Don't go all mediant on me!

$\dfrac{a+c}{b+d}$ = $\dfrac{a}{b+d}$ + $\dfrac{c}{b+d}$

If a,b,c,d are positive integers then $\dfrac{a}{b}$ > $\dfrac{a}{b+d}$ and $\dfrac{c}{d}$ > $\dfrac{c}{b+d}$ .

This implies

$\dfrac{a}{b}$ + $\dfrac{c}{d}$ > $\dfrac{a}{b+d}$ + $\dfrac{c}{b+d}$

and

$\dfrac{a}{b}$ + $\dfrac{c}{d}$ > $\dfrac{a+c}{b+d}$

$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{a + c}{b + d} \Longrightarrow \dfrac{ad + bc}{bd} = \dfrac{a + c}{b + d} \Longrightarrow (ad + bc)(b + d) = (a + c)bd \Longrightarrow adb + ad^{2} + b^{2}c + bcd = abd + cbd \Longrightarrow ad^{2} +b^{2}c = 0$ .

But as $a,b,c,d \gt 0$ we can only have $ad^{2} + b^{2}c \gt 0$ , in which case the answer is $\boxed{\text{Yes}}$ , there are no positive integer solutions to the given equation.

Note: We can have a solution if precisely one of $a,c$ is negative, e.g., $(a,b,c,d) = (2,5,-8,10)$ or $(2,3,-8,6)$ .

Let us assume that $\dfrac ab + \dfrac cd = \dfrac{a+c}{b+d}$ is possible for positive integers and derive a contradiction.

First we simplify the LHS by adding the factions (the correct way!) to get

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

Now 'cross multiply'

$(ad+bc)(b+d)=abd + cbd$

Now multiply out the LHS, and some very obvious 'cancelling' gives

$ad^2 + cb^2 = 0$

Since this is impossible in positive integers, we can invoke the principle of proof by contradiction and say that the bold expression is false, so

$\dfrac ab + \dfrac cd = \dfrac{a+c}{b+d}$ is NOT possible for positive integers.

Suppose that there are positive integers $a, b, c. d$ such that $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$ . We know that $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$ . Then, by equivalence of fractions, $ad+bc=a+c$ $bd=b+d$ The second equation only allows for one solution: $b=d=2$ . ( Proof: Suppose that there are two positive integers $b,d$ such that $b+d=bd$ . Subtract $d$ from both sides and divide by $b-1$ to obtain $\frac{b}{b-1}=d$ ; simplifying the left hand side yields $1+\frac{1}{b-1}=d$ . Now, $1+\frac{1}{b-1}\in \mathbb{Z}\implies\frac{1}{b-1}\in \mathbb{Z}\implies(b-1)|1$ : this requires $b-1=1\implies b=2$ . Substituting this into our original equality gives $d+2=2d\implies d=2$ . For more solutions, see here ) Substituting these values of $b,d$ into the first equation gives $2a+2c=a+c\implies2(a+c)=a+c\implies a+c=0$ However, as $a$ and $c$ are both positive, this cannot be true. Thus, the answer is yes : there are no positive integers $a,b,c,d$ such that $\frac{a}{b}+\frac{c}{d}=\frac{a+b}{c+d}$ .

a/b+c/d=(a+c)/(b+d) Reduces to (ad+cb)*(b+d)=abd+cbd Reduces to (b/d)^2=-(a/c). ! Impossible

Please note that 0 is positiv too! So we have one solution: a=c=0.

Nothing won’t work bcuz that isn’t a rule for fractions

Multiply both sides with (b+d) and you get a+ad/b+c+cb/d=a+c or, ad/b+cb/d=0, which is not possible for positive a,b,c,d.

AFSOC there exist such (nonzero) positive integers. Just take the partial of both sides wrt a or c. WLOG, choose c and then we find after a little algebra that d=b+d, so b is zero, which is a contradiction.

Multiply both sides by a + c, and the solution is immediately clear