Real number multiplicative inverses expressed in another form I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.
I understand that for some rational numbers $c,d$ we must have:
$$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$
I was able to answer for the multiplicative inverse of $2 +\sqrt{3}$. We find that $1 = (2 +\sqrt{3})(c + d\sqrt{3})$ where $c = 2$ and $d = -1$. However this seems to be related to the original $2 +\sqrt{3}$; and in the problem at hand this is not the case. So I'm rather confused.
How could i go about solving this? I'd like steps without the answer IF possible; moreover, how could i prepare for more general questions in this form?

We have $a+b\sqrt c$, where $a,b,c$ are known, and $c$ is not a perfect square, and we want to find $x$ and $y$ to make $$(x +y\sqrt c)(a + b\sqrt c) = 1.$$
Multiplying out, we get $$(ax + bcy) + (ay+bx)\sqrt c = 1.$$
The first term, $ax+bcy$, is an integer and its sum with $(ay+bx)\sqrt c$  is $1$, another integer, so $(ay+bx)\sqrt c$ must be an integer as well. The only way this can happen is if $ay+bx=0$, in which case  we must have $ax+bcy = 1$.  So we have two equations in $x$ and $y$:
$$\begin{align}
ay&+bx&=0 \\
ax&+bcy&=1
\end{align}$$
We can solve these by any of the usual methods and get $$\begin{align}
x&=\frac{a}{a^2-b^2c} \\
y&=\frac{-b}{a^2-b^2c}
\end{align}$$
so the solution is $$\frac{a}{a^2-b^2c} +  \frac{-b}{a^2-b^2c}\sqrt c.$$
Checking with your $2+\sqrt3$ example, we have $a=2, b=1, c=3$, so $a^2-b^2c = 1$ and the inverse should be simply  $a-b\sqrt 3 = 2 -\sqrt 3$, as you said.
