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Let $c\in \mathbb F_p^\times $ (where $p$ is prime). I would like to prove that $x^m-c=0$ has at most $m$ solutions in $\mathbb F_p^\times $.

  • Since $c\in \mathbb F_p^\times $, there is $d\in \mathbb F_p^\times $ s.t. $cd=1$. Therefore $$x^m-c=0\implies dx^m=1,$$ but this doesn't really help.

  • Also, I tried to prove that if $A=\{x\in \mathbb F_p^\times \mid x^m-c=0\}$ then $|A|\leq m$, but I don't know how to do this.

  • Finally, if $m=kp+n$ where $n<p$, then in $\mathbb F_p^\times $, by Ferma theorem, $$x^m=c\implies x^{n+k}=c,$$ but same, I can't conclude.

Any idea ?

I can't use that a polynomial of degree $m$ in $\mathbb F_p[X]$ has at most $m$ roots since I didn't proved it yet.

Bruce
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1 Answers1

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Well, its a general fact that any polyomial $f\in K[x]$ of degree $m$ with coefficients in a field $K$ has at most $m$ roots in $K$.

One can show that for each $a\in K$ using the division theorem/algorithm:

$f(a) = 0$ iff $f(x) = (x-a)g(x)$,

where $g\in K[x]$.

This can be iterated (apply to $g$ next) by induction such that

$f(x) = (x-a_1)\cdots (x-a_k)g(x)$

where $g\in K[x]$ and $g$ has no root in $K$.

By comparing degrees, its clear that $k\leq $ degree of $f$.

Wuestenfux
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