[SOLVED] Problem involving Fermat's Little Theorem

**ikarous** · April 19th 2009, 12:08 PM

The solution to this problem is probably painfully obvious to most people, but I'm not making the leap for some reason. The problem is:

Show that $1^p+2^p+3^p+...(p-1)^p \cong 0(mod\; p)$ when p is an odd prime.

So, by Fermat's Little Theorem, we know that:

$a^p \cong a(mod\; p)$ and $a^{p-1} \cong 1(mod\; p)$ .

And thus, $1^p+2^p+3^p+...(p-1)^p \cong 1+2+3+... (p-1)\;(mod\; p).$

So, by my thinking, I need to show that $1+2+3+... (p-1) \cong 0\;(mod\; p)$ , but I'm not sure how to proceed.

Any help at all would be greatly appreciated.

**ikarous** · April 19th 2009, 12:58 PM

Solved. I'm dumb.

Since $(p-1)$ is even,

$1+2+...(p-1) \cong 1+2...+ (p-2) + (p-1) \cong 1+2 ... + -2 + -1 \cong 0 \; (mod\; p)$ .

Thread: [SOLVED] Problem involving Fermat's Little Theorem

LinkBack

Thread Tools

Display

[SOLVED] Problem involving Fermat's Little Theorem

Similar Math Help Forum Discussions

abstract algebra [primitive root mod n, euler theorem, fermat theorem]

[SOLVED] Fermat’s Little Theorem

[SOLVED] Difficult binomial theorem question(involving greatest integer function)?

A Problem involving the Intermediate Value Theorem

Fermat’s Little Theorem Problem

Search Tags