1. Find all possible values of the product of prime numbers $p, q, r$, for which

$$
p^{2}-(q+r)^{2}=637
$$
SOLUTION. We factor the left side of the given equation according to the formula for $A^{2}-B^{2}$. In the thus modified equation

$$
(p+q+r)(p-q-r)=637
$$

we can easily examine all possibilities for the two integer factors on the left. The first one is larger and positive, so the second one must also be positive (since their product is positive), so according to the prime factorization of the number $637=7^{2} \cdot 13$ it is one of the pairs $(637,1),(91,7)$ or $(49,13)$. The prime number $p$ is apparently the arithmetic mean of both factors, so it must equal one of the numbers $\frac{1}{2}(637+1)=319, \frac{1}{2}(91+7)=49, \frac{1}{2}(49+13)=31$. However, the first two are not prime numbers $\left(319=11 \cdot 29\right.$ and $\left.49=7^{2}\right)$, the third one is. Therefore, it is necessarily $p=31$ and the corresponding equalities $31+q+r=49$ and $31-q-r=13$ hold precisely when $q+r=18$. Such pairs of prime numbers $\{q, r\}$ are only $\{5,13\}$ and $\{7,11\}$ (it is sufficient to examine all possibilities, or to realize that one of the prime numbers $q, r$ must be at least $18: 2=9$, but at most $18-2=16$). The product $pqr$ thus has exactly two possible values, namely $31 \cdot 5 \cdot 13=2015$ and $31 \cdot 7 \cdot 11=2387$.

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Determine all natural numbers $a$ and $b$ for which the difference $a^{2}-b^{2}$ is the square of some prime number. $\left[a=\left(p^{2}+1\right) / 2\right.$ and $b=\left(p^{2}-1\right) / 2$, where $p$ is any odd prime number. $]$

D1. Find all pairs of non-negative integers $a, b$ for which $a^{2}+b+2=a+b^{2}$. $[\mathrm{C} 59-\mathrm{S}-3]$

D2. Find all pairs of prime numbers $p$ and $q$ for which $p+q^{2}=q+145 p^{2}$. [C55-II-4]