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I came across this question while solving a compendium of previous year questions of regional Mathematics Olympiads.

If $f : \mathbb{R} \to \mathbb{R}$ and $f(f(x))=x^2-x+1$, then find $f(2021)+f(1971)+f(50)$.

I have proceeded till finding out $$f(0)=f(1)=1$$ $$f(1-x)=f(x)$$

I proceeded like this: $$f(x^2-x+1) = f(f(f(x))) = f(x)^2-f(x)+1$$ This implies that $f(1) = f(1)^2-f(1)+1$ which is why $f(1)=1$.

Next I proved that $f(f(1-x))=f(f(x))$ using the fact that $$f(f(x))=\left(x-\frac{1}{2}\right)^2+\dfrac{3}{4}$$

Then I proved that $f(x)$ is symmetric along $\dfrac{1}{2}$.
Next we put $x\leftarrow\dfrac{1}{2}-x$ in the equations $f\left(\dfrac{1}{2}-x\right) = f\left(\dfrac{1}{2}+x\right)$ since it is valid for all values of $x$.

That is how I got $f(0)=f(1)=1$ and $f(1-x)=f(x)$.
But how do I proceed further to find the value of $f(2021)+f(1971)+f(50)$?

Alma Arjuna
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WizardGamer44
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    It seems you have deleted your previous question and asked the same question again... – Shivansh Jaiswal Jul 17 '23 at 11:44
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    yeah because it was closed and i could not understand why. – WizardGamer44 Jul 17 '23 at 11:59
  • I don't understand why $f(x) = f(1-x)$ – nonuser Jul 17 '23 at 13:23
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    Deleting and reposting is against the site rules. You should have edited that first post. May be adding the reasoning you give here to it would have sufficed? I cannot tell for sure because you deleted the question, but may be not having such details was the reason it was closed? – Jyrki Lahtonen Jul 17 '23 at 15:41
  • Anyway, having too many closed (other than duplicates) and deleted questions may lead to a (non-negotiable) post-ban. It is better to prevent that from ever happening, surely. Check out our abridged guide for new askers to learn about the expectations. – Jyrki Lahtonen Jul 17 '23 at 15:44
  • @SwagBeastSKJJ I do not know the answer. Would you mind sharing what you have done and how you have proceeded with the sum? – WizardGamer44 Jul 21 '23 at 13:27
  • Related: https://math.stackexchange.com/questions/1158619/do-there-exist-functions-f-such-that-ffx-x2-x1-for-every-x?noredirect=1, https://math.stackexchange.com/questions/1996141/if-ffx-x2-x1-what-is-f0?noredirect=1. – Gonçalo Feb 12 '25 at 23:15
  • Also: https://math.stackexchange.com/questions/2254594/how-much-does-f-circ-fx-x2-x-1-determine-f?noredirect=1&lq=1. – Gonçalo Feb 12 '25 at 23:29

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