Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
Can't really use Fermat's Theorem, since $p+q$ is not prime. Hint: Let $m$ be the number in the middle. –  André Nicolas Mar 19 '12 at 21:54
2  
Make sure to use $q=p+2$. –  lhf Mar 19 '12 at 22:03

3 Answers 3

Hint $\rm\ mod\ p+1\!:\:\ p\:\equiv\:\! {-}\!1\ \Rightarrow\ p^m + (p+2)^n\equiv\: (-1)^m + 1^n\equiv\: 0\:$ when $\rm\:m\:$ is odd

Alternatively let $\rm\:p = 2k-1,\ q = 2k+1.\:$ If $\rm\:m,n\:$ are odd then by the binomial theorem

$$\begin{eqnarray}\rm p^m + q^n &=&\rm\: (2k-1)^m + (2k+1)^n \\ &=&\rm\: (-1)^m + (-1)^{m-1} 2mk + 4k^2(\cdots)\: +\: 1 + 2nk + 4k^2(\cdots) \\ &=&\:\rm 2\:(m+n)\:k\ +\ 4k^2(\cdots) \end{eqnarray}$$

which is divisible by $\rm\:p+q = 4k\:$ since $\rm\:m+n\:$ is even.

share|improve this answer

Let our numbers be $p$ and $q$, where $q=p+2$. Then $p\equiv -q \pmod {p+q}$, and therefore $$p^p+q^q \equiv (-q)^p +q^q=q^p((-1)^p +q^2)=q^p(q^2-1)\pmod{p+q}.$$ So it is enough to show that $p+q$ divides $q^2-1$. But $p+q=2q-2$, and $q^2-1=(q-1)(q+1)$, so since $q+1$ is even the result follows.

share|improve this answer

It really has to do with "twin odd numbers", not just primes. When $n \in \mathbb N$ is odd, as polynomials we have $t^n-t = (t-1)(t^{n-1}+\ldots+t)$ and $t^n-t = (t+1)(t^{n-1}-t^{n-2}+\ldots-t)$, where the second factors have $n-1$ terms. So for any odd integer $x$, $x^n - x$ is divisible by $2(x-1)$ and by $2(x+1)$. In particular if $y = x+2$, $x^n + y^m \equiv x + y \equiv 0$ mod $x+y=2(x+1)=2(y-1)$ for any positive odd integers $m$ and $n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.