<s>[INST] Find all integer values of $m$ such that the equations $x+y+\sqrt{4-x^2}+\sqrt{4-y^2}=4$ and $x\sqrt{4-x^2}+y\sqrt{4-y^2}=m+2$ have a solution. 
Consider the system of equation $\left\{ \begin{aligned} x + y + \sqrt{4 - x^2} + \sqrt{4 - y^2} &= 4\\ x\sqrt{4 - x^2} + y\sqrt{4 - y^2} &= m + 2 \end{aligned} \right. (x; y \in \mathbb R)$. How many integer values of $m$ are there such that there exists a solution to the above system?

[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Of course, it's definitely not $$x\sqrt{4 - x^2} + y\sqrt{4 - y^2} \le \dfrac{(x + \sqrt{4 - x})^2 + (y + \sqrt{4 - y^2})^2}{2}$$, nor is it $$\left(x\sqrt{4 - x^2} + y\sqrt{4 - y^2}\right)^2 \le (x^2 + y^2)[(4 - x^2) + (4 - y^2)]$$
Let's just try to solve this the typical way.
Consider the function $f(z) = z + \sqrt{4 - z^2}$. The domain and codomain of the function are respectively $[-2; 2]$ and $[-2; 2\sqrt 2]$.
(This is an unnecessary step, but I'm going to include it anyway.
Since $f(x) + f(y) = 4$, this means $$f(x), f(y) \in [4 - 2\sqrt 2; 2\sqrt 2] \iff x, y \in \left[2 - \sqrt 2 - 2\sqrt{\sqrt 2 - 1}; 2\right]$$ This is just to further strengthen the domains of $f(x)$ and $f(y)$ specifically. Strengthen's definitely not the right word.)
Consider function $g(z) = z\sqrt{4 - z^2}$. For $\displaystyle x, y \in \left[2 - \sqrt 2 - 2\sqrt{\sqrt 2 - 1}; 2\right]$, the codomain of $g(x)$ and $g(y)$ are $[10 - 8\sqrt{2}; 2]$. Therefore, $g(x) + g(y) \in [20 - 16\sqrt{2}; 4]$, right?
If only it was that easy.
How about this? Let $\left\{ \begin{aligned} x = 2\cos a, \sqrt{4 - x^2} &= 2\sin a\\ y = 2\cos b, \sqrt{4 - y^2} &= 2\sin b \end{aligned} \right.$, (since $x^2 + (\sqrt{4 - x^2})^2 = y^2 + (\sqrt{4 - y^2})^2 = 4$). Afterwards, the system of equation could be rewritten as $$\left\{ \begin{aligned} \sin\left(\dfrac{a + b}{2} + \dfrac{\pi}{4}\right)\cos\left(\dfrac{a - b}{2}\right) &= \dfrac{\sqrt 2}{2} &&(1)\\ \sin(a + b)\cos(a - b) &= \dfrac{m + 2}{4} &&(2) \end{aligned} \right.$$
Combining $(1)$ and $(2)$, we have $\sin(a + b)\left[\dfrac{1}{\sin^2\left(\dfrac{a + b}{2} + \dfrac{\pi}{4}\right)} - 1\right] = \dfrac{m + 2}{4}$.
Let $\dfrac{a + b}{2} = z$ and function $$f(z) = 2\sin z\cos z\left[\dfrac{2}{(\sin z + \cos z)^2} - 1\right] \implies f'(z) = 2 \times \left[\dfrac{2(\cos z - \sin z)}{(\cos z + \sin z)^3} - \cos 2z\right]$$
How would one go about actually solving $f'(z) = 0$?
Hmmm~ well, it's obvious that $\cos z - \sin z = 0 \iff z = n\pi + \dfrac{\pi}{4}, \forall n \in \mathbb Z$ is one of the subset of roots.
Actually, let's graph the function.

(seeing that the $y$-intercept extends to $-\infty$) How about we don't graph the function?
Aside from the apparent fact that $z\sqrt{4 - z^2} = \dfrac{(z + \sqrt{4 - z^2})^2 - 4}{2}$, I don't know what to do next. Also, it's past midnight already. So as always, thanks for reading, (and even so if you could help), and have a great tomorrow, everyone~
By the way, the options were $1, 2, 3$ and $4$, and there are seven integers between $20 - 16\sqrt{2}$ and $4$, so my first attempt would have led me to a dead end.
 [/INST] If you solve the system as if $\sqrt{4-x^2}$ and $\sqrt{4-y^2}$ were the variables in a system of two linear equations, then you get:
$$\sqrt{4-x^2} = \frac{-4y + xy + y^2 + m + 2}{x - y}$$
$$\sqrt{4-y^2} = \frac{4x - x^2 - xy - m - 2}{x - y}$$
Note that $\sqrt{4-x^2} + \sqrt{4-y^2}$ simplifies to just $x + y + 4$, so the first equation gives $2x + 2y + 4 = 4$, or $y = -x$.  Substituting this into the second equation gives:
$$x \sqrt{4-x^2} - x \sqrt{4-x^2} = m + 2$$
$$0 = m + 2$$
$$m = -2$$
So, we now have a unique solution for $m$.  We just need to verify that there exists at least one real solution to $x \sqrt{4-x²} + y \sqrt{4-y²} = 0$.  And clearly, anything with $y = -x$, $|x| \le 2$, and $|y| \le 2$ will work.
Therefore, a solution exists with $m = -2$.
I don't know what you did all that trig for ;-)
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