3. Determine all ordered triples of numbers $(x, y, z)$ for which: $x-y=y-z=$ 96, where $x, y, z$ are squares of natural numbers.
3. Let $x=a^{2}, y=b^{2}, z=c^{2}$. Since $x>y>z$, it follows that $a>b>c$.

$$
\begin{array}{lll}
a^{2}-b^{2}=96 & \Rightarrow & (a+b)(a-b)=96 \\
b^{2}-c^{2}=96 & \Rightarrow & (b+c)(b-c)=96
\end{array}
$$

We have the following possibilities:

$a+b, \quad b+c=96, \quad 48, \quad 32, \quad 24, \quad 16, \quad 12$;

$a-b, \quad b-c=1, \quad 2, \quad 3, \quad 4,6, \quad 8 . \quad$ (5 points)

Since $(a+b)+(a-b)$ is an even number, the first and third possibilities are eliminated.

(5 points)

The remaining possibilities are

$(a, b), \quad(b, c)=(25,23), \quad(14,10), \quad(11,5), \quad(10,2)$.

Since the pair $(a, b)$ ends with the element that $(b, c)$ starts with, it can only be $a=14, \quad b=10, \quad c=2$. The only solution is $x=196, \quad y=$ $100, \quad z=4$.

(10 points)