# Number Theory

1. Nov 16, 2009

### icystrike

Find all odd primes $$p$$, if any, so that $$p$$ divides $$\sum_{n=1}^{103} n^{p-1}$$

2. Nov 16, 2009

### icystrike

By Fermat's Little Theorem,
$$n^{p-1} \equiv 1 (mod p)$$

$$\sum_{n=1}^{103} n^{p-1} \equiv 103 (mod p)$$ ,

Whence $$p \mid 103$$

Since 103 is prime , therefore 103 it is the only prime.

It seems like my proof is wrong please correct me. (=

3. Nov 16, 2009

### willem2

Your proof is valid for p>103. If p<=103 then you can have p | n and $n^{p-1} (modp)$ will be 0