A difficult problem on functions

[spaghetti3451]

Homework Statement

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 $f(3)$?

Homework Equations

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.

 

[micromass]

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

 

[RUber]

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

 

[Fredrik]

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).

 

[pasmith]

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

 

[RUber]

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

x-------------------f(x)-------------------f(f(x))
1-------------------??-------------------3

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

 

[Fredrik]

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.

 

[Mentallic]

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.

You're absolutely right!

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

 

[spaghetti3451]

Thanks! I've got it.

 

[epenguin]

Thanks! I've got it.

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

 

[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 !

 

[willem2]

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

The fact that f(x) is positive and strictly increasing already implies f(a) >= a for all a.

, by that fact f(3) > 3,

you haven't eliminated the possibility that 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 were left with f(1) =2

I don't really understand this.

and we re left with f(1) =2 by that fact f(3) = 5 !

[/QUOTE]
You better check your arithmetic here.