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

Question

Please help me to find $a,b,c,d$ such that $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$ ? I take $a,b$ for example $2,3$ to find $c,d$ but I get stuck on this. I know that is diophantine equation , but I am unable to find suitable $a,b,c,d$ . $$a,b,c,d \in \mathbb{Z} ,a,c\neq 0$$ Thanks in advance.

Answer 1

This is the best I could get, hope it helps :)

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

For any $h,\;k,\;j\in\mathbb{Z},\;j\ne 0,h\ne-1$

Let $\left\{a= k,b= j,c=- h^2 -k,d= h j\right\}$

we get $\dfrac{k}{j}-\dfrac{h k}{j}=\dfrac{k-h^2 k}{h j+j}$

indeed $\dfrac{k-hk}{j}=\dfrac{k(1+h)(1-h)}{j(h+1)}$

and finally $\dfrac{k-hk}{j}=\dfrac{k(1-h)}{j}$

Edit.

I am not sure that these are ALL the solutions

Answer 2

Hint: $$ \frac ab+\frac cd-\frac{a+c}{b+d}=\frac{ad^2+b^2c}{bd(b+d)} $$