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Question: Find the smallest non-negative integer solution (if any exist) to the equation

$$ x^{2017^2} + x^{2017} + 1 \equiv 0\mod\ 2017 $$

Hint: You may use the fact that 2017 is a prime number. Fermat’s Theorem can help with this problem.

I have attempted to apply some addition rules to get

$$ x^{2017^2} \equiv 0\mod 2017 $$ $$ x^{2017}\equiv 0\mod\ 2017 $$ $$ 1 \equiv 0\mod\ 2017 $$

and have gotten stuck. I have also tried to think about how $ x^{2017^2} + x^{2017} + 1 = 2017n $ to satisfy the "$ 0\mod \ 2017$" portion but cannot get much further. I also am not quite sure how to successfully utilize Fermat's Theorem to solve this. I played around a bit but didn't get far.

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  • $\begingroup$ If $n$ is an integer, then $n^{2017}$ is congruent to what mod $2017$? $\endgroup$ – quasi Apr 1 '17 at 22:35
  • $\begingroup$ Also, if my "system" of equations is correct, then x must be zero to satisfy a Corollary coming from Fermat's Theorem stating that if $ a\in Z, a^p \equiv a\mod \p$ $\endgroup$ – M Paul Apr 1 '17 at 22:35
  • $\begingroup$ User fermats theorem: $x^{2017} \equiv x \mod 2017$ $\endgroup$ – fleablood Apr 2 '17 at 1:28
  • $\begingroup$ If $x \equiv 0$ then $1 \equiv 0 \mod 2017$ so that isn't a solution. So $\gcd(x,2017) = 1$. So $x^{2016} \equiv 1 \mod 2017$. $\endgroup$ – fleablood Apr 2 '17 at 1:30
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Fermat's little theorem tells us that $x^{2017} \equiv x \pmod{2017}$. (It's usually stated as $x^{2016} \equiv 1 \pmod{1027}$ for $x$ not divisible by $2017$, but the first form is more useful in this case.)

But probably the first stumbling block in this problem is noticing that $x^{2017^2} = (x^{2017})^{2017}$, and therefore this equation can be rewritten as $$(x^{2017})^{2017} + x^{2017} + 1 \equiv 0 \pmod{2017} \implies x^{2017}+x+1 \equiv 0 \pmod{2017}.$$ (Even if Fermat's little theorem did not apply, the substitution $y = x^{2017}$ would be a good one at this point, but as it is we'd just have $y \equiv x \pmod{2017}$ if we did that.)

Another round of Fermat's little theorem, and we get $$x^{2017}+x+1 \equiv 0 \pmod{2017} \implies x + x + 1 \equiv 0 \pmod{2017}$$ so $2x+1 \equiv 0 \pmod{2017}$, which we can solve to get $x \equiv 1008 \pmod{2017}$.

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