Are there any examples of $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$?

Question

I am writing a short discussion of Farey numbers and was wondering if there are any examples of when the mediant function is ever actually equal to the sum of the two fractions in the usual sense? (Not to produce Farey numbers, I just thought it might be an amusing way to introduce Farey addition).

Explicitly: Are there any examples of fractions $\frac{a}{b}$ and $\frac{c}{d}$ where $$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$ for positive integers?

Currently I have found the example $$\frac{1}{1}+\frac{1}{i} =\frac{1+1}{1+i}$$ if we remove the integers requirement but it would be nice to find a case that didn't involve complex numbers though!

Answer 1

Let $k=\frac{a}{c}$ and $l=\frac{b}{d}$. Divide numerators on both sides by $c$ and denominator on both sides by $d$. Then you want to show, $$\frac{k}{l}+1=\frac{k+1}{l+1}$$ $$\Longrightarrow (k+l)(l+1)=(k+1)(l)$$ $$\Longrightarrow l^2+k=0$$

This is the required necessary and sufficient condition.

Observe that you don't need to do calculations to get that there are no positive solutions, Cauchy Schwarz inequality will do. $$\frac{a}{b}+\frac{c}{d}\geq \frac{(\sqrt{a}+\sqrt{c})^2}{b+d}>\frac{a+c}{b+d}$$ Last inequality holds because if either of $a$ or $c$ is zero, then it cannot hold trivially, so $ac>0$.

Answer 2

We can work it out as ordinary algebra. Render the LHS as

$\dfrac{ad+bc}{bd}$

and then put both sides over the common denominator $bd(b+d)$. Thereby the corresponding numerators are equal:

$abd+ad^2+b^2c+bcd=abd+bcd$

$ad^2+b^2c=0$

There are no solutions with all variables positive but, for instance, $d=b\not=0$ and $c=-a$ would satisfy this last equation and trivially give zero for the original sum.

Answer 3

Seems like straight forward algebra manipulation

$\frac ab + \frac cd = \frac {a+c}{b+d}$

$(ad+bc)(b+d) = bd(a+c)$

$abd + b^2c + ad^2+bcd = abd +bcd$

$b^2c +ad^2 = 0$

$b^2c = -ad^2$

Wolog $b,d$ are positive. Wolog $c$ is positive an $a$ negative (I assume you don't want $a=c =0$.)

If we let $c = c'(M^2)$ where $c'$ is square free then $a =-c'N^2$ for some $n$. We can shuffle the square factors of $b,c,M,N$ around anyway we want.

$c =12$ and $a=-75$ and $b=5$ and $d=2$ would be a solution.

$\frac {-75}5 + \frac {12}{2} =\frac {-75+12}{5+2}=-9$.