Are they related?

We begin with Problem 2 :

$\tan{(15^{\circ})}=\tan{(45^{\circ}-30^{\circ})}=\frac{\tan{(45^{\circ})}-\tan{(30^{\circ})}}{1+\tan{(45^{\circ})}\tan{(30^{\circ})}}$

$...=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$

So $B=2-\sqrt{3}$

Now we move on to Problem 1 :

The key is noticing this looks rather like the tan subtraction formula.

We convert our $N$ reals to $N$ values of $-90^{\circ}<\theta<90^{\circ}$ where $\tan{(\theta)}$ is equal to each of the reals.

$\frac{x-y}{1+xy}=\frac{\tan{(\theta_{1})}-\tan{(\theta_{2})}}{1+\tan{(\theta_{1})}\tan{(\theta_{2})}}=\tan{(\theta_{1}-\theta_{2})}$

So the problem is equivalent to finding $\theta_{1}-\theta_{2}$ between $tan^{-1}{(0)}=0^{\circ}$ and $tan^{-1}{(\frac{1}{\sqrt{3}})}=30^{\circ}$

$N>6$ because we can select $\theta=-89^{\circ},-58^{\circ},-27^{\circ},-4^{\circ},-35^{\circ},-66^{\circ}$ . All of these differ by more that $30^{\circ}$ .

We say $N=7$ . Consider the 6 intervals: $[-90^{\circ},-60^{\circ}],[-60^{\circ},-30^{\circ}],[-30^{\circ},0^{\circ}],[0^{\circ},30^{\circ}],[30^{\circ},60^{\circ}],[60^{\circ},90^{\circ}]$

By pigeon-hole principle, 2 values of $\theta$ lie in one of the ranges and therefore differ by less that $30^{\circ}$ .

So $A=7$

$B^2+(A-4B)=(2-\sqrt{3})^2+(7-4(2-\sqrt{3})=7-4\sqrt{3}+(7-8+4\sqrt{3})=6$