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Suppose a function $f:\mathbb{R}\to\mathbb{R}$ that satisfies a relation $$f(a^2+b^2)=f(a)^2+f(b)^2$$ Find all such functions. It is also given that $f(1)>0.$

Putting $a=1$ and $b=0$ yields
$f(0)^2=f(1)[1-f(1)]$
$\implies 0<f(1)\le1$

Let's say I do $f(1)=1$ then we immediately have $f(0)=0$ and as $f(2)=f(1^2+1^2)$ we will also have $f(2)=2$.

This suggests that $f(k)=k$ can be a possible solution. Now if I put this in my original equation, it is indeed satisfying the relation, but I don't think that this is the correct way to prove that $f(k)=k$ is a solution to this functional equation.

How can I prove it mathematically without presuming that it is a possible solution? Also how can I find other solutions, if they exist.

Any help is greatly appreciated.

jd27
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Blue Cat Blues
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    Direct verification is an absolutely admissible way of rigorously showing that $f(k)=k$ is a possible solution to the functional equation. – PierreCarre Jan 13 '25 at 11:36
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    Note that the question requires you to find all such functions. You have produced one example, though you have certainly done that successfully. – lulu Jan 13 '25 at 11:52
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    Might be a duplicate of https://math.stackexchange.com/q/3332821/1031068? – Y.D.X. Jan 13 '25 at 12:07
  • @Y.D.X. For future reference, if you believe that something is a duplicate, please flag it as such. This ensures that other users on the site see the question, and have an opportunity to either agree or disagree with you. Putting it only in a comment makes it harder for other users to make any kind of judgement. – Xander Henderson Jan 13 '25 at 13:31
  • Oh, I didn't know it can be suggested by clicking Flag. Thank you! @XanderHenderson – Y.D.X. Jan 13 '25 at 16:29

1 Answers1

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For any real number $x$, we have $$ f \left( x^2 \right) = f \left( x^2 + 0^2 \right) = f(x)^2 + f(0)^2, $$ and also $$ f \left( 2x^2 \right) = f \left( x^2 + x^2 \right) = f(x)^2 + f(x)^2 = 2 f(x)^2. $$

And if $x \geq 0$, then we have $$ f(x) = f \left( \left( \sqrt{x} \right)^2 \right) = f \left( \sqrt{x} \right)^2 + f(0)^2, $$ and also $$ f (x) = f \left( \frac{x}{2} + \frac{x}{2} \right) = f \left( \left( \sqrt{ \frac{x}{2} } \right)^2 + \left( \sqrt{ \frac{x}{2} } \right)^2 \right) = f \left( \sqrt{ \frac{x}{2} } \right)^2 + f \left( \sqrt{ \frac{x}{2} } \right)^2 = 2 f \left( \sqrt{ \frac{x}{2} } \right)^2, $$ and thus we also have $$ 2 f \left( \sqrt{ \frac{x}{2} } \right)^2 = f \left( \sqrt{x} \right)^2 + f(0)^2. $$

Saaqib Mahmood
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