Minimum $a+b$ such that $ab=M$ This question is related to the fleablood's answer to another question.
Let $ab=M$ where $0<a,b,M\in\mathbb N$. He wanted to show that the sum $a+b$ is the smallest when $a,b$ are closest together. But he defined $a=\sqrt{M}r$, $b=\sqrt{M}/r$ with $r=1+e$ (where I suppose that $0<e\in\mathbb R$).
My question. Is it always possible to define the positive integers in $ab=M$ as $a=\sqrt{M}r$ and $b=\sqrt{M}/r$ with $r=1+e$? How would you show it?
Thanks in advance.

Fleablood is ignoring the integer constraint there. They are looking at the problem of minimising $a + b$, where $a, b \in (0, \infty)$ such that $ab = M$. This is well known to be $2\sqrt{M}$, achieved at $a = b = \sqrt{M}$.
However, Fleablood is going a little further, given that there are extra constraints currently being (semi)-ignored. The point of the argument is to show that, under these constraints, the sum $a + b$ is a function of the distance between $a$ and $b$. Normally, this would be $|a - b|$, but if you assume $b \le a$ (which is not unreasonable), then it would be $a - b$.
Not only that, the sum is an increasing function of $a - b$. So, minimising $a + b$ would be equivalent to minimising $a - b$ according to the constraints of the question.
I find Fleablood's calculations a little convoluted for my tastes. I would make a substitution of $a + b = u$ and $a - b = v$ (both positive). Then $a = (u + v) / 2$ and $b = (u - v)/2$, and
$$M = ab = \frac{(u + v)(u - v)}{4} = \frac{u^2 - v^2}{4},$$
hence
$$u = \sqrt{v^2 + 4M}.$$
So, $u$ is an increasing function of $v$. That is, the larger the difference $a - b = v$, the larger the sum $a + b = u$.
As with Fleablood's answer, there's no mention of $a$ and $b$ being integers. We don't need $\sqrt{M}$, $r$, or $e$ (or $u$ or $v$, for that matter) to be integers either. But, once we add in the rest of the constraints, we now know we need to search for integers $a > b$, still multiplying to $M = 2 \cdot 25 \cdot 49 \cdot 61$, where $a - b$ is minimised.
