# For which integers $a,b,c,d$ does $\frac{a}{b}+\frac{c}{d} = \frac{a+c}{b+d}$?

For which $a,b,c,d \in \mathbb{Z}$ does $\frac{a}{b}+\frac{c}{d} = \frac{a+c}{b+d}$?

This is actually the question I meant to ask in a previous question that I asked here.

What about $a,b,c,d \in \mathbb{R}?$

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

We will keep careful track of which quantities may be zero. To start, we know that $b, d, b+d \neq 0$ because they are in the denominator.

Let's multiply both sides by $b\cdot d \cdot (b+d)$:

\begin{align*}ad(b+d) + bc(b+d) &= (a + c)bd\\ abd + ad^2 + b^2c + bcd &= abd + bcd\\ abd + ad^2 + b^2c + bcd - abd - bcd &= 0\\ \fbox{ad^2 + b^2 c = 0} \end{align*}

Here's an example assignment that satisfies this equation: \begin{align*}a &= 5\\b&=3\\c&=-20\\d&=6\\\end{align*}

We can parameterize this set of solutions with three parameters, $\alpha, \beta_1, \beta_2 \in \mathbb{R}$ and three minimal constraints $\beta_1 \neq 0, \beta_2 \neq 0$, and $\beta_1 \neq \beta_2$. For any such triple, a solution is given by \begin{align*} a &\equiv \alpha\\ b &\equiv \frac{1}{\beta_1}\\ c &\equiv -\alpha\frac{\beta_1^2}{\beta_2^2}\\ d &\equiv -\frac{1}{\beta_2}\\ \end{align*}

Conversely, given any solution $\langle a, b, c, d\rangle$ with $\frac{a}{b} + \frac{c}{d} = \frac{a+c}{b+d}$, we can solve uniquely for parameters $\langle \alpha, \beta_1, \beta_2\rangle$ meeting these constraints.

For fun, here is a nomogram that allows you to find such collections of numbers:

You can draw any line passing through, say, the B and C axes, and any line passing through the A and D axes. If the two lines meet at the same point on the central vertical axis, then the corresponding points $(a, b, c, d)$ satisfy the equation.

In the picture, the dashed lines show an example. Here, we see that the equation is satisfied when $a = 4$, $b=3$, $c=-1$, and $d=1.5$.

Also, if you specify any four of the following five variables—a value for $a$, a value for $b$, a value for $c$, a value for $d$, a spot on the central axis— you can "solve" for the remaining point.

• A parametrization of all the solutions would be nice however :-p. – YoTengoUnLCD Jul 17 '16 at 5:55
• How did you draw the nomogram? Did you use pynomo.org ? – Rodrigo de Azevedo Jul 17 '16 at 12:56
• @RodrigodeAzevedo Yes! I used Pynomo. – user326210 Jul 17 '16 at 17:44
• Looks like you also fell under Ron Doerfler's spell ;-) – Rodrigo de Azevedo Jul 17 '16 at 17:56

If you clear the fractions you get $abd+ad^2+b^2c+bcd=abd+bcd$ or $$ad^2+b^2c=0$$ As long as none of the denominators are zero, you are set.

Multiply both sides by $bd(b+d)$ and you get:

\begin{align}ad(d+b) + cb (d+b) = bd(a+c)&\iff\\ ad^2 + abd + cbd + cb^2 = b d(a+c)&\iff\\ ad^2 +cb^2 = 0&\iff\\ ad^2=-cb^2 \end{align}

Given $b,c$, you need $d^2$ to be a factor of $cb^2$ and then $a=-\frac{cb^2}{d^2}$.

Given non-zero integers $m,n,p,q$, with you have:

$$(a,b,c,d)=(pm^2,mq,-pn^2,nq)$$

This gives all solutions, excluding the case where $m=-n$.

The only solution with $\gcd(a,b)=1$ and $\gcd(c,d)=1$ are the cases where $m,n=\pm 1$, and then you are dealing with the solutions:

$$\frac{a}{b}+\frac{-a}{b}=\frac{a+(-a)}{b+b}$$