# If the sum of two irreducible fractions is an integer, then the denominators are equal

I have to show the following:"If the sum of two irreducible fractions with positive denominators is an integer, then the denominators are equal." $$\frac{a}{b}+\frac{c}{d}=k, \text{ where k an integer }$$ Since the fractions are irreducible, $(a,b)=1$ and $(c,d)=1$. Right? But how can I continue??

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$ad+bc=kbd$, what does it say about divisibility of $d$ by $b$ and $b$ by $d$? – Marcin Łoś Feb 24 '14 at 0:03
"Fractions" can be tricky in number-theoretic reasoning. So an almost automatic first step is to get rid of them, in this case by multiplying through by $bd$. – André Nicolas Feb 24 '14 at 0:05
@MarcinŁoś André Nicolas Could you check if my answer is correct? – Mary Star Feb 24 '14 at 0:19

The equation means $ad+bc=bd k$. It follows that $b$ divides $ad$, hence also $d$. The rest is for you ...

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you must mean $ad+bc=bdk$ :-) – robjohn Feb 24 '14 at 0:08
@MartinBrandenburg Could you check if my answer is correct? – Mary Star Feb 24 '14 at 0:18

Here is a proof using Bezout's Identity.

Bezout's Identity says that since $(a,b)=1$ $$ax+by=1\tag{1}$$ and that since $(c,d)=1$ $$cu+dv=1\tag{2}$$ Multiplying your equation by $bd$ gives $$ad+bc=bdk\tag{3}$$ Multiply $(1)$ by $d$ to get $$\color{#C00000}{adx}+bdy=d\tag{4}$$ Multiply $(3)$ by $x$ to get $$\color{#C00000}{adx}+bcx=bdkx\tag{5}$$ Solving $(5)$ for $adx$ and plugging that into $(4)$ yields $$d=b(dy+dkx-cx)\tag{6}$$ Multiply $(2)$ by $b$ to get $$\color{#C00000}{bcu}+bdv=b\tag{7}$$ Multiply $(3)$ by $u$ to get $$adu+\color{#C00000}{bcu}=bdku\tag{8}$$ Solving $(8)$ for $bcu$ and plugging that into $(7)$ yields $$b=d(bku-au+bv)\tag{9}$$ Equations $(6)$ and $(9)$ should finish things off.

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Ok!Thank you for your answer!!! – Mary Star Feb 24 '14 at 0:42
It is along the same lines as the others, but I think of Bezout as a low level idea. – robjohn Feb 24 '14 at 0:45

$$\frac{a}{b}+\frac{c}{d}=k \text{ with } (a,b)=1, (c,d)=1$$ $$\Rightarrow ad+bc=kbd$$ $$>ad=kbd-cb \Rightarrow ad=b(kd-c) \Rightarrow b|ad \xrightarrow{(a,b)=1} b|d (1)$$ $$>cb=kbd-ad \Rightarrow cb=d(kb-a) \Rightarrow d|cb \xrightarrow{(c,d)=1} d|b (2)$$ $$(1) \Rightarrow b \leq d$$ $$(2) \Rightarrow d \leq b$$ So $b=d$.

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Yes, that seems correct. – Marcin Łoś Feb 24 '14 at 0:20
Ok!Thanks a lot!!! :-) – Mary Star Feb 24 '14 at 0:20