So I was bored, and decided to do a functional equation, which I have never done before. After a while, I came up with this:$$f(a+2b)-f(a+1)=f(b)$$ which I thought that I might be able to solve. Here is my attempt at solving the functional equation:
Step $1$: Plug in $a=0$:$$f(2b)-f(1)=f(b)$$$$\implies f(b)=f(1)-f(2b)$$Step $2$: Plug in $f(b)=f(1)-f(2b)$:$$f(a+2b)-f(a+1)=f(1)-f(2b)$$$$\implies f(2b)=f(1)+f(a+1)-f(a+2b)$$$$\text{When we plug in }a=0\text{ again we get} f(2b)=2f(1)-f(2b)$$$$\implies f(1)=f(2b)$$$$\implies f(b)=f(1)=0$$$$\implies f(a+2b)=f(a+1)$$Now, to plug in $b=0$ for Step $3$:$$f(a)=f(a+1)$$This can only be true if the following conditions are met:$$f(a,b)=\begin{cases}c\text{ (where }c\text{ is an unspecified constant)}&,a+2b\ne1\\ 0&,a+2b=1\\ 0&,a=0\end{cases}$$
My question
Is my solution to the functional equation correct, or what could I do to solve it more easily?
Mistakes I might have made
- Everything (I've never done one of these before, so I honestly don't know what I'm doing here)
If $a=2b,$ then $f(4b)=f(2b+1)+f(b)$. But by your equality, $f(2n)=f(1)+f(n),$ we get:
$$f(4b)=f(1)+f(2b)=2f(1)+f(b).$$
So we get $f(2b+1)=2f(1),$ for all $b.$
So the function is constant on odd numbers.
If $a=2k-1, b=1$ then $a+2$ is odd, so:
$$2f(1)-f(2k)=f(a+2)-f(2k)=f(1),$$ or $f(2k)=f(1),$ so $f$ must be constant on the even numbers, as well.
But then $$f(1)=f(4)=f(2)+f(1)=2f(1),$$ which means $f(1)=0.$ Then $f(n)=0$ for all $n.$