If $f(f(x))=x^2-x+1$, what is $f(0)$?

Question

Suppose that $f\colon\mathbb{R}\to\mathbb{R}$ without any further restriction. If $f(f(x))=x^2-x+1$, how can one find $f(0)$?

Thanks in advance.

Same question on AOPS: "2019 Nepal National Mathematics Olympiad", "Determine f(0) if given f(f(x))" and "hard functional equation". It seems that the problem has originally appeared as Problem 5 of Baltic Way 2011 contest. — Mohsen Shahriari, Jan 23 '22 at 19:46

score 17 · Accepted Answer · edited Dec 09 '17 at 11:04

We have that

$$f(f(f(x)))=f(x)^2-f(x)+1.$$ Since $f(f(0))=1$ we get that $$f(1)=f(0)^2-f(0)+1.$$ That is

$$f(0)^2-f(0)+1-f(1)=0.$$ Repeating the process, since $f(f(1))=1$ we get that $$f(1)=f(1)^2-f(1)+1.$$ That is

$$(f(1)-1)^2=f(1)^2-2f(1)+1=0.$$ Thus, $f(1)=1.$ So it is $$f(0)^2=f(0),$$ from where $f(0)=0$ or $f(0)=1.$

If $f(0)=0$ then $f(f(0))=f(0)=0$ which contradicts $f(f(0))=1.$ So, it must be $f(0)=1.$

Jack D'Aurizio · Answer 2 · 2016-11-02T15:25:32.510

8

$f(f(0))=f(f(1))=1$. Apply $f$ once again: $f(f(f(0)))=f(f(f(1)))=f(1)=f(0)^2-f(0)+1=f(1)^2-f(1)+1$.
That leads to $f(1)=1$, hence $f(0)^2-f(0)=0$ and $f(0)$ can only be $0$ or $1$.
But $f(0)=0$ leads to $f(f(0))=0$, contra $f(f(0))=1$, so $\color{red}{f(0)=1}$.

edited Nov 02 '16 at 15:25

answered Nov 02 '16 at 15:20

Jack D'Aurizio

361,689

Mike Pierce · Answer 3 · 2016-11-03T06:00:22.110

5

If such a function $f$ exists, then $f(0) = 1$, ~~but such a function $f$ does not exist~~. See the paper:

When is $f(f(x)) = az^2 +bz+c$?
R. E. Rice, B. Schweizer and A. Sklar
The American Mathematical Monthly
Vol. 87, No. 4 (Apr., 1980), pp. 252-263
(Link to PDF not behind the JSTOR paywall)

Edit: Such a function does not exist in $\mathbb{C}$! Or in any algebraically closed field of characteristic zero. But you can have such a function in the reals. See the epilogue of the paper (page 262).

edited Nov 03 '16 at 06:00

answered Nov 03 '16 at 05:16

Mike Pierce

19,406

2

Actually the result there is "heavily domain dependent" and in fact they show a quadratic that has arbitrary iterative roots over the reals. – Jacob Wakem Nov 03 '16 at 05:48
1

Oh wow, you're right. I just read the introduction and assumed that if it was true in $\mathbb{C}$ then it must be true in $\mathbb{R}$. – Mike Pierce Nov 03 '16 at 05:57
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Not true because the way the subset relation works. – Jacob Wakem Nov 03 '16 at 05:59

score 4 · Answer 4 · answered Nov 02 '16 at 15:21

I think it is more easy from what it seems. You know that $$f(f(0))=1,$$ then $$f(1)=f(f(f(0)))=f(0)^2-f(0)+1,\;(*)$$ but also $$f(f(1))=1^2-1+1=1,$$ so $$f(1)=f(f(f(1)))=f(1)^2-f(1)-1$$ and you can calculate $f(1)$ from here, and then substitute its value in $(*)$

If $f(f(x))=x^2-x+1$, what is $f(0)$?

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