Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $f(f(x))=x+2$ does it necessarily follow that $f(x)=x+1$? This question comes from a precalculus algebra student.

share|cite|improve this question
Are you working with functions from $\mathbb R$ to $\mathbb R$? From $\mathbb Z$ to $\mathbb Z$? – Martin Sleziak Nov 6 '12 at 11:05
I am very impressed by the ingenuity of the answers. Many thanks. Perhaps I should add that the question was raised by a student of mine. – user48432 Nov 6 '12 at 19:38
you should accept an answer – miracle173 Jan 4 '13 at 19:46

The answer to your question is no. For example, if you are working with function $f\colon\mathbb Z\to\mathbb Z$ then $$f(x)= \begin{cases} x-1; & x\text{ is even}, \\ x+3; & x\text{ is odd}; \end{cases} $$ is an example of a different function such that $f(f(x))=x+2$.

If you want the function $f\colon\mathbb R\to\mathbb R$, you can simply extend this one by putting $f(x)=x+1$ for $x\notin\mathbb Z$.

If you would like to see a continuous solution different from $x+1$, you could use the piecewise linear function obtained by putting $0\mapsto2/3, 2\mapsto2+2/3, 4\mapsto2+2/3, \dots$ and $2/3\mapsto 2, 2+2/3\mapsto 4, 4+2/3\mapsto 6,\dots$

The composition $f\circ f$ will be again piecewise linear and to check that $f(f(x))=x+2$ you only need to verify this for $x\in2\mathbb Z$ and $x\in\frac23+2\mathbb Z$.

continuous monotone solution

The intuition behind this is that the interval $[0,2/3]$ is stretched to the interval $[2/3,2]$ (which has twice the length of the original interval) and $[2/3,2]$ contracted to $[2,2+2/3]$. The same thing is done on other intervals $[2k,2(k+1)]$, $k\in\mathbb Z$.

share|cite|improve this answer
Would the claim $f(f(x))=x+2 \,\rightarrow \, f(x)=x+1$ be true in case $f$ is continuous ? how about when $f$ is a monotonic function ? – Teddy Nov 6 '12 at 11:47
@Teddy I've added an example which is monotone and continuous. – Martin Sleziak Nov 6 '12 at 11:55
I think what he's really asking about is a function that is infinitely differentiable. – MatsT Nov 6 '12 at 17:02

Answering the question in the topic: definitely not for all f. Consider f(f(x))) = x; is f(x) = x or f(x) = -x?

share|cite|improve this answer


In fact this belongs to a functional equation of the form

Let $\begin{cases}x=u(t)\\f=u(t+1)\end{cases}$ ,

Then $u(t+2)=u(t)+2$


For $u_c(t+2)-u_c(t)=0$ ,

$u_c(t)=\theta(t)$ , where $\theta(t)$ is an arbitrary periodic functions with period $2$

For $u_p(t)$ ,

Let $u_p(t)=At$ ,

Then $A(t+2)-At\equiv2$




$\therefore u(t)=\theta(t)+t$ , where $\theta(t)$ is an arbitrary periodic functions with period $2$

Hence $\begin{cases}x=\theta(t)+t\\f=\theta(t+1)+t+1\end{cases}$ , where $\theta(t)$ is an arbitrary periodic functions with period $2$

In fact $f(x)=x+1$ is only a paticular solution when taking $\theta(t)=0$ .

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.