## Task $7 / 65$

We are looking for three square numbers $a, b$, and $c$ such that $a-b=b-c=24$.
If we set $a=x^{2}, b=y^{2}$ and $c=z^{2}$, it follows from the problem that $x^{2}-y^{2}=y^{2}-z^{2}=24$. This leads to

$$
x^{2}-y^{2}=(x+y)(x-y)=24 \quad(1) ; \quad y^{2}-z^{2}=(y+z)(y-z)=24
$$

For $x+y$ and $x-y$ as well as for $y+z$ and $y-z$, only the integer divisors of 24 are possible. This results in the following pairs:

$$
\begin{array}{lcccc} 
& 1. & 2. & 3. & 4. \\
x+y \text { or } y+z & 24 & 12 & 8 & 6 \\
x-y \text { or } y-z & 1 & 2 & 3 & 4
\end{array}
$$

Pairs 1. and 3. are eliminated because the sum and difference of both are odd, which does not yield integer solutions for $x, y$ and $z$. From the remaining two pairs, the following solutions are derived:

$$
\text { 2. } x=7, y=5 \text { or } y=7, z=5 \quad ; \quad \text { 4. } x=5, y=1 \text { or } y=5, z=1
$$

From this (by combining both solutions), it follows that $x=7, y=5$ and $z=1$.

The numbers sought are thus $a=7^{2}=49, b=5^{2}=25, c=1^{2}=1$.