$\underline{\text { Folklore }}$

Find all integer solutions of the equation $y^{k}=x^{2}+x$ ( $k$ is a natural number greater than 1$)$.

#
$y^{k}=x(x+1)$. If in the prime factorization of $y$ there is $p^{n}$, then $y^{k}$ is divisible by $p^{n k}$; hence (since $x$ and $x+1$ are coprime), exactly one of the numbers $x, x+1$ is divisible by $p^{n k}$. Therefore, $|x|$ and $|x+1|$ are $k$-th powers. For $k \neq 1$, this is possible only when one of the numbers is zero, and the other is one. In any case, $y=0$.

## Answer

$(0,0)$ and $(-1,0)$.