## Task $9 / 86$

We are looking for all triples $(x ; y ; z)$ of natural numbers $x ; y ; z$ that satisfy the following conditions:

1. The sum of the three numbers is 107.
2. At least two of the three numbers are squares of natural numbers.
3. At least two of the three numbers decreased by 13 are squares of natural numbers.
According to the Dirichlet drawer principle, for at least one of the three numbers (w.l.o.g. let this be $x$): $x=k_{1}^{2}$ and $x-13=k_{2}^{2}$ with $k_{1} ; k_{2} \in \mathbb{N}$, it follows that $k_{1}^{2}-k_{2}^{2}=13$, thus $k_{1}^{2}=49=x ; k_{2}^{2}=36$ as the only solution (it is easily verified that for $k_{1}>6$ the difference between adjacent square numbers is always greater than 13, even more so for non-adjacent ones).

Now let $y$ be the second number, which, when decreased by 13, is also a square number. Then, because $x+y+z=107$ and $x=49$,

$$
y+z=58 ; \quad y=58-z \rightarrow y-13=k_{3}^{2}
$$

with $k_{3} \in \mathbb{N}$, so $y-13=k_{3}^{2}=58-13-z=45-z$, i.e., $z=45-k_{3}^{2}$. Therefore, $k_{3}^{2} \leq 36 ; k_{3} \leq 6$ because $z \geq 0$. Thus, the 7 numbers $k_{3}=0 ; 1 ; 2 ; \ldots ; 6$ need to be checked to see if $y$ or $z$ is a square number. Clearly, this is only the case for $k_{31}=3 ; z_{1}=36 ; y_{1}=22$ and for $k_{32}=6 ; z_{2}=9, y_{2}=49$. Therefore, there are exactly two such triples (up to the order): $(x ; y ; z)=(49 ; 22 ; 36),(x ; y ; z)=(49 ; 49 ; 9)$. In the second triple, all three numbers are even squares.