Let $n \in \Bbb N$. Let $p>2$ a prime number. Show that $1^n+2^n+...+(p-1)^n \equiv 0 \pmod {p}$

Question

This is an exercise in my abstract algebra reader, in the chapter about polynomial rings.

Let $n \in \Bbb N$. Let $p>2$ a prime number. And let $n$ not divisble by $p-1$. Show that

$$1^n+2^n+...+(p-1)^n \equiv 0 \pmod {p}$$

Our teacher assistant gave a hint to solve this with induction. We have tried to solve this using induction, but after some hours tring, I don't believe anyomre that this is possible. At least, I don't think I'm able to do that with my current knowlegde about mathematics. If someone has an idea how to solve this with induction (or maybe an other way), then I'm all ears.

You can't prove it, because as stated, it's wrong. Consider for example, $n = 4$, and $p = 5$. Generally, for all primes $p$, with $n = p-1$ you get the sum $p-1$ (modulo $p$). — Daniel Fischer, Oct 02 '13 at 19:41
Very similar to http://math.stackexchange.com/questions/511444/1n-2n-cdots-p-1n-mod-p/511470#511470 — abiessu, Oct 02 '13 at 19:49

score 1 · Answer 1 · answered Oct 02 '13 at 19:55

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Hint: Let $g$ be a primitive root of $p$ (a generator of the multiplicative group).

Then our sum $S$ is congruent to $\sum_{k=0}^{p-2} (g^n)^k$. The formula for the sum of a geometric progression gives $$S(g^n-1)\equiv (g^n)^{p-1}-1\pmod{p}$$

answered Oct 02 '13 at 19:55

André Nicolas

507,029

score 0 · Answer 2 · answered Oct 02 '13 at 19:43

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Hints:

If $\;n=p^k\;,\;\;k\in\Bbb N\;$ , then $\;a^n=a\pmod p\;,\;\;\;\forall\,a\in \{0,1,2,...,p-1\}\pmod p$

If $\;(n,p)=1\;$ , then the map $\;\Bbb Z_p\to \Bbb Z_p\;,\;\;a\to a^n\;$ , is onto

answered Oct 02 '13 at 19:43

DonAntonio

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Let $n \in \Bbb N$. Let $p>2$ a prime number. Show that $1^n+2^n+...+(p-1)^n \equiv 0 \pmod {p}$

2 Answers2