<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] $$\begin{align} 
x + y + \sqrt{4-x^2} + \sqrt{4-y^2} &= 4 \tag 1\\
x\sqrt{4-x^2} + y\sqrt{4-y^2} &= m + 2 \tag 2
\end{align}$$
Obviously, $x,y\in[-2,2]$, that is we can represent $x$ and $y$ as
$$\begin{align}
x &= 2\sin a \\
y &= 2\sin b
\end{align}$$
which transforms (1) and (2) into
$$\begin{align}
\sin a + \sin b + \cos a  + \cos b &= 2 \tag{1a}\\
\sin a \cos a + \sin b \cos b &= m + 2 \tag{2a}\\
\end{align}$$
where $a,b\in[-\pi/2, \pi/2]$ because the square roots are non-negative, and similar must apply to $\cos$.
We can rewrite (2a) as
$$\sin 2a + \sin 2b = 4+2m \tag{2b}$$
which constrains $m$ to
$$
-3\leqslant m\leqslant -1 \implies m\in\{-3,-2,-1\}
$$
Now let's try to find $a$, $b$ for these $m$:

*

*$m=-1 \implies a=b=\pi/4$ because the only way to get the right-hand side of (2b) to equal 2 is to have both sines equal to 1.  However for (1a) this means $4\cdot 1/\sqrt 2 = 2$.  Thus, $m=-1$ has no solution.


*$m=-3 \implies a=b=-\pi/4$ because the only way to get the right-hand side of (2b) to equal -2 is to have both sines equal to −1.   However for (1a) this means $0 = 2$.  Thus, $m=-1$ has no solution.


*$m=-2 \implies \sin2a + \sin2b = 0$ which has solutions $a=-b$ and $a + b=\pm\pi/2$, yielding the sub-cases:

*

*$a=-b$ cancels the sines from (1a) and yields $\cos a+ \cos b=2$, i.e. $a=b=0$, i.e. $x=y=2$.


*$a =\pm\pi/2 -b$ is left as an exercise :-)
As the question just asks for the number of $m$'s such that the equations posess solution(s), the correct answer is "1"; no need to work out the last sub-case as it won't change the number of $m$'s.
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