# Deceptively simple divisibility problem

Suppose we are given integers $a,b$ with the condition that there exists a prime $k$ such that

$$2a+b\mid (a+b)^k$$

What can we say about $\gcd(a,b)$?

So far, I can see that for all primes $p:p\mid 2a+b\implies p\mid a+b\implies p\mid a\implies p\mid b$. My conjecture is that these conditions force $a\mid b$, or at the very least for all primes $p:p\mid a\implies p\mid b$. I don't know that $k$ must be prime, but this sub-problem related to FLT shows up with these conditions.

• I might miss something, but I think that $p \mid 2a + b \not \Rightarrow p \mid a + b$. For example $p = 3, a = 2, b = 5 \Rightarrow p \mid 9 = (2\cdot a + b) \Rightarrow p \nmid 7 = a + b$. Am I wrong? – tmrlvi Jan 3 '15 at 2:36
• @tmrlvi but there is no $k$ such that $9=2a+b\mid (a+b)^k=7^k$. – user121880 Jan 3 '15 at 2:39
• @tmrlvi But in this case, the premise that $2a+b\mid(a+b)^k$ is not true - $9$ never divides $7^k$. – Peter Woolfitt Jan 3 '15 at 2:39

This is not true, take $a=6,b=15$, $a+b$ is $21$ and $2a+b=27$. These numbers fit the criterion, what must happens is that there must be a prime $p$ such that the largest power of the prime dividing both $a$ and $b$ is equally high (call it $n$), and the largest power of $2a+b$ is larger than $n$, in this case that prime was $3$