(I apologize for any english inconsistencies! This was translated from a portuguese original question).
A drawer contains red and white socks. When two socks are picked randomly (without replacement), the probability of both being red is equal to $1/2$.
a) What is the lowest number of red and white socks in the drawer? (since this was translated from portuguese, maybe “lowest” isn’t the best word to use in such a mathematical context)
b) What should the lowest number of red and white socks be if the number of white socks is even?
I actually don't know if I can solve this with basic statistics (which is what I barely grasp). My attempt didn't go very far:
Let's say the number of socks is given by $V + B$. In such case, my first pick has a probability of ${V \over {V + B}}$ of coming red; my second pick has a probability of ${{{V - 1} \over {\left( {V - 1} \right) + B}}}$ of coming red.
So, the information I have would be that:
$$\eqalign{ & {V \over {V + B}}\left( {{{V - 1} \over {\left( {V - 1} \right) + B}}} \right) = {1 \over 2} \cr & {{{V^2} - V} \over {{V^2} - V + 2VB - B + {B^2}}} = {1 \over 2} \cr} $$
But I can't really continue, nor do I know if I was supposed to be there in the first place. Any directions would be really appreciated.
Your attempt was on the right track . . .
If you solve for the quadratic equation for $v$, you get $$v=\frac{2b+1\pm\sqrt{8b^2+1}}{2}$$ which is an integer if and only if $8b^2+1$ is a perfect square.
By inspection, $b=1$ works, yielding the solution pair $(v,b)=(3,1)$, with $v+b=4$.
For the case where $b$ is required to be even, by trial and error, we find that $8(6)^2+1=17^2$, so we get the solution pair $(v,b)=(15,6)$, with $v+b=21$.