Task 1. Determine all pairs of prime numbers whose difference of squares is equal to 120.
Solution. Let $(x, y), x>y$ be a pair of prime numbers. From the condition of the problem, it follows that $x^{2}-y^{2}=120$, i.e., $\quad(x-y)(x+y)=120$. Obviously, $x-y+x+y=2 x$ is an even number, so both factors must have the same parity. Since the product of two odd numbers is an odd number, we conclude that both factors must be even. Therefore, the following cases are possible:

$$
\left\{\begin{array}{l}
x+y=60 \\
x-y=2
\end{array}, \quad\left\{\begin{array}{l}
x+y=30 \\
x-y=4
\end{array}, \quad\left\{\begin{array} { l } 
{ x + y = 2 0 } \\
{ x - y = 6 }
\end{array} \text { and } \left\{\begin{array}{l}
x+y=12 \\
x-y=10
\end{array} .\right.\right.\right.\right.
$$

The solutions to the first three systems are the required ordered pairs of prime numbers $(x, y)$, i.e., $(31,29),(17,13)$ and $(13,7)$, while the last system does not provide a solution, as it yields $x=11$ and $y=1$, and 1 is not a prime number.