Equation $(a+b)^a=a^b$ How can we find the positive integer solutions to $(a+b)^a=a^b$?
Since $a+b>a$, it is necessary that $a<b$, otherwise the left-hand side is less than the right-hand side. So let $b=a+x$. The equation turns into $(2a+x)^a=a^{a+x}$, or equivalently, $\left(2+\dfrac{x}{a}\right)^a=a^x$.

Your equation means that $(2+\frac xa)^a$ is an integer. This is only possible if $2+\frac xa$ is itself an integer. In particular, $x$ is divisible by $a$. Therefore we may write $b=ka$ for some integer $k\geq2$. Using this in the original equation, we obtain $$((k+1)a)^a = a^{ka}$$ or equivalently $$(k+1)^a = a^{(k-1)a}$$ which further simplifies to $$k+1=a^{k-1}.$$ This implies $k+1\geq 2^{k-1}$, leaving us with the possibilities $k=2,3$. These give us the solutions $(k,a)\in\{(2,3),(3,2)\}$, which gives us the solutions of the original equation: $$a=2,b=6,\\a=3,b=6.$$
