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

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.

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

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 $$101$$.
$$\displaystyle \frac{x^{100} - 1}{x - 1} ~: ~x \neq 1.$$
Therefore, for any $$x$$ that is not congruent to either $$1$$ or $$0$$ mod($$101$$), $$x$$ will satisfy the equation.