1. Find all pairs of prime numbers $p, q$ for which there exists a natural number $a$ such that

$$
\frac{p q}{p+q}=\frac{a^{2}+1}{a+1}
$$
SOLUTION. We will first deal with the case where the sought prime numbers $p$ and $q$ are different. Then the numbers $p q$ and $p+q$ are coprime: the product $p q$ is divisible only by the two prime numbers $p$ and $q$, while the sum $p+q$ is not divisible by any of these primes.

We will determine which natural number $r$ can be a common divisor of the numbers $a+1$ and $a^{2}+1$. If $r \mid a+1$ and simultaneously $r \mid a^{2}+1$, then $r \mid(a+1)(a-1)$ and also $r \mid\left(a^{2}+1\right)-\left(a^{2}-1\right)=2$, so $r$ can only be one of the numbers 1 and 2. The fraction $\frac{a^{2}+1}{a+1}$ is thus either in its simplest form, or the simplest form is obtained by canceling by two, depending on whether the number $a$ is even or odd.

In the case of even $a$, it must hold that

$$
p q=a^{2}+1 \quad \text { and } \quad p+q=a+1
$$

The numbers $p, q$ are thus the roots of the quadratic equation $x^{2}-(a+1) x+a^{2}+1=0$, whose discriminant

$$
(a+1)^{2}-4\left(a^{2}+1\right)=-3 a^{2}+2 a-3=-2 a^{2}-(a-1)^{2}-2
$$

is clearly negative, so the equation has no solutions in the set of real numbers.

In the case of odd $a$, with regard to the cancellation by two, it must hold that

$$
2 p q=a^{2}+1 \quad \text { and } \quad 2(p+q)=a+1 .
$$

The numbers $p, q$ are thus the roots of the quadratic equation $2 x^{2}-(a+1) x+a^{2}+1=0$, whose discriminant is also negative. No pair of different prime numbers $p, q$ therefore satisfies the problem.

It remains to consider the case where $p=q$. Then

$$
\frac{p \cdot q}{p+q}=\frac{p \cdot p}{p+p}=\frac{p}{2}
$$

so it must hold that

$$
p=\frac{2\left(a^{2}+1\right)}{a+1}=2 a-2+\frac{4}{a+1} ;
$$

this is an integer precisely when $a+1 \mid 4$ i.e., $a \in\{1,3\}$, so $p=2$ or $p=5$.

The problem is thus satisfied only by the pairs $p=q=2$ and $p=q=5$.

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