Z2) Find all triples $(a, b, p)$ of natural numbers such that $p$ is a prime number and the equation

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

is fulfilled.
Solution: $(a, b, p)=(1,1,2)$ is the only solution. We divide the problem into two cases.

- Case 1: $a=b$

The equation becomes $2^{p} \cdot a^{p}=2 \cdot p^{a}$, and since $4\left|2^{p}, 2\right| p^{a}$ it follows that $p=2$. If we insert this new information into the original equation, we get $4 a^{2}=2^{a+1}$, i.e. $a^{2}=2^{a-1}$. If $a>1$, then $a$ must be even because $2^{a-1}$ is even. However, since $a^{2}$ is a square number, $2^{a-1}$ must also be a square number, so $a$ must be odd, a contradiction. The only case is therefore $a=1$, which actually fulfills all conditions. In this case, we therefore obtain a solution: $(a, b, p)=(1,1,2)$.

- Case 2: $a \neq b$

Because the equation is symmetric in $a$ and $b$, we assume $b>a$ without restriction of generality. Now we write the equation as

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

This gives us $p \mid(a+b)^{p}$, i.e. also $p \mid a+b$. Now we look at the prime factorization of both sides of the equation and look at the exponent of $p$. The right-hand side of the equation is $p^{a}$ times a number that is not divisible by $p$ because $b-a>0$. If $(a+b)^{p}=p^{x} \cdot y$ where $y$ is not divisible by $p$, then we have $p \mid x$. Thus we also have $p \mid a$ because $x=a$. If we combine this with $p \mid a+b$, we get $p \mid a, b$. Now we notice that $p^{a}\left(1+p^{b-a}\right)$ is a $p$th power, and since $p^{a}$ is a $p$th power divisor of $1+p^{b-a}$, we know that $1+p^{b-a}$ is also a $p$th power. We write $1+p^{b-a}=z^{p}$, and since $p \mid b-a$ and $b-a>0$, we get $z^{p}-p^{b-a}=1$, a contradiction since the difference of two $p$th powers is at least $2^{p}-1>1$. So in this case there are no solutions.