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$\begingroup$

(When I use an $=$, I am referring to a congruence relation.)

I am curious is anybody can point me in the right direction for solving the following congruence relation ($a$,$b$ of course both integers)

$$a^b=b^a \mod (a+b)$$

I have tried rewriting this in a few forms, such as let $b=a+n$ and finding more likely scenarios for this to occur. However, I dont even know really how to approach writing a solution!

If anybody can just point me in the right direction, if I'm looking for abstract algebra terms, or some other form of math all together, that's all I really need!

Although, if you do have a solution, I would love to see that as well!

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    You can rewrite the equality as $a^b \equiv (-b)^b \equiv (-1)^b \cdot b^b \equiv b^a \mod (a+b)$ or $b^{a - b} \equiv (-1)^b \mod (a+b)$. I'm not sure if that helps.2012-10-22
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    @TMM To cancel $\rm\:b\:$ as you did requires $\rm\: 1 = (b,a+b) = (b,a).\ \ $2012-10-22
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    @Bill Yes you are right. So you can only do that if $(a,b) = 1$. But even then, it's not clear how to continue (or what to do if $(a,b) = d > 1$).2012-10-22

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