5. Find all natural numbers $k$ for which the number $k^{2}-101 k$ is a perfect square, i.e., the square of an integer.
Answer: 101 or 2601.

Solution. Let $k^{2}-101 k=(k-m)^{2}, m \geqslant 1$, then $k^{2}-101 k=k^{2}-2 m k+m^{2}$, from which $m(2 k-m)=101 k$. Since the number 101 is prime, one of the numbers $m$ or $2 k-m$ is divisible by 101.

If $m=101 s$, then we get $s(2 k-101 s)=k$, from which

$$
k=\frac{101 s^{2}}{2 s-1}=101\left(\frac{s}{2}+\frac{1}{4}+\frac{1}{4(2 s-1)}\right) \quad \Rightarrow \quad 4 k=202 s+101+\frac{101}{2 s-1}
$$

This means that 101 is divisible by $2 s-1$, which is only possible if $2 s-1=1$ or $2 s-1=101$, i.e., respectively $s=1$ or $s=51$. In the first case, we get $m=k=101$, and in the second case $m=101 \cdot 51, k=51^{2}=2601$.

The case where $2 k-m$ is divisible by 101 reduces to the previous one by replacing $m_{1}=2 k-m$, $m=2 k-m_{1}$.