Minimum sum of factors of a natural number Let's say I have a natural number $N$. $a$ and $b$ are two factors of $N$. How can I find $a$ and $b$ such that $a + b$ is minimum.
Examples:


*

*$N = 12$, $a = 3$, $b = 4$

*$N = 13$, $a = 1$, $b = 13$

consider the factors of 36 and their sums
 1  +  36 = 37
 2  +  18 = 20
 3  +  12 = 15
 4  +   9 = 13
 6  +   6 = 12

So $a = b = 6 = \sqrt{36}$
Consider the factors of 12 and their sums
 1  +  12 = 13 
 2  +   6 =  8
 3  +   4 =  7

So $a = 3 \lt \sqrt{12} \lt b = 4$
It should be fairly clear that the two factors that are "closest to" $sqrt N$ are the two factors that you are looking for.
To prove this, you would have to show that
$1 \le a_1 \lt a_2 \le \sqrt N$ implies that
$\sqrt N \le \dfrac{N}{a_2} \lt \dfrac{N}{a_1} \le N$
