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??
 A: The equation means $ad+bc=bd k$. It follows that $b$ divides $ad$, hence also $d$. The rest is for you ...
A: $$\frac{a}{b}+\frac{c}{d}=k \text{ with } (a,b)=1, (c,d)=1$$
$$\Rightarrow ad+bc=kbd$$
$$>ad=kbd-cb \Rightarrow ad=b(kd-c) \Rightarrow b|ad \xrightarrow{(a,b)=1} b|d (1)$$
$$>cb=kbd-ad \Rightarrow cb=d(kb-a) \Rightarrow d|cb \xrightarrow{(c,d)=1} d|b (2)$$
$$(1) \Rightarrow b \leq d$$
$$(2) \Rightarrow d \leq b$$
So $b=d$.
A: 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.
