We must have $b=0$ therefore $a$ can be anything. So the solution is $(k,0), k\in \mathbb{N^*}$, am I wrong?

you sure it's not something like: $$a^b = (a+b)^a$$

WeakMathemetician wrote: you sure it's not something like: $$a^b = (a+b)^a$$ that is correct, thanks for noticing, my blame for the typo, I had posted $a^b = (a+b)^b$ at start

Here's an observation that helps in this problem. If for naturals $a, b, c, d$, $a^b = c^d$ then there exists a natural number $t$ such that $a = t^x, c = t^y$. (For a proof, write out the prime factorization of $a$ and $c$, and consider the ratio of exponents of corresponding primes) Coming back to the problem, let $x = a, y = a+b$. Then we have $x^{y-x} = y^x$. So $x = \alpha ^ \beta, y = \alpha ^ \gamma$. Then we have that since $a \ge 1, b \ge 1$ so $y > 1$, so $\alpha \neq 1$. Also $x \neq 0$ because otherwise, since $a+b \neq 1$, $a = 0$ which is not a natural number. Also note that $\gamma > \beta$ as $y = x+b > x$. Comparing the exponents, we have $1 + \frac{\gamma}{\beta} = \alpha^{\gamma - \beta}$. Since everything except the fraction is guaranteed to be an integer, $\beta | \gamma$. Let $\gamma / \beta = z$. Then $1+z = \left( {\alpha^\beta} \right) ^ {z-1}$. Since $\alpha ^ \beta \ge 2$, we have $1+z \ge 2^{z-1}$. This means that $z \le 3$. $z$ is lower bounded by 1 as well. We make cases on $z$. If $z=1$, $2 = \alpha ^ 0$, contradiction. If $z=2$, $3 = \alpha ^ \beta$. So $\alpha = 3, \beta = 1$, so $x = 3, y = 9$ so $(a,b) = (3, 6)$ which works. If $z=3$, $4 = \alpha ^ {2 \beta}$. So $\alpha = 2, \beta = 1, \gamma = 3, x = 2, y = 8, (a, b) = (2, 6)$ which also works.

easy to see that $a+b=ka$ $$a^b = k^a \cdot a^a$$$$a^{b-a} = k^a$$in naturals It's well known that the only such solution in naturals is $(n,n)$ and $(2,4)$ and $(4,2)$ $a=2$... $2^{b-2} = 16$. ==> $b=6$ $4^{b-4} = 16$ no such solutions $a=n=k$ hence $a+b=a^2$. $b=a^2-a$ $$a^{a^2-a} = a^{2a}$$$$a^2-a=2a$$$$a^2=3a$$$a=3, b=6$ so solutions are $(a,b) = (2,6), (3,6)$

It's easy to prove that the only solution to $$m^n = n^m$$with $m,n \in \mathbb{N}^2$ is $(n,n),(4,2),(2,4)$ using calculus.

WeakMathemetician wrote: It's easy to prove that the only solution to $$m^n = n^m$$with $m,n \in \mathbb{N}^2$ is $(n,n),(4,2),(2,4)$ using calculus. Doesn't everyone know that already?