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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.

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  • $\begingroup$ 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? $\endgroup$ – tmrlvi Jan 3 '15 at 2:36
  • $\begingroup$ @tmrlvi but there is no $k$ such that $9=2a+b\mid (a+b)^k=7^k$. $\endgroup$ – user121880 Jan 3 '15 at 2:39
  • $\begingroup$ @tmrlvi But in this case, the premise that $2a+b\mid(a+b)^k$ is not true - $9$ never divides $7^k$. $\endgroup$ – Peter Woolfitt Jan 3 '15 at 2:39
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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 $3$

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  • $\begingroup$ In particular, my conjecture is false. Thank you! $\endgroup$ – abiessu Jan 3 '15 at 3:21

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