Determine the integers $m \geqslant 2, n \geqslant 2$ and $k \geqslant 3$ having the following property: $m$ and $n$ each have $k$ positive divisors, and if we denote $d_{1}<\ldots<d_{k}$ the positive divisors of $m$ (with $d_{1}=1$ and $d_{k}=m$) and $d_{1}^{\prime}<\ldots<d_{k}^{\prime}$ the positive divisors of $n$ (with $d_{1}^{\prime}=1$ and $d_{k}^{\prime}=n$), then $d_{i}^{\prime}=d_{i}+1$ for every integer $i$ such that $2 \leqslant i \leqslant k-1$.

## - Statements intended for high school students -
First, note that the smallest divisors strictly greater than 1 of \( m \) and \( n \) are \( d_{2} \) and \( d_{2}+1 \), which are therefore necessarily prime. Since these two numbers are consecutive, one of them must be even, hence it must be 2. If \( d_{2}+1=2 \), then \( d_{2}=1 \), which contradicts the statement. Therefore, \( d_{2}=2 \) and \( d_{2}+1=3 \).

If \( k=3 \), the only strict divisor of \( m \) is 2, so \( m=4 \). The only strict divisor of \( m \) being 3, we must have \( n=9 \). Conversely, since the only strict divisor of 4 is 2 and the only strict divisor of 9 is 3, the triplet \( (4,9,3) \) satisfies the statement.

Suppose \( k \geqslant 4 \). Since the smallest strict divisor of \( m \) is 3, \( m \) is not divisible by 2, so it is odd. Therefore, \( d_{3}+1 \) must be odd, so \( d_{3} \) is even. Let \( d_{3}=2l \). Since \( d_{3}>d_{2}=2 \), we have \( l>1 \), so \( l \geqslant 2 \). If \( l \geqslant 4 \), then \( l \) divides \( d_{3} \), so \( l \) divides \( m \) and \( l \neq m \) because \( 2l \) divides \( m \). In particular, since \( l<2l \) is a divisor of \( m \), this contradicts the statement. Therefore, we must have \( l=2 \), so \( d_{3}=4 \). We deduce that \( d_{3}+1=5 \). If \( k=4 \), the strict divisors of \( m \) are 2 and 4, so \( m=8 \), and since the strict divisors of \( n \) are 3 and 5, we have \( n=15 \). Conversely, the only strict divisors of 8 are 2 and 4, and those of 15 are 3 and 5, so the triplet \( (8,15,4) \) satisfies the statement.

If \( k \geqslant 5 \), since 3 and 5 divide \( n \), 15 divides \( n \). Since \( k \geqslant 3 \), we cannot have \( n=15 \) otherwise \( n \) would have only 2 strict divisors. In particular, 15 is a strict divisor of \( n \), so \( 15-1=14 \) is a strict divisor of \( m \). Therefore, \( 7+1=8 \) is a strict divisor of \( n \), so \( n \) is even, which leads to a contradiction.

In particular, for \( k \geqslant 5 \), there are no solutions, so the only solutions are \( (4,9,3) \) and \( (8,15,4) \).

Alternative solution: In the same way as in the previous solution, we obtain that \( d_{2}=2 \). If \( m \) is divisible by an odd number \( l \), then \( l \) is a strict divisor because \( m \) is divisible by 2, so it is even. In particular, \( l+1 \) is even and divides \( n \), so 2 divides \( n \), which contradicts the fact that \( d_{2}+1 \) is the smallest divisor of \( n \). Therefore, \( m \) must be a power of 2. If \( m \geqslant 16 \), then 8 and 4 are strict divisors of \( m \). Thus, 9 and 5 divide \( n \), so 45 divides \( n \). In particular, 15 is a strict factor of \( n \), so \( 15-1=14 \) is a strict factor of \( m \). Thus, \( m \) is divisible by 7, which contradicts the fact that \( m \) is a power of 2.

Therefore, \( m \) is a power of 2 such that \( m<16 \) and 2 is a strict divisor of \( m \). We have \( m=4 \) or \( m=8 \). If \( m=4 \), the only strict divisor of 4 is 2, so the only strict divisor of \( n \) is 3, hence \( n=9 \). Since the only strict divisor of 4 is 2 and the only strict divisor of 9 is 3, the triplet \( (4,9,3) \) satisfies the statement.

If \( m=8 \), the strict divisors of \( m \) are 2 and 4, so the strict divisors of \( n \) are 3 and 5, hence \( n=15 \). Conversely, the only strict divisors of 8 are 2 and 4, and those of 15 are 3 and 5, so the triplet \( (8,15,4) \) satisfies the statement.

The solution triplets are therefore \( (4,9,3) \) and \( (8,15,4) \).

Comment: The exercise was difficult, but a good number of students managed to make significant progress on the exercise. Be careful when considering the possible parities for \( m \) and \( n \) to treat all cases. Also, remember to verify the solutions obtained!