Switzerland — Second Round, 2023

Solution:
$(a, b, p) = (1, 1, 2)$ is the only solution. Let's split the problem into two cases.

- Case 1: $a = b$

The equation simplifies into $2^{p} a^{p} = 2 p^{a}$, and since $4 \mid 2^{p}$, $2 \mid p^{a}$ which implies that $p = 2$. Plugging this new piece of information in the initial equation yields $4 a^{2} = 2^{a+1}$ so $a^{2} = 2^{a-1}$. If $a > 1$, then $a$ has to be even since $2^{a-1}$ is even. However, on the other hand, since $a^{2}$ is a perfect square, $2^{a-1}$ has to be a perfect square as well, so $a$ must also be odd, which is a contradiction. Thus, the only option is $a = 1$ which indeed satisfies the initial equation. Thus, out of this case, we obtain one solution: $(a, b, p) = (1, 1, 2)$.

- Case 2: $a \neq b$

Since the equation is symmetric in $a$ and $b$, assume wlog $b > a$. Now, the equation can be rewritten as
$$(a+b)^{p} = p^{a} (1 + p^{b-a})$$
Therefore, $p \mid (a+b)^{p}$, so $p \mid a+b$. Now, look at the prime factorisation of both sides of the equation and consider the powers of $p$. The right side of the equation is $p^{a}$ times a number not divisible by $p$, because $b-a > 0$. If $(a+b)^{p} = p^{x} \cdot y$ where $y$ is not divisible by $p$, then $p \mid x$. Thus, we have $p \mid a$, because $x = a$. Combining this with $p \mid a+b$, we deduce that $p \mid a, b$. Now, to conclude, observe that $p^{a}(1 + p^{b-a})$ is a perfect $p$-th power, and since $p^{a}$ is a perfect $p$-th power and is coprime to $1 + p^{b-a}$, we get that $1 + p^{b-a}$ is a $p$-th power. Write $1 + p^{b-a} = z^{p}$, and since $p \mid b-a$ and $b-a > 0$, we get that $z^{p} - p^{b-a} = 1$ which is a contradiction because the difference of 2 different perfect $p$-th powers is at least $2^{p} - 1 > 1$. Thus, in this case, there are no solutions.