How to show $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$ has $p-2$ incongruent solutions: $2, 3,..., p-1.$ when $p$ is an odd prime?

Question

Let $p$ be an odd prime. How would I prove that the congruence $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$ has exactly $p-2$ incongruent solutions, and they are the integers $2, 3,..., p-1?$

I found this question posted, but the answer doesn't quite make sense to me.

Here is my attempt:

Proof As $p$ is an odd prime, by Fermat's Theorem, \begin{align*} x^{p-1}&\equiv1\pmod{p}\\ &\implies x^{p-1}-1\equiv0\pmod{p} \end{align*} and by Lagrange's Corollary, with $d=p-1, p-1\mid p-1$, this congruence has exactly $p-1$ solutions. \begin{align*} x^{p-1}-1 \pmod{p}&\equiv0\pmod{p}\\ &\equiv (x-1)(x^{p-2}+\cdots+x^2+x+1) \pmod{p}\\ \end{align*} Assume $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$...

And then I get stuck.

The proof I linked above used:

Now, $x^{p-1}-1\equiv 0\pmod{p}$ has exactly $p-1$ incongruent solutions modulo $p$ by Lagrange's Theorem.

Note that $g(1)=(1-1)f(1)=0\equiv 0\pmod{p}$, so $1$ is a root of $g(x)$ modulo $p$. Hence, the incongruent roots of $g(x)$ modulo $p$ are $1,2,3,\dots,p-1$.

But every root of $g(x)$ other than $1$ is also a root of $f(x)$ (This is the part I'm concerned about. Is it clear that this is the case?), hence $f(x)$ has exactly $p-2$ incongruent roots modulo $p$, which are $2,3,\dots,p-1$. $\blacksquare$

But I don't understand the part about roots. Nor do I quite get the use of creating $g(x)$ and $f(x)$. Could someone explain the proof using a different method?

Answer 1

To phrase the yellow block differently; the equation $x^{p-1}-1\equiv0\pmod{p}$ has precisely $p-1$ incongruent solutions by Lagrange's theorem. These are precisely the nonzero residues modulo $p$, i.e. the congruence classes $1$, $2$,..., $p-1$. We can factor the left hand side of this equation as follows: $$x^{p-1}-1=(x-1)(x^{p-2}+x^{p-1}+\cdots+x^2+x+1).$$ In particular this shows that if $x$ is a solution and $x-1\not\equiv0\pmod{p}$ then we must have $$x^{p-2}+x^{p-1}+\cdots+x^2+x+1\equiv0\pmod{p}.$$ This tells us that $2$, $3$, ..., $p-1$ are solutions to the congruence.

To see that there are no other (incongruent) solutions, note that the only remaining congruence classes are $0$ and $1$. And plugging them in yields $$0^{p-2}+0^{p-1}+\cdots+0^2+0+1\equiv1\pmod{p},$$ $$1^{p-2}+1^{p-1}+\cdots+1^2+1+1\equiv p-1\pmod{p},$$ so these are not solutions.

Answer 2

Multiply by $x-1$: $$x^{p-2}+\cdots+x^2+x+1\equiv 0\implies (x-1)(x^{p-2}+\cdots+x^2+x+1)=x^{p-1}-1\equiv 0\mod p. $$ Lil' Fermat says the latter congruence is satisfied by every $x\not\equiv 0\mod p$. this maked $p-1$ congriuence classes. However, we must remove $1$, which does not satisfy the given congruence equatio, since it would imply $(p-1)\cdot 1\equiv 0\mod p$.

Answer 3

You can avoid congruences by working in the $p$-element field $\mathbb{F}_p$. The polynomial $f(x)=x^{p-1}-1$ has distinct roots (in the algebraic closure), because its derivative is $f'(x)=(p-1)x^{p-2}$, which has no common root with $f(x)$ (in any extension).

On the other hand, the group of nonzero elements in the field has order $p-1$, so for every $a\in\mathbb{F}_p$, $a\ne0$, $a^{p-1}=1$. Therefore every element in $\mathbb{F}_p\setminus\{0\}$ is a root of $f(x)$.

Since $f(x)=(x-1)(x^{p-2}+\dots+x+1)$, we see that $2,3,\dots,p-1$ (as elements of $\mathbb{F}_p$) are roots of $x^{p-2}+\dots+x+1$.

This precisely amounts to the statement you have to prove, when we go back to the integers, taking into account that $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$.