Finding period of $f$ from the functional equation $f(x)+f(x+4)=f(x+2)+f(x+6)$

Question

How can I find the period of real valued function satisfying $f(x)+f(x+4)=f(x+2)+f(x+6)$?

Note: Use of recurrence relations not allowed. Use of elementary algebraic manipulations is better!

Answer 1

Observe, since $$ f(x)+f(x+4)=f(x+2)+f(x+6), $$ we can substitute $x+2$ for $x$ to get $$ f(x+2)+f(x+6)=f(x+4)+f(x+8). $$

Equating these, we know that $f(x)=f(x+8)$.

Answer 2

You are given that $f(x)+f(x+4) = f(x+2)+f(x+6)$ for all $x \in \mathbb{R}$.

Replace $x$ with $x+2$ to get $f(x+2)+f(x+6) = f(x+4)+f(x+8)$ for all $x \in \mathbb{R}$.

Thus, $f(x)+f(x+4) = f(x+2)+f(x+6) = f(x+4)+f(x+8)$ for all $x \in \mathbb{R}$.

Subtract $f(x+4)$ from the left and right side of the last equation to get: $f(x) = f(x+8)$ for all $x \in \mathbb{R}$.

Important note: This tells you that $f$ is $8$-periodic, but $8$ may not necessarily be the minimum period. For instance, $f(x) = \sin(\pi x)$ satisfies $f(x)+f(x+4) = f(x+2)+f(x+6)$ for all $x \in \mathbb{R}$, but has period $2$.