<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] Since $2a-1, 2b-1$ are primes, then $a,b\neq 1$. Since $a+b$ is a prime, then we can suppose $a$ is even and $b$ is odd.
Suppose $p=a+b|a^a+b^b$. We have: $$a^a+b^b\equiv a^a-a^b=a^b(a^{a-b}-1) \pmod {a+b}$$
Since $\gcd(a^b,a+b)=1$, then $p|a^{a-b}-1$. Let $h=ord_p(a)$, then $h|p-1,h|a-b$, then $h|p-1-(a-b)=2b-1$, thus $h=1$ or $h=2b-1$. 
If $h=1$, then $a+b|a-1$, but this is impossible since $a+b>a-1$. 
Then $h=2b-1$, which means $p|a^{2b-1}-1$. Hence $p|a^{2a-1}(a^{2b-1}-1)=a^{2a+2b-2}-2^{2a-1}$, then $p|2^{2a-1}-1$. Thus $2b-1|2a-1$, or $2a-1=2b-1$ (since they are both primes), or $a=b$, which is impossible.
Proving $a+b \not|a^b+b^a$ is similar.
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