Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $\,x \not\equiv15\pmod{17},\text{ then }x^5 \not\equiv2\pmod{17}.$

I tried to take the contrapositive:

If $\,x^5 \equiv2\pmod{17},\text{ then }x \equiv15\pmod{17}$ and

then I assume that $x^5=17y+2$ for some integer $y$

But I am not what to do after this step.

How do I continue?

share|cite|improve this question

Proving the contrapositive is a good idea.

The key observation is that if $x$ is not divisible by $17$, then $x^{16} \equiv 1(mod\ 17)$ by Euler's theorem. If we assume $x^5 \equiv 2(mod\ 17)$, then it follows that $$ 15 \equiv -2 \equiv -2 \cdot x^{16} \equiv -2\cdot(x^5)^3\cdot x \equiv -2 \cdot 2^3 \cdot x \equiv -16x \equiv x\ (mod\ 17). $$

share|cite|improve this answer

Will Fermat's Little Theorem do? If $x^5\equiv2\pmod{17}$, then $x^{20}\equiv-1\pmod{17}$ and $x^{40}\equiv1\pmod{17}$. We then have

$$x^{65}\equiv x\equiv2(-1)(1)\equiv-2\equiv15\pmod{17}$$

share|cite|improve this answer


Its easier to follow :

$$a\equiv\pm1, a^5\equiv\pm1$$

$$a\equiv\pm2, a^5\equiv\pm32\equiv\mp2$$ and so on

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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