Problem 990

Solution. It is sufficient to show that at least one of the two numbers $d_{a}=\operatorname{GCD}(a, c)$ and $d_{b}=\operatorname{GCD}(b, c)$ is greater than 1. Indeed, if, for example, $d_{a}>1$, then $a+c$ is divisible by $d_{a}$ and $a+c>d_{a}$, which means that $a+c$ is a composite number.

From the equality $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ it follows that $c(a+b)=a b$, so $a b$ is divisible by $c$. But then, if $d_{a}=d_{b}=1$, then $c=1$, which is impossible by the condition. Thus, one of the numbers $d_{a}$ and $d_{b}$ is greater than 1, which was to be proved.

Remark. Note that if natural numbers $a, b, c$ satisfy the equality $1 / a+1 / b=1 / c$, then the number $a+b$ is also composite.

Comment. It has been proved that one of the numbers $d_{a}$ and $d_{b}$ is greater than one - 5 points.

If the problem is reduced to the statement that one of the numbers $d_{a}$ and $d_{b}$ is greater than one, but this statement itself is not proved 2 points.