The number and sum of positive integer divisors of the no 12500 including 1 and itself
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Answer:
positive integer with only four positive divisors has its prime factorization in the form of $a*b$, where $a$ and $b$ are both prime positive integers or $c^3$ where $c$ is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of $a*b$. The four factors of this number would be $1$, $a$, $b$, and $ab$. The sum of these would be $ab+a+b+1$, which can be factored into the form $(a+1)(b+1)$. Easily we can see that now we can take cases again.
Case 1: Either $a$ or $b$ is 2.
If this is true then we have to have that one of $(a+1)$ or $(b+1)$ is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either $\frac{2016}{3} - 1$ or $\frac{2010}{3} - 1$ is a prime. We see that in this case none of them work.