Exercise 8. 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 $2 \leqslant i \leqslant k-1$.
Solution to Exercise 8 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 is even, so it must be 2. If $d_{2}+1=2, d_{2}=1$ which contradicts the statement. Therefore, we have $d_{2}=2$ 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 1 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 satisfying $m<16$ and 2 is a strict factor 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, so $n=9$. 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 $m=8$, $m$ has strict divisors 2 and 4, so $n$ has strict divisors 3 and 5, so $n=15$. 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 $(8,15,4)$.

Graders' Comments The exercise was difficult, but a good number of students made 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!

High School Exercises