Austrian Mathematical Olympiad

The numbers $8=2^3$ and $9=3^2$ form a pair of consecutive powerful numbers.

We now show that for each pair $(k,k+1)$ of powerful positive integers we can find a new pair, namely the pair $(4k(k+1),(2k+1{)}^2)$. Obviously, $4k(k+1)+1=(2k+1{)}^2$. Since $k$ and $k+1$ are powerful, the product $4k(k+1)=2^{2}k(k+1)$ is also powerful. And a square is certainly powerful, so in particular $(2k+1{)}^2$. Finally, $4k(k+1)>k$ for positive integers $k$. This means that there are infinitely many pairs of powerful consecutive positive integers.