# Find all natural numbers that satisfy the condition of the given problem

Find all natural numbers whose maximum proper divisor is $$2$$ more than the square of the minimum proper divisor. The divisor of a natural number is called proper if it is not equal to this number and one.

A solution was proposed: The smallest proper divisor of any natural number is a prime number, otherwise it is not the smallest. If $$a$$ is the largest proper divisor of $$n$$, and $$b$$ is the smallest proper divisor of $$n$$, then $$n=a⋅b$$. By hypothesis, the maximum proper divisor is $$2$$ more than the square of the minimum proper divisor, i.e. $$a=b^2+2$$. Substitute and get $$n=b⋅(b^2+2)$$, where $$b$$ is a prime number. It turns out that this will be the answer, i.e. all numbers that can be expressed by this formula, not any finite set of numbers. Or am I not understanding something and making a mistake?

• The only solutions I could find are $12$ and $33$. Apparently, $b$ is not arbitary, there is some additional condition. Maybe that $b^2+2$ is prime as well except in the case $b=2$ ? If so, the list would be complete since $b^2+2$ is divisible by $3$ if $b$ is a prime greater than $3$. Commented Nov 29, 2021 at 18:05

The solution you found works formally, but not in practice. It works for $$b=2,3$$ to give $$12,33$$, but larger primes ($$b>3$$) have the form $$b=6k\pm 1$$ so their squares have the form $$b^2=6m+1 \Rightarrow b^2+2=6m+3$$.
Hence, for $$b>3$$, we have $$3\mid (b^2+2)$$ and $$b$$ is no longer the smallest proper divisor.