How much can a sum of fractions reduce?

Question

Assume $x/a$ and $y/b$ are reduced, i.e. $\,\gcd(x,a) = 1 = \gcd(y,b)$.

$${\rm if}\ \ \dfrac{x}a+\dfrac{y}b=\dfrac{z}c\ \ {\rm does}\ \dfrac{ab}{\gcd(a,b)^2}\mid c\ ?\qquad$$

Answer 1

Yes, $\,\rm c\,$ must be divisible by $\,\rm \dfrac{ab}{(a,b)^2} = \dfrac{lcm(a,b)}{(a,b)}.\,$ Below are a few proofs.

$\rm(1)\ \ abz = bcx + acy\ \Rightarrow\ a,b\ |\ ac,bc\ \Rightarrow\ lcm(a,b)\ |\ (a,b)\ c\ \Rightarrow\ lcm(a,b)/(a,b)\ |\ c\ \ $ QED

Note $\rm\ lcm(a,b)/(a,b) =\: ab/(a,b)^2\ $ by the LCM $\cdot$ GCD law $\rm\ lcm(a,b)\ (a,b)\ =\ a\:b\:.$

We used lcm & gcd universal laws: $\rm\ a,b\ |\ x,y \iff lcm(a,b)\ |\ x,y \iff lcm(a,b)\ |\ (x,y)\:. $

Below are two more proofs, the first a bit more conceptual and more detailed.

$(2)\ $ Suppose $\rm\ c\ $ is a denominator for $\rm\ \dfrac{x}a\: +\: \dfrac{y}b\ $ where $\rm\ (a,x) = 1 = (b,y)\:.\ $ Then we infer

$\rm\phantom{\quad\Rightarrow\quad}\: c\ \left(\dfrac{x}a\: +\: \dfrac{y}b\right) =\: z \in \mathbb Z\ \ \Rightarrow\ \ abz\ =\ cbx + cay\ \Rightarrow\ \ a\ |\ cbx,\ \ b\ |\ cay$

$\rm\quad\Rightarrow\quad c\ \left\{\dfrac{b}a,\ \dfrac{a}b\right\}\subset\: \mathbb Z\ \ $ by $\rm\ (a,x) = 1,\ \ a\ |\ cbx\ \Rightarrow\ a\ |\ cb\:.\ $ Similarly $\rm\ b\ |\ ca\:.$

$\rm\quad\Rightarrow\quad c\ \left\{\dfrac{\beta}\alpha,\ \dfrac{\alpha}\beta\right\}\subset\: \mathbb Z\ \ $ by cancelling out $\rm\ (a,b),\:$ with $\rm\ \alpha = a/(a,b),\ \ \beta = b/(a,b)\:.\ $

$\rm\quad\Rightarrow\quad c\ \left\{\dfrac{1}\alpha,\ \dfrac{1}\beta\right\}\subset\: \mathbb Z\ \ $ by $\rm\ (\alpha,\beta) = 1,\ \ \alpha\ |\ c\:\beta\ \Rightarrow\ \alpha\ |\ c\:.\ $ Similarly $\rm\ \beta\ |\ c\:. \ $

$\rm\quad\Rightarrow\quad c\ \:\left(\dfrac{1}\alpha\ \:\dfrac{1}\beta\right)\ \in\ \mathbb Z\ $ by $\rm\: \alpha,\beta\ |\ c\ \Rightarrow\ lcm(\alpha,\beta)\ |\ c\:.\: $ $\rm lcm(\alpha,\beta) = \alpha\: \beta\ $ by $\rm\: (\alpha,\beta) = 1. $

$(3)\ $ Finally, here is another proof based upon squaring a gcd and applying basic gcd laws.

$\rm\ \ abz\ =\ bcx\: +\: acy\ \ \Rightarrow\ \ a\ |\ bcx\ \ \Rightarrow\ \ a\ |\ bc\ $ by $\rm\ (a,x) = 1\:.\ $ Similarly $\rm\ b\ |\ ac\:.$

$\rm Thus\quad \rm a\ |\ bc,\ b\ |\ ac\ \ \Rightarrow\ \ ab\ |\ a^2c,\ abc,\ b^2 c $

$\rm \phantom{Thus\quad \rm a\ |\ bc,\ b\ |\ ac\ \ } \Rightarrow\ \ ab\ |\ (a^2c,\ abc,\ b^2 c)\ =\ (a,b)^2\:c\quad\ $ QED

Note that the above proof uses only basic gcd laws (associative, commutative, distributive, etc) therefore it holds true in any GCD domain. Below is some further detail using said laws

$$\rm (a,b)\,(a,b)\,=\,(a(a,b),b(a,b))\,=\,((a^2,ab),(ab,b^2))\,=\,(a^2,ab,ab,b^2)\,=\,(a^2,ab,b^2) $$

For further discussion of arithmetic using such laws see here.

Remark $\ $ Further $\rm\ c \mid \dfrac{ab}{\gcd(a,b)}\ $ if $\,z/c\,$ is reduced, i.e. $\,\rm (c,z)=1,\,$ by $$\rm\ c\:(a,b)\mid c\ (b x + a y) = zab\ \Rightarrow\ c \mid \dfrac{zab}{(a,b)}\ \Rightarrow\ c\mid \dfrac{ab}{(a,b)}\ \ {\rm by}\ \ (c,z)=1$$

Combining the above we obtain the following

Theorem $\ \ \rm (x,a)\!=\!(y,b)\!=\!(z,x)\!=\!1,\ \ \dfrac{x}a+\dfrac{y}b = \dfrac{z}c\ \Rightarrow\ \dfrac{ab}{\gcd(a,b)^2}\, {\Large \mid }\ c\ {\Large\mid}\ \dfrac{ab}{\gcd(a,b)}$

Answer 2

Yes, that is true. To see this, let $g = \gcd(a,b)$ so we can rewrite the sum as $$ \dfrac x {ga} + \dfrac y { gb} = \dfrac z c $$ where $\gcd(a, b) = \gcd(a,x) = \gcd(b,y) = 1$. Then $$ \dfrac{ bx + ay } {gab} = \dfrac z c. $$ But $\gcd(a,bx) = 1$ so the factor of $a$ cannot cancel, and similarly neither can the factor of $b$. Thus $$ ab = \dfrac{ ga \cdot gb }{g^2} | c $$ as desired.