62nd Czech and Slovak Mathematical Olympiad

First, we will deal with the case when the wanted primes $p$ and $q$ are distinct. Then, the numbers $pq$ and $p+q$ are relatively prime: the product $pq$ is divisible by two primes only (namely $p$ and $q$), while the sum $p+q$ is divisible by neither of these primes. We will look for a positive integer $r$ which can be a common divisor of both $a+1$ and $a^2+1$. If $r\mid a+1$ and, at the same time, $r\mid a^2+1$, then $r\mid (a+1)(a-1)$ and also $r\mid (a^2+1)-(a^2-1)=2$, so $r$ can only be one of the numbers $1$ and $2$. Thus the fraction $\frac{a^2+1}{a+1}$ either is in lowest terms, or will be in lowest terms when reduced by two, depending on whether the integer $a$ is even, or odd.

If $a$ is even, we must have $$pq=a^2+1\text{and}p+q=a+1.$$ The numbers $p$, $q$ are thus the roots of the quadratic equation $x^2-(a+1)x+a^2+1=0$, whose discriminant $$(a+1{)}^2-4(a^2+1)=-3a^2+2a-3=-2a^2-(a-1{)}^2-2$$ is apparently negative, so the equation has no solution in the real numbers.

If $a$ is odd, we must have (taking into account the reduction by two) $$2pq=a^2+1\text{and}2(p+q)=a+1.$$ The numbers $p$, $q$ are thus the roots of the quadratic equation $2x^2-(a+1)x+a^2+1=0$, whose discriminant is negative as well.

Therefore, there is no pair of distinct primes $p$, $q$ satisfying the conditions. It remains to analyze the case of $p=q$. Then, $$\frac{p\cdot q}{p+q}=\frac{p\cdot p}{p+p}=\frac{p}{2}$$ so we must have $$p=\frac{2(a^2+1)}{a+1}=2a-2+\frac{4}{a+1};$$ this is an integer if and only if $a+1\mid 4$, i.e. $a\in \{1,3\}$, so $p=2$ or $p=5$.

To summarize, there are exactly two pairs of primes satisfying the conditions, namely $p=q=2$ and $p=q=5$.