Find all continuous functions $f:\mathbb R\to \mathbb R$ such that : $$f(\sqrt 2 \, x)=2f(x)$$ $$f(x+1)=f(x)+2x+1$$ for all $x\in \mathbb R$.
Clearly $x^2$ is a valid solution but how can i find other solutions or prove that there is no other solution? And please give me ressources for studying functional equations and some advanced techniques cause the techniques I use are still for a beginner level
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Let $\,G\,$ be the function defined as $\,G(x)=f(x)-x^2$ for all $\,x\in\Bbb R\,.\,$ Without using continuity, we can prove that $\,G(r)=0\,$ for any $\,r\in\Bbb Q\,.$
Let $\,\varphi\,$ be the Euler’s totient function.
For any $\,r\in\Bbb Q\,$ there exist $\,a,d\in\Bbb Z\,,\,$ $b,c\in\Bbb N_0=\Bbb N\cup\{0\}\,$ such that
$r\!=\!\dfrac a{2^b(2c\!+\!1)}\;\;,\qquad$ $2^{\varphi(2c+1)}\!=\!1\!+\!(2c\!+\!1)d\;.$
Since $\,G(2^nx)\!=\!4^nG(x)\,$ and $\,G(x\!+\!m)\!=\!G(x)\,$ for any $\,n,m\in\Bbb Z\,$ and for any $\,x\in\Bbb R\,,\,$ it follows that$$4^{\phi(2c+1)+b}G(r)=G\!\left(2^{\phi(2c+1)+b}r\right)=G\left(2^br[1\!+\!(2c\!+\!1)d]\right)=G\left(2^br\!+\!ad\right)=G\left(2^br\right)=4^bG(r)$$Consequently , $\;G(r)=0\,.$
It means that $\,f(x)=x^2\,$ for any $\,x\in\Bbb Q\,$ even if the function $\,f:\Bbb R\to\Bbb R\,$ is not continuous.
To summarize the discussion in the comments:
We note that $x\mapsto x^2$ is a possible solution for $f(x)$.
Working on the theory that this is the only possible solution, we define $$G(x)=f(x)-x^2$$ and note that $$G(\sqrt 2\,x)=2G(x)\quad \&\quad G(x+1)=G(x)$$
We'd like to show that $G(x)$ was identically $0$.
Now, $G(x)$ is continuous and periodic. That means that it attains a global maximum, $M$. Let $G(x_0)=M$. We remark that $M≥0$ since $G(0)=0$. But then $$G(\sqrt 2\,x_0)=2M\implies 2M≤M\implies M=0$$ A similar argument shows that the global minimum is also $0$ and we are done.