Find integers $p$ and $q$ such that ${\sqrt{5} \over \sqrt{5} - 2} = {p + q\sqrt{5}}$ My child brought home a problem that I'm stumped with helping them with. It's basically finding integers p and q such that:
$${\sqrt{5} \over \sqrt{5} - 2} = {p + q\sqrt{5}}$$
The reason I'm stumped is because I seem to remember from many years ago that you need two distinct equations to solve for two variables. With one equation, all you can do is define one variable in terms of the other. Obviously, I can do that bit with:
$$p = {\sqrt{5} \over \sqrt{5} - 2} - q\sqrt{5}$$
then further simplifying from there.
However, I'm pretty certain that the fact the solutions are integers are meant to limit the solution space somehow but I'm unsure how to proceed.
Any advice would be gratefully accepted. I have no intention passing the solution on to my child but, if I could get some information on how to proceed (or even the outright solution should you wish to be generous), I can at least nudge them in the right direction.
 A: Hint: Multiply the fraction by
$$1=\frac{\sqrt{5}+2}{\sqrt{5}+2}.$$
Full solution:
Applying the hint shows that
\begin{eqnarray*}
 p+q\sqrt{5}&=&\frac{\sqrt{5}}{\sqrt{5}-2}\\
 &=&\frac{\sqrt{5}}{\sqrt{5}-2}\frac{\sqrt{5}+2}{\sqrt{5}+2}\\
 &=&\frac{\sqrt{5}(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)}\\
 &=&\frac{5+2\sqrt{5}}{\sqrt{5}^2-2^2}\\
 &=&5+2\sqrt{5},
 \end{eqnarray*}
and so $p=5$ and $q=2$.
A: Just cross-multiply.
${\sqrt{5} \over \sqrt{5} - 2} = {p + q\sqrt{5}}
$
becomes
$\sqrt{5}
=(\sqrt{5} - 2)(p + q\sqrt{5})
=\sqrt{5}(p-2q)-2p+5q
$.
If $p$ and $q$ are rational,
we must have
$p-2q=1$ and $2p=5q$
so
$5q=2(2q+1)
=4q+2
$
so
$q=2, p=5$.
Check:
$(5+2\sqrt{5})(\sqrt{5}-2)
=\sqrt{5}(\sqrt{5}+2)(\sqrt{5}-2)
=\sqrt{5}
$.
A: Multiply both the numerator and denominator by ${\sqrt{5}}+2$
We get $\frac{\sqrt{5}*(\sqrt{5}+2)}{(\sqrt{5}-2)*(\sqrt{5}+2)}$
Then the denominator can be simplified as $\sqrt{5}\sqrt{5}+2\sqrt{5}-2\sqrt{5}-4$
Then, after expanding $\frac{5+2\sqrt{5}}{1}$ is the result 
Then we get $p=5, q=2$
