## Task 15/79

We are looking for all natural numbers $n$ with the property that both $n+100$ and $n-100$ are squares of natural numbers.
From the required property, the equations $n+100=x^{2}$ and $n-100=y^{2}$ with $x ; y ; n \in N$ follow. By subtracting the second equation from the first, we get

$$
200=x^{2}-y^{2}=(x-y)(x+y)
$$

Thus, all possible numbers $x$; $y$ can be found by decomposing the number 200 into all possible products of two natural factors $f_{1}$ and $f_{2}$ and solving the system of equations

$$
x-y=f_{1} \quad ; \quad x+y=f_{2}
$$

Since the solutions are $x=0.5\left(f_{1}+f_{2}\right), y=0.5\left(f_{2}-f_{1}\right)$, it follows that $f_{1}$ and $f_{2}$ must both be even or both be odd (otherwise $x ; y \notin N$ would hold). Therefore, only the decompositions

$$
200=2 \cdot 100=4 \cdot 50=10 \cdot 20
$$

are valid. From these, the following solutions arise:

| $f_{1}$ | $f_{2}$ | $\mathrm{x}$ | $\mathrm{y}$ | $\mathrm{n}$ |
| :---: | :---: | :---: | :---: | :---: |
| 2 | 100 | 51 | 49 | 250 |
| 4 | 50 | 27 | 23 | 62 |
| 10 | 20 | 15 | 5 | 12 |

The verification confirms the correctness, and the solution method excludes further solutions.