Solution $x^{73}+x^{48}+x^{26} \equiv 0$ mod $13$

Question

I am trying to solve $x^{73}+x^{48}+x^{26} \equiv 0$ mod $13$. Can I say that By Fermat’s Little Theorem I have $x^{12} \equiv 1$ mod $13$? If I am correct remaining part will be simple but otherwise I have no idea how to deal with it.

another solution is $$x=3$$ or $$x=9$$ – Dr. Sonnhard Graubner Dec 10 '16 at 14:32 — Dr. Sonnhard Graubner, Dec 10 '16 at 14:32

Jack's wasted life · Accepted Answer · 2016-12-10T14:47:23.610

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Using Euler's theorem (assuming $x$ is a positive natural number less than $13$ :-)(as the comments below point out $x\equiv 0\mod13\;$ is a solution) $$ x^{a}\equiv x^{a\mod \phi(13)}\equiv x^{a\mod 12}\mod 13 $$ Therefore our equation turns into $$ x+1+x^2\equiv 0\mod 13 $$ A little experimentation leads to the solution $x=3$ and this leads to the factorization $$x^2+x+1=(x-9)(x-3)\mod13$$

edited Dec 10 '16 at 14:47

answered Dec 10 '16 at 14:35

Jack's wasted life

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or $x$ is 0 of course – Confuse Dec 10 '16 at 14:37
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Don't forget $x \equiv 0 \bmod 13$. – Fly by Night Dec 10 '16 at 14:44

Solution $x^{73}+x^{48}+x^{26} \equiv 0$ mod $13$

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