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$f$ is a function on real numbers: $$f(x)^2=f(2x)+2f(x)-2$$ and $$f(1)=3$$

What is the value of $f(6)$?

I find a solution $f(x)=2^x+1$. But, I don't know is there more solutions?

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    $\begingroup$ I get that $f(2^x) = 2^x + 1$ However, there is no reason that I see why this function need be continuous, and I am not convinced that $f(3)$ cannot be whatever you like. $\endgroup$
    – Doug M
    Apr 5 '17 at 21:08
  • $\begingroup$ God knows that the equation has discontinious $f$ solutions. You may right. $\endgroup$
    – scarface
    Apr 5 '17 at 21:16
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Let's look at $g(x)=f(x)-1$. Then $$f(x)^2-2f(x)+1=f(2x)-1\\ g(x)^2=g(2x)$$ There are infinitely many solutions, even continuous ones.

Knowing $f(x)$ gives us $f(2x)$, then $f(4x)$ and so on, but nothing else.

Since $6$ is not a power of $2$ times $1$, you can make $f(6)$ whatever you want.

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  • $\begingroup$ Typo: $-1$ at LHS is $+1$. $\endgroup$ Apr 5 '17 at 21:16
  • $\begingroup$ Got it, thanks. $\endgroup$ Apr 5 '17 at 21:19
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$f(6)=65$ is not the only solution
Define f as follows:
$f(x)=2^x+1$, for $x=2^n$ or $x=\frac{1}{2^n}$ for $n\in \mathbb{N}$ and 1 otherwise.
In similar way you can define to be 2 otherwise.

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The way this question is put, there is a relation to be satisfied between $f(x)$ and $f(y)$ only if $$\frac{x}{y}=2^n$$ for some $n\in\mathbb{Z}$. In particular, the numbers $f(1)$ and $f(6)$ have nothing to do with one another, and you can choose $f(6)$ as you please.

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If we write $f(x) = g(x)+1$, the functional equation simplifies to $$ g(2x) = g(x)^2$$ $g(2x)$ must be nonnegative if you want real values, but otherwise $g(x)$ can be arbitrary on, say, $(-2,1] \cup [1,2)$, with $g(2^n x) = g(x)^{2^n}$ for integers $n$. In particular, $f(6)$ is not determined by $f(1)$.

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