Minimum sum of factors of a natural number

Question

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:

• $N = 12$, $a = 3$, $b = 4$

• $N = 13$, $a = 1$, $b = 13$

Answer 1

Look at the set of all your factors $\{a_1,a_2,...,a_r\}$. Now find $a_i$ such that $|a_i-\sqrt{N}|$ is minimal. Then the pair you are looking for is $a_i,\frac{N}{a_i}$.

Can you prove that?

Answer 2

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 $\sqrt N \le \dfrac{N}{a_2} \lt \dfrac{N}{a_1} \le N$