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

Are these condition correct? My derivation is given below.
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 $g$.

Greg Martin · Answer

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 $Fmid f$. 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 $Bmid b$. 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 $hmid(aF+eB)$ and $hmid g$ by definition, and therefore $h^2mid(aF+eB)g = (af+eb)$; since $af+eb$ is squarefree, we must have $h=1$, proving the weak claim.
A counterexample to the strong claim is
$$
frac ab=frac5{12times29}, quad frac ef=frac7{18times29},
$$
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.

Condition for the sum of two fractions to be irreducible

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