<s>[INST] Find constant in system of function equations 
Question: Let $a \in \mathbb{R}$. Find all possible values of $a$ such that there exists a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x + 2) = -f(x)$ and $f(x + 3) = f(x) + a$ for any real number $x \in \mathbb{R}$.

My (maybe wrong) solution:
Start listing down some values of $f(x)$ with $f(2) = -f(0)$ and $f(3) = f(0) + a$.
$f(2) + f(3) = - f(0) + f(0) + a = a$.
Similarly, you can write $f(1) = -(f(-1))$ and $f(2) = f(-1) + a$.
Hence, $f(1) + f(2) = -(f(-1)) + f(-1) + a = a$.
So, $f(1) + f(2) = f(2) + f(3) \Longrightarrow f(1) = f(3).$
But, according to the problem, $f(x + 2) = -f(x) \Longrightarrow f(3) = -f(1)$.
Since $f(1) = f(3)$ and $f(1) = -f(3) \Longrightarrow f(1) = f(3) = 0.$
Doing similar steps, we find out that $f(0) = f(2) = 0$. Hence, $a = f(2) + f(3) = 0 + 0 = 0$.
The only possible value of $a$ is $\boxed{0}$.
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If I'm doing something wrong, can someone please explain? Thanks in advance!
(Thanks to the answers, I now know I only need to prove that $a = 0$ can satisfy a valid function :))
 [/INST] This makes sense. Here is a different approach leading to the same place. Note that using the first relationship,
$$
f(x+6) = -f(x+4) = f(x+2) = -f(x)
$$
and using the second relationship,
$$
f(x+6) = f(x+3)+a = f(x) + 2a,
$$
so we have $f(x)+2a = -f(x) \iff f(x) = -a$, so $f$ is constant.
Now from the first relationship,
$$
-f(x) = f(x+2) = f(x) \iff f(x) = 0 \ \forall x \in \mathbb{R}.
$$
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