# A difficult problem on functions

1. Jun 20, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

I've been trying to solve the following problem but can't wrap my head around it.

Let $x$, $f(x)$, $a$, $b$ be positive integers. Furthermore, if $a > b$, then $f(a) > f(b)$. Now, if $f(f(x)) = x^2 + 2$, then what is $f(3)$?

2. Relevant equations

3. The attempt at a solution

Well, the table illustrates my current understanding of the problem:

x-------------------f(x)-------------------f(f(x))
1-------------------??-------------------3
2-------------------??-------------------6
3-------------------??-------------------11
4-------------------??-------------------18
5-------------------??-------------------27
6-------------------??-------------------38
7-------------------??-------------------51
8-------------------??-------------------66
9-------------------??-------------------83
10-------------------??-------------------102

$f(x)$ is a monotonic function, and is allowed to take integer values, so this means that the values in the third column must be interspersed in the second column in the order in which they appear in the third column.

That's how far I can take my intuition.

2. Jun 20, 2015

### micromass

What is $f(1)$? There is only one possibility

3. Jun 22, 2015

### RUber

In other words, what is f(f(1))?

4. Jun 22, 2015

### Fredrik

Staff Emeritus
It's also relevant that the formula $f(f(x))=x^2+2$ makes sense for all x only if the range of f is a subset of the positive integers. So f(1) must be a positive integer. Try a few guesses: f(1)=1, f(1)=2, ... You should find that all but one of the assumptions you try lead to a contradiction. Then you should try to prove that all other values of f(1) lead to a contradiction. Once you have found the only possible value of f(1), it's pretty easy to use it to find f(3).

5. Jun 22, 2015

### pasmith

Hint: Show that $f(1) = 1$ is impossible, so that $1 < f(1)$. Why does it follow that $f(1) < f(f(1))$?

6. Jun 22, 2015

### RUber

if $f(f(1))=3$, then $f(3) = f(f(f(1))) = (f(1))^2+2$

7. Jun 22, 2015

### Fredrik

Staff Emeritus
Now we have told him the entire solution to the problem except for how to prove that f(1)=1 is impossible and how to compute $2^2+2$. Both of these steps are pretty trivial.

8. Jun 22, 2015

### Mentallic

You're absolutely right!

Hint: 2^2 = 2*2 = 4.

9. Jun 23, 2015

### spaghetti3451

Thanks! I've got it.

10. Jun 23, 2015

### epenguin

Don't tell us you have got the answer, tell us the answer you have got!

11. Jun 26, 2015

### Noctisdark

Suppose that f(a) < a for some admissible a, then f(a-1)<f(a) < a, then f(f(a-1)) < f(a), (a-1)^2 + 2 < a so a^2 - 2a + 1 +2 < a, a^2 - 3a +2 < 0, some algebra and end up with (a-3/2)^2 < 1/4, a < 2.15 so a <2, by that fact f(3) > 3, then 3 > f(1), which leaves f(1) = 0 which will make f(0)<0, f(1) = 1 so will yield to 3 = f(1) which is non sense and we re left with f(1) =2, by that fact f(3) = 5 !

12. Jun 27, 2015

### willem2

The fact that f(x) is positive and strictly increasing already implies f(a) >= a for all a.
you haven't eliminated the possibility that f(3) = 3.
I don't really understand this.

[/QUOTE]
You better check your arithmetic here.