# A Couple Problems with Primes

• Sep 30th 2010, 08:32 AM
Janu42
A Couple Problems with Primes
1) Show that $1^p + 2^p + 3^p + .... + (p-1)^p = 0$ (mod p) when p is an odd prime

2) Show that if p is a prime and p > 3, then $2^{p-2} + 3^{p-2} + 6^{p-2} = 1$ (mod p)

For 1), I really don't understand which theorem or identity I'm supposed to put into use to prove this. Obviously it works every time I just don't understand how this works in terms of technicality.

2) is the same story. I don't see how it works and therefore I can't prove it formally.
• Sep 30th 2010, 08:43 AM
undefined
Quote:

Originally Posted by Janu42
1) Show that $1^p + 2^p + 3^p + .... + (p-1)^p = 0$ (mod p) when p is an odd prime

2) Show that if p is a prime and p > 3, then $2^{p-2} + 3^{p-2} + 6^{p-2} = 1$ (mod p)

For 1), I really don't understand which theorem or identity I'm supposed to put into use to prove this. Obviously it works every time I just don't understand how this works in terms of technicality.

2) is the same story. I don't see how it works and therefore I can't prove it formally.

For #1, use Fermat's Little Theorem.

For #2, I'm not sure if the following is valid, but if not then I think it's close. Using Fermat's Little Theorem,

$2^{p-2} + 3^{p-2} + 6^{p-2} \equiv 2^{-1} + 3^{-1} + 6^{-1} \equiv \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \equiv 1\pmod{p}$
• Sep 30th 2010, 08:54 AM
melese
The form of the terms in both sums suggests using Fermat's Theorem.
Let $S=2^{p-2}+3^{p-2}+6^{p-2}$. Then $6S=3\cdot2^{p-1}+2\cdot3^{p-1}+6^{p-1}$.
Now Fermat's Theorem implies that $6S\equiv3+2+1=6\bmod p$ - show this!. Use the fact that $(p,6)=1$- again, why?
Therefore, you can cancell to get $S\equiv1\bmod p$.