0

If $\gcd(a,b) = \gcd(c,d) = 1$, then $\frac{a}{b} + \frac {c}{d}$ an integer implies $b = d,\,$ for $a,b,c,d\in \Bbb N_{> 0}$

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?

Bill Dubuque
  • 282,220
user216004
  • 37
  • 2
  • 3

3 Answers3

4

$$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}.$$ 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.

  • The proof essentially "inlines" a proof of the uniqueness of lowest term denonimators (vs. invoking that theorem by name). See my answer for that viewpoint. – Bill Dubuque Feb 17 '15 at 02:31
1

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 $b=d$.

robjohn
  • 353,833
-3

Conceptually it follows immediately by $\:\!\overbrace{\small \text{uniqueness of reduced fractions}}^{\color{black}{\large \text{"unique fractionization"}}}$ $\, \color{#0af}{\small \rm (URF)},\,$ viz.

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

Remark $ $ A typical application of this basic result is:

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 $\,qrs\in \Bbb Z\Rightarrow \color{#c00}{d^2\mid s\Rightarrow d\!=\!1}\,$

Further exploiting innate $\rm\color{#0a0}{symmetry}$, the prior 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 $\,a_i\in\Bbb Z.\,$

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

Bill Dubuque
  • 282,220
  • 1
    The three downvotes here almost surely have nothing to do with math - they arrived quickly after I used this thread as a target for dupe closure. In any case, exploiting uniqueness of reduced fractions and symmetry is the best way to do problems like this, and one should master such methods early on (it's just a fractional variant of the fundamental uniqueness of prime factorizations, and (vengeful) downvotes will never destroy the innate beauty of such) – Bill Dubuque Jan 14 '23 at 21:14