33 (1275). Find all prime numbers $p$ and $q$ for which $p^{2} - 2 q^{2}=1$.
Solution. From the equality $p^{2}-2 q^{2}=1$ we have $p^{2}-1=2 q^{2}$, from which $\frac{(p-1)(p+1)}{2}=q \cdot q$. Since in the right-hand side of the equality we have the product of equal prime numbers ( $q$ has no other divisors except $q$ and 1), then in the left-hand side of the equality we also have the product of equal numbers, i.e., $p-1=\frac{p+1}{2}$, from which $p=3$. Therefore, $9-2 q^{2}=1, q^{2}=4, q=2$. Answer: $p=3, q=2$.

Remark. Another way to solve this problem is given in $\$ 4$, p. 28.