## Task 4 - 190924

Determine all natural numbers $x$ for which the following holds!

(1) $x$ is the square of a natural number.

(2) If $x$ is increased by 24, the result is the square of a natural number.

(3) If $x$ is decreased by 24, the result is the square of a natural number.
Let it be according to (1) $x=n^{2}$ and according to (2) $x+24=n^{2}+24=(n+s)^{2}=n^{2}+2 s n+s^{2}$ with positive integers $n$ and $s$. It follows that $24=2 s n+s^{2}$ and $s^{2}=24-2 s n=2(12-s n)$, so $s^{2}$ and thus $s$ is even and $s \geq 2$.

Furthermore, according to (3) $x \geq 24$, so $n \geq 5$ and thus $s^{2} \leq 24-2 \cdot 2 \cdot 5=4$, so also $s \leq 2$. We obtain that $s=2$ and thus $n=\frac{24-s^{2}}{2 s}=\frac{20}{4}=5$ must be. It is $x=n^{2}=25$ the only possible solution.

Indeed, not only is $x=25=5^{2}$ the square of a natural number (1), but also $x^{2}+24=$ $59=7^{2}$ (2) and $x^{2}-24=1=1^{2}(3)$, so $x=25$ is indeed also a solution (and thus the only one).