Find all integer solutions of the equation $y^{k}=x^{2}+x$, where $k-$ is a fixed natural number, greater than 1.
Factorize the right-hand side and use the mutual simplicity of these factors.

## Solution

$y^{k}=x(x+1)$. The numbers $x$ and $x+1$ are coprime, so $x$ and $x+1$ are $k$-th powers of integers. It is clear that there are exactly two pairs of consecutive integers that are $k$-th powers for $k>1$: $(-1,0)$ and $(0,1)$. Therefore, $x$ can only take two values: 0 and -1. It remains to check that both these values fit, and in both cases $y=0$.

## Answer

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