<s>[INST] Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$ Let a and b be natural numbers so that $2a-1,2b-1$ and $a+b$ are prime numbers. Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$.
I get that $gcd(a,b)=1$.
I haven't got anything special for now but if I do I will update the question.
 [/INST] Let $p=a+b$, and suppose that $p$ divides at least one of $a^a+b^b$ or $a^b+b^a$.
Then $p$ divides the product
$$(a^a+b^b)(a^b+b^a)=a^p+b^p+(ab)^a+(ab)^b $$
By Fermat's little theorem

*

*$a^p= a\mod p$

*$b^p= b\mod p$

*$a^p+b^p= a+b = 0 \mod p$
So (remember we suppose that $p$ divides the product) we have :
$$ (ab)^a+(ab)^b=0\mod p$$
But $b=-a\mod p$.
$$ (-a^2)^a+(-a^2)^b=0 \mod p$$
As $a+b$ is odd, one of $a$ and $b$ is odd and the other is even. So
$$ a^{2a}=a^{2b} \mod p $$
it means that $2b-2a$ is multiple of the order $r$ of $a$, and that order $r$ divides $p-1=a+b-1$. But if $r$ divides $(2b-1)-(2a-1)$ and $2(a+b-1)=(2a-1)+(2b-1)$, then either $r=2$ or it divides $2a-1$ and $2b-1$ (they are both different prime numbers) so $r=1$.
So $a=1\mod p$ or $a=-1\mod p$. But $a+b=p$. So $a<p$, so $a=1$ (but then $2a-1$ is not prime) or $a=p-1$ (but then $2b-1$ is not prime), this is not possible.
A contradiction. So, our first hypothesis that $p$ divides the product is false.
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