Exercise 14. Find the integers $n$ greater than or equal to 2 such that, if we denote $a$ as the smallest prime divisor of $n$, we can find a positive divisor of $n$ denoted $d$ such that $n=a^{3}+d^{3}$.
Solution to Exercise 14: Let's note that $d$ divides $n$ and $d^{3}$, so $d$ divides $a^{3}$. Since $a$ is prime, we get that $d=1, a, a^{2}$ or $a^{3}$.

If $d=1$, the equation becomes $n=a^{3}+1$. Since $a$ divides $n$, $a$ divides $n-a^{3}=1$ so $a=1$, which is a contradiction. Otherwise, $d=a, a^{2}$ or $a^{3}$, and thus $d$ has the same parity as $a$, so $d^{3}$ has the same parity as $a^{3}$. In particular, $n=a^{3}+d^{3}$ is the sum of two numbers of the same parity, so it is even. Since $n$ is divisible by 2, we must have $a=2$. Thus $d=2,4,8$.

Let's check the potential values of $n$:

$\triangleright$ If $a=2$ and $d=8$, we get $n=8+512=520=8 \times 65$ so $d$ divides $n$ well, the smallest divisor of $n$ different from 1 is $2: n=520$ is a solution.

$\triangleright$ If $a=2$ and $d=4, n=8+64=72=4 \times 18$ so $d$ divides $n$ well, the smallest divisor of $n$ different from 1 is $2: n=72$ is a solution.

$\triangleright$ If $a=2$ and $d=2, n=8+8=16=2 \times 8$ so $d$ divides $n$ well, the smallest divisor of $n$ different from 1 is $2: n=16$ is a solution.

The solutions are therefore $n=16, n=72, n=520$.