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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?

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    $\begingroup$ 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 $3$. $\endgroup$
    – Peter
    Commented Nov 29, 2021 at 18:05

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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.

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