5. Find all pairs of natural numbers ( $m, n$ ), for which $\operatorname{LCM}(m, n)=3 m+2 n+1$. (Recall that LCM is the least common multiple of the numbers.)
Answer: $(3,10)$ and $(4,9)$.

## Solution:

## First method.

Note that $m$ does not divide $n$ (otherwise $LCM(m, n)=m$). In this case, $3 m+1=7$, which is not equal to $2 n$, since it is odd.

(2) Let $m>n$ (i.e., $n \leq m-1$). Then $2 n+1=k m$, but $2 n+1 \leq 2(m-1)+11$, so $LCM(m, n)$ is divisible by $d$, but $LCM(m, n)-(3 m+2 n)=1$ - the left side of the equation is divisible by $d$, but the right side is not. Therefore, $m$ and $n$ are coprime (do not have common divisors other than 1), and then $LCM(m, n)=m n$.

The original equation can be rewritten as: $m n=3 m+2 n+1$.

Move everything to the left side, factor where possible:

$m n-3 m-2 n-1=m(n-3)-2(n-3)-7=(m-2)(n-3)-7$.

Thus, we have the equation in natural numbers: $(m-2)(n-3)=7$.

Since $m$ is a natural number, $(m-2)$ is an integer, not less than -1. The number 7 has only three integer divisors not less than $-1: -1, 1$, and 7. If $m-2=-1$, then $n-3=-7$, so $n<0$ (not natural). When $m-2=1$, we get $n-3=7$, hence $m=3$ and $n=10$, and when $m-2=7$, we get $n-3=1$, hence $m=9$ and $n=4$ - both these pairs are valid.

## Criteria:

Both (!) solutions are found - 1 point.

In the first method of solving, one of the two cases is considered - plus 2 points.

The problem is solved under the unproven assumption that the numbers are coprime - the proof is worth 1 point.

It is proven that the numbers are coprime - plus 2 points.