Possible values of $\gcd(a+b, a\times b)$

Question

Main Question: Let $N \in \mathbb{N}$.

What are the possible values of $\gcd(a+b, a\times b)$ given that $\gcd(a,b) = N$?

Fact 0. If $\gcd(a,b) = N$, then $N \leq \gcd(a+b, a\times b) \leq N^2$.

More substantially:

Fact 1. (D. Fischer below).

Let $N \in \mathbb{N}$. For every $d$ a divisor of $N$, there are natural numbers $a, b$ for which $\gcd(a,b) = N$ and $\gcd(a+b, a\times b) = dN$. Moreover, these are precisely the values that $\gcd(a+b, a\times b)$ can take on provided that $\gcd(a,b) = N$.

Answer 1

Let us write $a = N\cdot \alpha$ and $b = N\cdot \beta$, where $N = \gcd(a,b)$. Then we have $\gcd(\alpha,\beta) = 1$ (for $N\cdot \gcd(\alpha,\beta)$ is a common divisor of $a$ and $b$). Further,

$$\gcd(a+b,a\cdot b) = \gcd(N(\alpha+\beta),N^2\alpha\beta) = N\cdot \gcd(\alpha+\beta,N\alpha\beta).$$

Now we note that $\gcd(\alpha+\beta,\alpha\beta) = 1$ (why?), and from that deduce

$$\gcd(a+b,ab) = N\cdot \gcd(\alpha+\beta,N).$$

It follows that $\gcd(a+b,ab)$ is always of the form $N\cdot d$, where $d$ is a divisor of $N$.

It remains to see that every $N\cdot d$ where $d\mid N$ can be such a $\gcd$. For $d = 1$, take $\alpha = 1$ and $\beta = N$, for $d > 1$, take $\alpha = 1$ and $\beta = d-1$.