Prove that ## 1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}\equiv -1 ##.

[Math100]

Homework Statement

Employ Fermat's theorem to prove that, if $p$ is an odd prime, then
$1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}\equiv -1\pmod {p}$.

Relevant Equations

None.

Proof:

Suppose $p$ is an odd prime such that $p\geq 3$.
Note that $p\nmid a$.
By Fermat's theorem, we have that $a^{p-1}\equiv 1\pmod {p}$.
Observe that there are $p-1$ terms in $1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}$.
This means
\begin{align*}
1^{p-1}&\equiv 1\pmod {p}\\
2^{p-1}&\equiv 1\pmod {p}\\
3^{p-1}&\equiv 1\pmod {p}\\
&\vdots \\
(p-1)^{p-1}&\equiv 1\pmod {p}.\\
\end{align*}
Thus $1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}\equiv (p-1)\equiv -1\pmod {p}$.
Therefore, if $p$ is an odd prime, then $1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}\equiv -1\pmod {p}$.

 

[fresh_42]

Homework Statement:: Employ Fermat's theorem to prove that, if $p$ is an odd prime, then
$1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}\equiv -1\pmod {p}$.
Relevant Equations:: None.

Proof:

Suppose $p$ is an odd prime such that $p\geq 3$.
Note that $p\nmid a$.

... for all $a\in\{1,2,3,\ldots,p-1\}.$

By Fermat's theorem, we have that $a^{p-1}\equiv 1\pmod {p}$.
Observe that there are $p-1$ terms in $1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}$.
This means
\begin{align*}
1^{p-1}&\equiv 1\pmod {p}\\
2^{p-1}&\equiv 1\pmod {p}\\
3^{p-1}&\equiv 1\pmod {p}\\
&\vdots \\
(p-1)^{p-1}&\equiv 1\pmod {p}.\\
\end{align*}
Thus $1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}\equiv (p-1)\equiv -1\pmod {p}$.
Therefore, if $p$ is an odd prime, then $1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}\equiv -1\pmod {p}$.