.$

Therefore the solution is $\{n|n \neq 0\pmod {13}\}$

Zau
  • 3,909
  • 12
  • 29
  • That won't help the OP since he said in the first comment that he does not yet have available Fermat's little Theorem. – Bill Dubuque Jul 05 '16 at 14:42