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I am trying to find all real functions that satisfy the property: $f(x+y) = (f(x))^2 + (f(y))^2$, $x$, $y$ real numbers. I tried to substitute $x$ and $y$ with $0$ but end up with nothing, then I tried substituting $y$ with $x$ and I get $f(2x) = 2(f(x))^2$, then I tried to prove that $f(0) = 0$ and I think I am missing an important step.

How can I proceed from here? If it isn't the wrong already.

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    $\begingroup$ Imagine you know $f(1)$. At how many other points can you get the value of $f$ knowing just $f(1)$? Use the given recurrence wisely. For example, you will know $f(2)$. Then using $y=1,x=2$, you'll know $f(3)$. SImilarly, you'll know $f(1/2), f(1/4)$ etc. Find the exact set of values at which $f$ becomes determined. Then, we can try to think about what happens "outside" that set. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Commented Mar 6, 2023 at 7:32
  • $\begingroup$ How can I determine the exact set? Is it the set of positive rational numbers? $\endgroup$
    – David399
    Commented Mar 6, 2023 at 7:43
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    $\begingroup$ If $f(0) = 0$ then: $$0 = f(0) = f(x + (-x)) = f(x)^2 + f(-x)^2$$ but since each term is positive that gives $f(x)^2 = f(-x)^2 = 0$, and therefore: $\forall x \in \mathbb{R}, f(x) = 0$. In general, if there's any $z$ for which $f(z) = 0$, then you can show that $f$ is the zero function by doing the same thing with $z = x + (z - x)$. Thus you may require the function to be strictly positive. $\endgroup$
    – Bruno B
    Commented Mar 6, 2023 at 7:54
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    $\begingroup$ Not sure if it helps, but this may be feasible with operator calculus, which is always an interesting avenue. Since $f(x+1)=f(x)^2+f(1)^2$, noting $f(x+1)=e^D f(x)$ where $e^D=e^\frac{d}{dx}$ is the shift operator, letting $y=f(x)$ and $c=f(1)$, then we have a differential equation of the form $e^Dy-y^2-c=0$. In other words, $y$ should obey $y+y'+\frac{y''}{2}+\cdots=y^2+c$. $\endgroup$
    – Graviton
    Commented Mar 6, 2023 at 7:58
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    $\begingroup$ Hint: first take $x=y=0$, and then take $y=0$ and $x$ general; that will give you very strong information about the possible values of $f$. Must $f$ be continuous? $\endgroup$
    – Greg Martin
    Commented Mar 6, 2023 at 8:00

2 Answers 2

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Setting $x=y=0$ tells us that $f(0)=0$ or $f(0)=\frac 1 2$.

Now, $y=0$ gives us $$f(x)=f(x)^2+f(0)^2\tag1$$

Write $z=f(x)$. This presents us with $z^2-z+f(0)^2=0$. Therefore, the image of $x$ under $f$ is always a root of this equation. Can you take it from here, case by case?

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  • $\begingroup$ So that would mean that $f(x) = 1$ or $f(x) = 0$ or $f(x) = \frac{1}{2}$ $\endgroup$
    – David399
    Commented Mar 6, 2023 at 8:09
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    $\begingroup$ And since $f(x) = 1$ can't possibly a solution because it doesn't satisfy the relation, the only real functions are: $f(x) = 0$, $f(x) = \frac{1}{2}$ $\endgroup$
    – David399
    Commented Mar 6, 2023 at 8:50
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    $\begingroup$ I misread your comment. My bad. Yes, $f(x)=0$ and $f(x)=1/2$ are the only solutions. $\endgroup$
    – Umesh Shankar
    Commented Mar 6, 2023 at 8:50
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$x = y = 0$ gives $f(0) = (f(0))^2 + (f(0))^2$, and since $f(x)$, which has the solutions $f(0) = 0, f(0) = \frac{1}{2}$. Case $1$ ($f(0) = 0$) gives that, if $y = -x$: $f(x + -x) = (f(x))^2 + (f(y))^2 \iff 0 = (f(x))^2 + (f(y))^2$, but since $f$ is a real valued function, both terms in the $RHS$ of the equation have to equal $0$, and since we made no assumptions about $x$, this means that $f(x) = 0 \ \forall x \in \mathbf{R}$. Case $2$ ($f(0) = \frac{1}{2}$) gives that, if $y = 0$, $f(x) = (f(x))^2 + \frac{1}{2}$, which has the only solution $f(x) = \frac{1}{2}$. This means that the valid solutions are $f(x) = 0, f(x) = \frac{1}{2}$.

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