18.3.9 ** Find four solutions to the equation $p^{2}+1=q^{2}+r^{2}$, where $p, q, r$ are prime numbers.
It is not difficult to find that $7^{2}+11^{2}=170=13^{2}+1$. This leads to the identity
$$
(5 x+13)^{2}+1=(3 x+7)^{2}+(4 x+11)^{2} \text {. }
$$

If $p=5 x+13, q=3 x+7, r=4 x+11$ are prime numbers, then they form one set of solutions. Thus, the four sets of solutions $(p, q, r)$ that satisfy the equation are: $13,7,11 ; 23,13,19 ; 53,31,43$; $73,43,59$.

Alternatively, from $11^{2}-1^{2}=10 \times 12=6 \times 20=4 \times 30=2 \times 60$, we can derive three sets of solutions: $13,7,11 ; 17,13,11 ; 31,29,11$.

From $13^{2}-1^{2}=168=14 \times 12=2 \times 84=4 \times 42=6 \times 28$, we can derive three sets of solutions: $43, 41,13 ; 23,19,13 ; 17,11,13$.