9.4. How many three-digit natural numbers $n$ exist for which the number $n^{3}-n^{2}$ is a perfect square
Answer: 22. Solution. Let $n^{3}-n^{2}=m^{2}$ for some natural number $m$. Then $n^{2}(n-1)=m^{2}$, and therefore $n-1$ must also be a perfect square: $n-1=a^{2}$. Thus, $n=a^{2}+1$ and $m=\left(a^{2}+1\right) a$. Therefore, we need to find all three-digit numbers $n$ that are one more than perfect squares. Such numbers are $101=10^{2}+1, 122=11^{2}+1, \ldots, 962=31^{2}+1$ (the next square $32^{2}=1024-$ is already too large).

## 9th grade