Find all the functions $f(x)$ where $f(x\sqrt{2}) = 2f(x)$ and $f(x+1) = f(x)+2x +1$ for all real $x$.

Question

Find all the functions $f: \mathbb{R}\to\mathbb{R}$ such that $$f(x\sqrt{2}) = 2f(x)$$ and $$f(x+1) = f(x)+2x +1$$ for all $x\in\mathbb{R}$.

So, the only function that I intuitively can imaging is $f(x) = x^{2}$, but I tried to plug $x=0$ and other different expressions and numbers without any success. Is there any other functions and how to prove that $f(x)=x^{2}$. Could you give me any hint?

Answer 1

Set $$f(x)=x^2+g(x)$$ and plug it into the equations. You will obtain $$g(x+1)=g(x) \tag{1}$$ $$g(\sqrt{2}x)=2g(x) \tag{2}$$ i.e. $g$ is periodic with period $1$. Now:

• From the second equation you get $g(0)=0$ and by the periodicity $g(n)=0$ at all integers.
• Furthermore $g(\sqrt{2})=0$ and so $g(\sqrt{2}+1)=0$. Inductively $g(m\sqrt{2}+n)=0$ for integers $m,n$.
• Similarly you can set $x=m/\sqrt{2}$ in equation (2) to find $g(m/\sqrt{2})=0$. Repeating this process will give $g(m/2^{k/2})=0$.
• You can repeatedly apply (1) and (2) to first find $g(m2^{k/2}+n2^{l/2})=0$ for $m,k,n,l\in \mathbb{Z}$ or more generally $$g\left(\sum_{k\in \mathbb{Z}} m_k 2^k + \sqrt{2} \sum_{k\in \mathbb{Z}} n_k 2^k\right)=0 \, .$$
• The last result implies (binary representation) $$g(x\sqrt{2}+y)=0$$ for $x,y$ being dyadic rationals.