Reduced fractions with integer sum have equal denominators

Question

Prove for positive integers $a,b,c$ and $d$ (where $b \neq 0$ , $d \neq 0$ and $b \neq d$), if $\gcd(a,b) = \gcd(c,d) = 1$, then $\frac{a}{b} + \frac {c}{d}$ is not an integer.

I understand that if $\gcd(a,b)$ and $\gcd(c,d) = 1$, at least one number in each pair is a prime or is $1$. As for after that, I'm totally stumped, could I get some tips, clues, help?

Answer 1

$$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}.$$ If this is an integer then in particular $b \mid ad \implies b\mid d$, and viceversa $d\mid b$. This condition implies that $b=\pm d$. But they are both positive, therefore they have to be equal. Contradiction.

Answer 2

Conceptually, it follows immediately from URF = $\overbrace{\small \text{uniqueness of reduced fractions}}^{\color{#c00}{\large \text{"unique fractionization"}}}$

$$(a,b)\!=\!1\!=\!(\color{#0a0}{c,d}), \ \overbrace{\dfrac{a}b + \dfrac{c}{d}}^{\large =\ n\ \in\ \Bbb Z}\, \Rightarrow\,\dfrac{a}{\color{#c00}b} \!=\! \dfrac{dn\!-\!c}{\color{#c00}d}, \ \overbrace{\small \text{both in least terms}}^{\small\textstyle{p\mid \color{#0a0}d,dn\!-\!c\Rightarrow p\mid\color{#0a0} c}} \,\overset{\rm URF}\Rightarrow\, \color{#c00}{b = d}\qquad\qquad\quad$$

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Remark $ $ Below is a typical application of this basic result

Theorem $\ $ If $\,q,r\in\Bbb Q\,$ then $\, q+r,\, qrs\in\Bbb Z\,\Rightarrow q,r\in\Bbb Z,\,$ if $\,\color{#c00}{{\rm squarefree}\ s\in \Bbb Z}\,$ (e.g. $\, s\!=\!1)$

Proof $\, $ By $\,q+r\in\Bbb Z\,$ they have equal least denominator $\,d\,$ so $\,qrs\in \Bbb Z\Rightarrow \color{#c00}{d^2\mid s\Rightarrow d\!=\!1}\,$

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Further exploiting innate $\rm\color{#0a0}{symmetry}$, this generalizes as below (for any number of $\,a_i)$

Theorem' $ $ If squarefree $\,q\in\Bbb Z,\,$ $\,a_i\in \Bbb Q,\, $ $\, e_i\,$ are elementary $\rm\color{#0a0}{symmetric}$ polynomials in $\,a_i,\,$ e.g. $\,(x\!-\!a_1)(x\!-\!a_2)(x\!-\!a_3) = x^3\!-e_1\:\! x^2 + e_2\:\! x - e_3,\,$ then all $\,\color{#0a0}{q^i e_{i+1}\in\Bbb Z}\,\Rightarrow\,$ all $\,a_i\in\Bbb Z.\,$

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More generally see How much can a sum of fractions reduce?

Answer 3

This is a expansion on arturocanguro's answer.

Bezout's Identity says that we have $x,y,z,w$ so that $ax+by=1$ and $cz+dw=1$. If $$ \frac ab+\frac cd=\frac{ad+bc}{bd}\in\mathbb{Z}\tag{1} $$ then $$ bd\mid ad+bc\implies b\mid ad\implies b\mid adx=d(1-by)\implies b\mid d\tag{2} $$ and $$ bd\mid ad+bc\implies d\mid bc\implies d\mid bcz=b(1-dw)\implies d\mid b\tag{3} $$ $(2)$ and $(3)$ imply that $b=d$.