Ultimate Resemblance

Observe that since the given system of equations have real solution only. It is possible only when, $|x| \leq 1$ and $|y| \leq 1$ .

So, Let $x= \sin (p)$ and let $y=\sin (q)$ .

So, $\sin(p) \cos(q) = a^2$ and $\cos(p) \sin(q) = a$

Adding the both we get, $\sin(p+q) = a^2 + a$

But $-1 \leq \sin(p+q) \leq 1$ .

So, $-1 \leq a^2 + a \leq 1$

Which implies that $a^2 + a + 1 \geq 0$ and $a^2 + a -1 \leq 0$

On solving these we get,

$-\left(\dfrac{\sqrt{5} + 1}{2} \right) \leq a \leq \left(\dfrac{\sqrt{5} - 1}{2} \right)$ .

Again on subtracting the two given inequalities, we obtain $\sin(p-q) = a^2 - a$ .

But, $-1 \leq \sin(p-q) \leq 1$ , which implies that $-1 \leq a^2 - a \leq 1$ .

Solving these we get,

$\left(\dfrac{1-\sqrt{5} }{2} \right) \leq a \leq \left(\dfrac{\sqrt{5} + 1}{2} \right)$

From the both inequalities, we get

$a \in {[{ \dfrac{1-\sqrt{5} }{2} }, \dfrac{\sqrt{5} -1}{2} ]}$

So, $\alpha + \beta = \boxed{0}$ .