74th Romanian Mathematical Olympiad

a) Without loss of generality, we may suppose $1\le m<n$. If $m$ and $n$ are odd, then $m\ge 1$ and $n\ge 3$. Taking $d_1=1$ and $d_2=n$ yields $d_1+d_2=n+1$, which is an even number at least $4$, so it is composite, contradiction. If $m$ and $n$ are even, taking $d_1=2$ and $d_2=2$ yields $d_1+d_2=4$, contradiction. In conclusion $m$ and $n$ have different parities, so $m+n$ is odd.

b) If $m=n$, then $d_1=m$ and $d_2=n$ leads to $d_1+d_2=2m$, which must be a prime, hence $m=n=1$. If $m<n$, then $m\ne n$ and, from a), $m$ and $n$ have different parities.

If $m$ is even and $n$ is odd, then $m\ge 2$ and $n\ge 3$. Then $d_1=1$ and $d_2=n$ yields $d_1+d_2=n+1\ge 4$, contradiction. If $m$ is odd and $n$ is even, then $m\ge 1$ and $n\ge 2$. Take $a\in {\mathbb{N}}^{\ast }$ and $b$ odd such that $n=2^a\cdot b$.

I. If $a\ge 3$, then $8$ divides $n$ and, taking $d_1=1$ and $d_2=8$ yields $d_1+d_2=9$, contradiction.

II. If $a=1$, then $n=2b$, with odd $b$.

i) If $b\ge 3$, then take $d_1=1$ and $d_2=b$ to get $d_1+d_2=b+1\ge 4$, contradiction.

ii) If $b=1$, then $n=2$ and, since $m<n$ and $m$ is odd, $m=1$. It is easy to check that the pair $(m,n)=(1,2)$ is a solution.

III. If $a=2$, then $n=4b$, with odd $b$.

i) If $b\ge 3$, then $d_1=1$ and $d_2=b$ yields $d_1+d_2=b+1\ge 4$, contradiction.

ii) If $b=1$, then $n=4$ and, since $m<n$ and $m$ is odd, $m\in \{1,3\}$. It is easy to check that only $(m,n)=(1,4)$ satisfies the statement.

Finally, the solutions are $(1,1)$, $(1,2)$, $(1,4)$.