Condition for the sum of two fractions to be irreducible

Question

Let $\frac{a}{b}$ and $\frac{e}{f}$ be two rationals where all parameters are positive integers and are in their lowest terms. Let $\gcd(b,f) = g$ so that $b = gB$, $f = gF$ and $\gcd(B,F) = 1$. As an intermediate step in one of the problems I am solving, I am interested in the following.

Weak Claim: If $af+eb$ is square-free then $\gcd(aF+eB, gFB) = 1$.

Strong Claim: If $ \min (\gcd(g,\frac{b}{g}),\gcd(g,\frac{f}{g})) > 1$ then $\gcd(aF+eB, gFB) = 1$.

Questions

1. Are these condition correct? My derivation is given below.
2. What are the necessary and sufficient conditions?

My solution

Let $\gcd(b,f) = g$ so that $b = gB$, $f = gF$ and $\gcd(B,F) = 1$. After eleminating $g$ from the numerator and the denominator, we get

$$ \frac{1}{g}\Big(\frac{a}{b} + \frac{e}{f}\Big) = \frac{aF + eB}{gBF} $$

We want to know if $\gcd(aF + eB, gBF) > 1$. Let us look at the different possibilities.

Case 1: $aF + eB$ and $B$ have a common factor. In this case since $B|eB$ hence we must have $\gcd(B,aF) > 1$. But by $\gcd(B,F) = 1$ hence we must have $\gcd(B,a) > 1$ which is impossible since $\frac{a}{b}$ is in its lowest terms. Hence this case is invalid.

Case 2: $aF + eB$ and $F$ have a common factor. In this case since $F|aF$ hence we must have $\gcd(F,eB) > 1$. But $\gcd(B,F) = 1$ hence we must have $\gcd(F,eB) > 1$ which is impossible since $\frac{e}{f}$ is in its lowest terms. Hence this case is also invalid.

Case 3: $aF + eB$ and $g$ have a common factor. Clearly $\gcd(g,a) = \gcd(g,e) = 1$ otherwise the fractions $\frac{a}{b}$ and $\frac{e}{f}$ will not be in their lowest terms. Also both $\gcd(g,B)$ and $\gcd(g,F)$ cannot be simultaneously $> 1$ otherwise $\gcd(b,f) > g$. Now

$$ \gcd(g, aF + eB) > 1 \implies \gcd(g^2, agF + egB) = \gcd(g^2, af + eb) > 1 $$

Hence $af+eb$ cannot be square free. This proves the weak claim.

We have two sub-cases:

Subcase 3.1 $aF + eB$ and $g$ have a common factor and $\gcd(g,F) > 1$. In this case since $g$ has a common factor with both $F$ and $aF + eB$ it must have a common factor with $eB$ which is impossible. Hence this case is invalid.

Subcase 3.2 $aF + eB$ and $g$ have a common factor and $\gcd(g,B) > 1$. In this case since $g$ has a common factor with both $B$ and $aF + eB$ it must have a common factor with $aF$ which is impossible. Hence this case is also invalid.

Hence the only possibility for $\gcd(aF + eB, gBF) > 1$ is when neither $F$ nor $B$ has a common factor with $g$ but $aF + eB$ always has a common factor with $g$.

Answer 1

The weak claim is correct but the strong claim is not. Here are streamlined arguments to that effect. We repeatedly use these facts (all variables are integers):

• If $\gcd(x,y)=1$ and $\gcd(x,z)=1$, then $\gcd(x,yz)=1$;
• $\gcd(x,y)=\gcd(x+wy,y)$.

Now note that:

• $\gcd(F,B)=1$ by construction, and $\gcd(F,e)=1$ follows from $\gcd(f,e)=1$ since $F\mid f$. Therefore $\gcd(F,eB)=1$, which implies that $\gcd(F,aF+eB)=1$.
• Similarly, $\gcd(B,F)=1$ by construction, and $\gcd(B,a)=1$ follows from $\gcd(b,a)=1$ since $B\mid b$. Therefore $\gcd(B,aF)=1$, which implies that $\gcd(B,aF+eB)=1$.

It follows that $\gcd(aF+eB,BF)=1$ and hence that $\gcd(aF+eB,gBF)=\gcd(aF+eB,g)$.

Suppose that $af+eb$ is squarefree. Setting $h=\gcd(aF+eB,g)$, we have $h\mid(aF+eB)$ and $h\mid g$ by definition, and therefore $h^2\mid(aF+eB)g = (af+eb)$; since $af+eb$ is squarefree, we must have $h=1$, proving the weak claim.

A counterexample to the strong claim is $$ \frac ab=\frac5{12\times29}, \quad \frac ef=\frac7{18\times29}, $$ for which $g=6$ and $B=2$ and $F=3$, and $\min (\gcd(g,\frac{b}{g}),\gcd(g,\frac{f}{g})) = \min (2,3) = 2 > 1$, but $\gcd(aF+eB, gFB) = 29$.

In general, I don't know that there's going to be any simpler necessary and sufficient condition than $\gcd(aF+eB,g)=1$ itself, as the above example illustrates.