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 $2 \leqslant i \leqslant k-1$.
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 equal to 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 \( d_{3} \) and thus \( l \) 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, since the only strict divisors of 8 are 2 and 4, and those of 15 are 3 and 5, 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 divisor of \( n \), so \( 15-1=14 \) is a strict divisor 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, since the only strict divisors of 8 are 2 and 4, and those of 15 are 3 and 5, the triplet \( (8,15,4) \) satisfies the statement.

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

Comment: This number theory problem was quite well solved, with nearly half of the students submitting a paper that scored 6 points or more. There were many solutions: while most students solving the problem used one of the solutions in the answer key, some found other more original solutions. For example, some noticed that if \( k \geqslant 5 \), then \( d_{2} d_{k-1}=d_{3} d_{k-2}=m \) and \( d_{2}^{\prime} d_{k-1}^{\prime}=d_{3}^{\prime} d_{k-2}^{\prime}=n \), and that this implies that since \( d_{2}^{\prime}=d_{2}+1 \) and \( d_{3}^{\prime}=d_{3}+1 \), \( d_{2}+d_{k-1}=d_{3}+d_{k-2} \), so in particular \( d_{2}=d_{3} \), which provides the contradiction. Others correctly noted that if \( k \geqslant 4 \), then \( d_{2}=2 \), \( d_{3}=4 \), so \( d_{2}^{\prime}=3 \), \( d_{3}^{\prime}=5 \), leading to the system of two equations \( \frac{m}{2}+1=\frac{n}{3} \) and \( \frac{m}{4}+1=\frac{n}{5} \). We nevertheless draw students' attention to the few problematic points or errors below:

- Some students have nearly complete proofs but forget to verify at the end whether the triplets found are indeed all solutions, which sometimes leads to the inclusion of a few extra triplets. Conversely, others forget one of the two solution triplets by not addressing what happens when \( k=3 \).
- Many students have well-founded intuitions (sometimes even intuiting the two solution triplets), but do not justify their claims. We remind students that any result stated in the papers must be at least justified by a sentence. For example, even if the explanation of this fact is very simple, simply stating that \( n \) cannot be even is not sufficient to earn the corresponding points. Writing a clear and precise paper not only allows the grader to know that you understand, but also allows you to verify and be convinced of your claims.
- Some case distinctions could be numerous in some solutions to the problem and sometimes lacked rigor. Some students assert that an integer with 4 divisors must necessarily be the cube of a prime number, but in fact it can also be the product of two distinct prime numbers. On the other hand, some provide a valid argument to obtain a contradiction as soon as \( k \geqslant 7 \) (the argument that 33 would divide \( n \) but 11 is not in the list of its divisors). In this case, it is important to explicitly treat the cases \( k=6 \) and \( k=5 \) separately.
- Some more technical remarks. Indeed, the statement implies that \( m \) or \( n \) is odd. But we lose generality by only treating the case where \( m \) is even, since the statement is not symmetric in \( m \) and \( n \). Moreover, while it is true that \( d_{2}^{\prime}=d_{2}+1 \) and \( d_{2} \) are primes and divide \( n \) and \( m \) respectively, it is not correct to say that if a prime \( p \) divides \( m \), then \( p+1 \) is also a prime that divides \( n \). Finally, the misuse of the terms "multiples" and "divisors" makes some proofs confusing.