Rationalising fractions proof 
Suppose that a, b, c and d are positive integers and c is not a square.
Given that $$\frac  a{b+\sqrt c}+\frac d{\sqrt c}\in \mathbb Q$$
prove that $b^2d = c(a + d)$

What I did was try and find the value of b in terms of a,c,d then sub it in, by rationalising the initial equation, then equating them, but ended up with a mess of numbers. I think I’m missing something really simple, any help would be appreciated. Thank you!
 A: So $$\frac  a{b+\sqrt c}+\frac d{\sqrt c} = q,\;\;\;\;\; q\in \mathbb Q$$
Let $x=\sqrt{c}\notin \mathbb Q$, then $$\frac  a{b+x}+\frac d{x} = q\;\;\;\;/\cdot x(b+x)$$
$$ax+bd+dx= cq+bxq$$
so $$ \boxed{x(a+d-bq) = cq-bd}$$
Now if $a+d\ne bq$ then $$ x  = {cq-bd\over a+d-bq}$$ which means that $x$ is rational. A contradiction. So $a+d=bq$ and then also $cq=bd$. Now the conclusion.
A: You start by making it a single fraction:
$$\frac pq = \frac{a}{b+\sqrt c} + \frac d{\sqrt c} = \frac{(d+a)\sqrt c + bd}{b\sqrt c + c}$$
with $p, q\in \mathbb Z$ and $q\neq 0$. Then rewrite as
$ p(b\sqrt c + c) = q((d+a)\sqrt c + bd)$ and finally isolate $\sqrt c$ to get 
$$ (pb-q(d+a))\sqrt c = qbd-pc.$$ 
Since $\sqrt c$ is irrational, both sides of the equation must be rational, and $(pb+q(d+a))$ is rational, we must have $pb-q(d+a)=0$ and $qbd-pc=0$. Rearranging the first equation yields $p=q(d+a)/b$ and inserting this into the second equation, we get $q(bd-(d+a)c/b)=0$ Since $q\neq 0$, we have $bd-(d+a)c/b=0$, which yields the conclusion $b^2d=(d+a)c$.
