- #1

Math100

- 767

- 207

- Homework Statement
- Employ Fermat's theorem to prove that, if ## p ## is an odd prime, then

## 1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p} ##.

[Hint: Recall the identity ## 1+2+3+\dotsb +(p-1)=p(p-1)/2 ##.]

- Relevant Equations
- None.

Proof:

Suppose ## p ## is an odd prime such that ## p\geq 3 ##.

By Fermat's theorem, we have that ## a^{p}\equiv a\pmod {p} ##.

Then

\begin{align*}

&1^{p}\equiv 1\pmod {p}\\

&2^{p}\equiv 2\pmod {p}\\

&3^{p}\equiv 3\pmod {p}\\

&\vdots \\

&(p-1)^{p}\equiv (p-1)\pmod {p}.\\

\end{align*}

This means ## 1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv [1+2+3+\dotsb +(p-1)]\pmod {p} ##.

Since ## 1+2+3+\dotsb +n=\frac{n(n+1)}{2} ##,

it follows that ## 1+2+3+\dotsb +(p-1)=\frac{p(p-1)}{2} ##.

Observe that ## p-1 ## is even, so ## p-1=2k ## for some ## k\in\mathbb{N} ##.

Now we have ## 1+2+3+\dotsb +(p-1)=pk ##.

Thus ## 1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv pk\equiv 0\pmod {p} ##.

Therefore, if ## p ## is an odd prime, then ## 1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p} ##.

Suppose ## p ## is an odd prime such that ## p\geq 3 ##.

By Fermat's theorem, we have that ## a^{p}\equiv a\pmod {p} ##.

Then

\begin{align*}

&1^{p}\equiv 1\pmod {p}\\

&2^{p}\equiv 2\pmod {p}\\

&3^{p}\equiv 3\pmod {p}\\

&\vdots \\

&(p-1)^{p}\equiv (p-1)\pmod {p}.\\

\end{align*}

This means ## 1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv [1+2+3+\dotsb +(p-1)]\pmod {p} ##.

Since ## 1+2+3+\dotsb +n=\frac{n(n+1)}{2} ##,

it follows that ## 1+2+3+\dotsb +(p-1)=\frac{p(p-1)}{2} ##.

Observe that ## p-1 ## is even, so ## p-1=2k ## for some ## k\in\mathbb{N} ##.

Now we have ## 1+2+3+\dotsb +(p-1)=pk ##.

Thus ## 1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv pk\equiv 0\pmod {p} ##.

Therefore, if ## p ## is an odd prime, then ## 1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p} ##.