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It's an exercise I found in a textbook and I don't know how to approach it. I tried applying Fermat's theorem but don't do much.

Solve for $x$: $x^{99}+x^{98}+\dots+x+1\equiv0 \pmod {101}$

This is the question I have trouble answering it. I think I have to use the fact that $101$ has a primitive root since its a prime but still I do not know how that helps. Any hint will help. Thanks.

something I tried was let $a$ be a primitive root, then $\left(a^j\right)^{99}+\dots+\left(a^k\right)\equiv0 \pmod {101}$ this gives us $a\left(a^{98j}\right)+\dots+a^{k-1}\equiv0 \pmod {101}$ but stills this doesn't seem to help at all.

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    $\begingroup$ Hint: multiply by $x-1$ $\endgroup$ Commented Sep 11, 2021 at 16:33

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Assumed that $x$ is an integer.

First, you can manually verify that $x \equiv 1\pmod{101}$ is not a solution.

By Fermat's Little Theorem $x^{100} - 1 \equiv 0\pmod{101}$, for any $x$ that is not a multiple of $101$.

The expression simplifies to
$\displaystyle \frac{x^{100} - 1}{x - 1} ~: ~x \neq 1.$

Therefore, for any $x$ that is not congruent to either $1$ or $0$ mod($101$), $x$ will satisfy the equation.

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