## Task 32/81

Provide a necessary and sufficient condition for the natural number $n$ that guarantees the equation $x+y+x y=n$ has exactly one solution in natural numbers $x ; y$ with $0 \leq x<y$.


If $x_{1} ; y_{1}$ is a solution, then with $x_{1}+y_{1}+x_{1} y_{1}=n$ it also holds that

$$
1+x_{1}+y_{1}+x_{1} y_{1}=\left(1+x_{1}\right)\left(1+y_{1}\right)=n+1
$$

Every decomposition of $n+1$ into a product of two factors provides a solution. A solution exists exactly when $n+1$ can be decomposed into a product of two different factors in exactly one way. This is necessary and sufficient if $n+1$ is a prime number or the square of a prime number: $n+1=1 \cdot(n+1)$.

Since $x<y$, then $\left(1+x_{1}\right)=1$ and $\left(1+y_{1}\right)=n+1$, so $x=0$ and $y=n$. The decomposition $\left(1+x_{1}\right)=p$ and $\left(1+y_{1}\right)=p$ ( $p$ prime, $\left.p^{2}=n-1\right)$ is not possible because $x<y$.