Given the axiom of choice, you can write the identity function as a sum of two periodic functions. An example of a way to do this can be found here: https://mathblag.wordpress.com/2013/09/01/sums-of-periodic-functions/
Today, we are not going to solve this for the function ℝ → ℝ, but a simpler variant. Suppose 𝒬 is the set of all numbers that can be written as $q_1+q_2\sqrt{2}$ with q1, q2 ∈ ℚ. This set is closed under addition and multiplication. Now, find to periodic function of the form 𝒬 → 𝒬 that sum to the identity function. This is a way simpler proof that does not need the axiom of choice, but the solution follows the same recipe.
First, we will prove that ∀q ∈ 𝒬, the decomposition into $q_1+q_2\sqrt{2}$ is unique, in other words, there exists only one pair q1, q2.
Proof:
Assume there are two pairs, q1, q2 and q′1, q′2 such that:
$q_1+q_2\sqrt{2}=q'_1+q'_2\sqrt{2}$
$q_1-q'_1=(q'_2-q_2)\sqrt{2}$
$\frac{q_1-q'_1}{q'_2-q_2}= \sqrt{2}$
This leads to a contradiction, as the left side is rational and the right side is irrational. The only way to solve this is if q′2 − q2 = 0, because then the last step was not allowed. In this case we get: q1 − q′1 = 0 Which means that both elements of the pair where equal.
Now that this is out of the way,
Define function $f: q_1+q_2\sqrt{2} \rightarrow q_1$
This function is well defined due to the previous theorem, and is periodic with period $\sqrt{2}$.
Proof:
$f(x + p\sqrt{2}) = f(q_1+q_2\sqrt{2} + p\sqrt{2}) = f(q_1+(q_2+p)\sqrt{2}) = q_1 = f(x)$
Analogous, define $g: q_1+q_2\sqrt{2} \rightarrow q_2\sqrt{2}$
g is for the same reason as above periodic with period 1
It is now easy to see that f + g is the identity function, so we found both our functions.