Is a function $f$ satisfying $f(x+1) = f(x) + 1$ and $f(x^2) = (f(x))^2$ odd or even?

Question

The problem statement, all variables and given/known data

1) $f(x+1)=f(x)+1$
2) $f(x^2) =(f(x))^2$

Let a function $f \colon \mathbb{R} \to \mathbb{R}$ satisfy the above statements. Then prove whether the fuction is odd or even.

The attempt at a solution:

using 2) we get
a) $f(0) = 0,1$ and
b) $f(1) = 0,1$
putting $x = 0$ in the 1st equation we get $f(0) = 0$ and $f(1) = 1$. From this we can prove $f(-1) = -1$ and for integers we get $f(-x) = -f(x)$. But how to prove for all real $x$?

I think the function is increasing from the first condition. So it may not be even. — eccstartup, Jun 15 '13 at 11:31
Tag also this question be [tag:systems-of-equations] may be better. — doraemonpaul, Jul 11 '13 at 21:15

score 2 · Answer 1 · edited Jun 12 '20 at 10:38

Proposition. $f(x)=x$ for all $x\in\mathbb R$.

Proof. Let $g(x)=x-f(x)$, $Y=\{g(x)\mid x\in\mathbb R\}$ and $s=\sup Y\in\mathbb R\cup\{+\infty\}$. If $0\le x <1$, we find from $(2)$ that $ f(x)=f(\sqrt x^2)=f(\sqrt x)^2\ge 0$ and hence $g(x)<1$. By $(1)$, $g$ is periodic with period $1$ so that we have $g(x)<1$ for all $x\in\mathbb R$ and hence $s\le 1$ (especially, $s$ is finite). As an intermezzo, we show a little

Lemma. If $y\in Y$ and $a\in\mathbb R$, then there exists $x\in[a,a+1)$ such that $(2x-y)y\in Y$.

Proof: Since $g$ is periodic with period $1$, there exists some $x\in[a,a+1)$ with $g(x)=y$. We compute $(2x-y)y=(x+f(x))(x-f(x))=x^2-f(x)^2=x^2-f(x^2)=g(x^2)\in Y$. $_\square$

Back to the proof of the proposition. Assume $Y\ne \{0\}$. Let $y\in Y\setminus\{0\}$. If $y>0$ then immediately $s\ge y>0$. If $y<0$, let $a=\frac y2-1$ in the lemma and obtain $2x-y<0$, hence again $s\ge (2x-y)y>0$. Therefore, we have $0<s\le 1$. Select $y\in Y$ with $y>\frac s2>0$ and let $a=1+\frac y2$ in the lemma, we obtain $2x-y\ge 2$, i.e. the contradiction $s\ge (2x-y)y\ge2y>s$. We conclude that $Y=\{0\}$, i.e. the proposition. $_\square$

f(x)=0, f(x)=x, f(x)=1 and f(x)={x1if x≠−1,if x=−1 are solutions to (2). how did u get this..? — maths lover, Jun 15 '13 at 11:47
@tomasz As it turns out, using both conditions simultaneously heavily reduces the number of solutions. :) — Hagen von Eitzen, Jun 15 '13 at 16:34

M.H · Answer 2 · 2013-06-15T12:50:47.183

0

Hint:from second condition $$\begin{cases} f((-x)^2)=f(x^2)=(f(-x))^2 \\ f(x^2)=(f(x))^2 \\ \end{cases}\color{red}{\Rightarrow } 0=f(x^2)-f(x^2)=(f(x))^2-(f(-x))^2\color{red}{\Rightarrow } $$ $$(f(-x))^2=(f(x))^2\to f(x)=\pm f(-x)$$$$\color{red}{\Rightarrow }\begin{cases} f(x)=f(-x) \color{green}{(1)}\\ f(x)=-f(-x) \color{green }{(2)}\\ \end{cases}$$ if we let $\color{green}{(1)}$ be true but there are counter example like $f(x)=x^2$ such that don't satisfy in $f(x+1)=f(x)+1$

and if we let $\color{green}{(2)}$ be true but there is odd function like $f(x)=x^3$ such that don't satisfy in $f(x+1)=f(x)+1$

edited Jun 15 '13 at 12:50

answered Jun 15 '13 at 12:08

M.H

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what if f(x) = -f(x)?? – maths lover Jun 15 '13 at 12:21
it works for both condiftions – chenbai Jun 15 '13 at 12:23
what do we get from this approach.? – maths lover Jun 15 '13 at 12:27
@chenbai f(1/2)=f(−1/2)+1≠f(−1/2). hence f(-x)≠f(x) – maths lover Jun 15 '13 at 12:28
so what is the final answer – maths lover Jun 15 '13 at 12:42
only f(x）=-f(x) can be possible satisfy both conditions. and f(x)=x is a perfect one. – chenbai Jun 15 '13 at 13:00
@ chenbai : notice $f(x)=x^2 $ don't satisfy in both condition – M.H Jun 15 '13 at 13:18

Is a function $f$ satisfying $f(x+1) = f(x) + 1$ and $f(x^2) = (f(x))^2$ odd or even?

2 Answers2