$g(x)=f(4^x,0)$, $f(a,b)=f(a+b,b-a)$. Show that $\exists$ a constant $c$ such that $g(x+c)=g(x)$ for all real $x$.

Question

Question :

Suppose $f$ is a function of two variables which satisfies : $$f(a,b)=f(a+b,b-a) \quad \forall a,b \in \mathbb{R}$$ Show that $\exists$ a constant $c$ such that $g(x+c)=g(x) ~ \forall x\in \mathbb{R}$ given that $g(x)=f(4^x,0)$

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The problem can be solved as:

Using $f(a,b)=f(a+b,b-a)$ repeatedly, we get:

$$g(x)=f(4^x,0)\tag1$$

$$=f(4^x, -4^x)\tag2$$

$$=f(0,-2\cdot 4^x)\tag3$$

$$=f(2 \cdot 4^x, -2 \cdot 4^x)\tag4$$

$$=f(0,-4.4^x)=f(0,-4^{x+1})\tag5$$

$$=f(4^{x+1}, -4^{x+1})\tag6$$

$$.$$

$$.$$

$$.$$

$$= f(4^{x+c}, -4^{x+c})\tag7$$

$$=f(4^{x+c}+0, 0 -4^{x+c})\tag8$$

$$=f(4^{x+c},0)=g(x+c)\tag9$$

Doesn't this imply that the problem is solved? And doesn't it imply that any $c\in \mathbb{Z^+}+\{0\}$ satisfy?

Answer 1

Let's stay with $(a,b)$ a little longer. $$f(a,b)=f(a+b,b-a)=\\ =f(2b,-2a)=f(2b-2a,-2a-2b)=\\ =f(-4a,-4b)=f(-4a-4b,4a-4b)=\\ =f(-8b,8a)=f(8a-8b,8a+8b)=\\ =f(16a,16b)$$ That's about the size of it. So the minimal period of $f(4^x,0)$ is $\bf2$, not 1.