Exercise 6. Determine all integers $n \geqslant 2$ such that for any integer $d \geqslant 2$, if $d$ is a divisor of $n$ then $d-1$ is a divisor of $n-1$.
Solution to Exercise 6 If $n$ is a prime number, then $d=n$ is the unique divisor $\geq 2$ of $n$, so $n$ works.

If $n=p^{2}$ is the square of a prime number, then $d=p$ or $d=n$. Now, $p-1$ divides $p^{2}-1=(p-1)(p+1)$ and $n-1$ divides $n-1$, so $n$ works.

Conversely, suppose that $n$ is neither a prime number nor the square of a prime number. Let $a$ be the smallest divisor $\geq 2$ of $n$. Then $n=a b$ with $10$, we have $a-1 \geq b-1$, which contradicts that $a<b$.

Conclusion: The integers that work are the prime numbers and the squares of prime numbers.