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I have an interesting problem I solved a while ago, and I was wondering if anyone had a different solution.

Let p and q be twin prime integers. Prove that $p^p+q^q\equiv 0 \pmod{p+q}$

I came up with:

Proof.

First, we are given that we want to prove $p^p+q^q\equiv 0 \pmod{p+q}$ We also know that p and q are twin primes and both odd. We know that an odd multiplied by an odd is odd, so $p^p$ and $q^q \equiv 1 \pmod 2$

We see that $(1+1 \bmod 2)\equiv 0 \pmod 2$

In order to prove that $p^p+q^q\equiv 0 \pmod{p+q}$, we need to prove $p^p\equiv p \pmod {p+q}$ and $q^q \equiv q \pmod {p+q}$

Using Fermat's Theorem where $a=p$, $$(p^{p-1} \bmod (p+q))\cdot(p \bmod (p+q)) \equiv p \pmod {p+q}$$

We do a similar procedure with $q$ to get $p^p+q^q\equiv 0 \pmod {p+q}$. Q.E.D

Thanks

  • 3
    Can't really use Fermat's Theorem, since $p+q$ is not prime. Hint: Let $m$ be the number in the middle.2012-03-19
  • 2
    Make sure to use $q=p+2$.2012-03-19

3 Answers 3