N2) Find all triplets $(a, b, p)$ of strictly positive integers where $p$ is prime and the equation

$$
(a+b)^{p}=p^{a}+p^{b}
$$

is satisfied.
Solution: $(a, b, p)=(1,1,2)$ is the only triplet that satisfies the equation. Let's break the problem into two cases.

- Case $1: a=b$

The equation simplifies to $2^{p} \cdot a^{p}=2 \cdot p^{a}$, and since $4\left|2^{p}, 2\right| p^{a}$, this implies that $p=2$. Substituting this into the base equation gives the equality $4 a^{2}=2^{a+1}$, so $a^{2}=2^{a-1}$. If $a>1$, then $a$ must be even as $2^{a-1}$ is even. However, since $a^{2}$ is a perfect square, $2^{a-1}$ must also be a perfect square, so $a$ must also be odd, which contradicts the previous statement. Thus, the only possibility is $a=1$, which satisfies the equation. Therefore, in this case, we obtain an exact solution: $(a, b, p)=(1,1,2)$.

- Case $2: a \neq b$

Since the equation is symmetric in $a$ and $b$, suppose without loss of generality $b>a$. The equation can therefore be written as

$$
(a+b)^{p}=p^{a}\left(1+p^{b-a}\right)
$$

Thus, $p \mid(a+b)^{p}$, so $p \mid a+b$. Now, let's look at the prime factorization of both sides and observe the powers of $p$. The right side of the equation is $p^{a}$ times a number coprime with $p$, since $b-a>0$. If $(a+b)^{p}=p^{x} \cdot y$ where $y$ is not divisible by $p$, then $p \mid x$. Thus, we have $p \mid a$, since $x=a$. Combining this with $p \mid a+b$, we can deduce that $p \mid a, b$. Now, to conclude, note that $p^{a}\left(1+p^{b-a}\right)$ is a $p$-th power, and since $p^{a}$ is a $p$-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 between different $2 p$-th powers is at least $2^{p}-1>1$. Thus, in this case, there are no solutions.