This is supposed to be one of the easiest problems from BdMO 2014. But I can't seem to find a rigorous way of solving this problem. Am I missing something?

\(x\), \(a\), \(b\) are positive numbers such that if \(a>b\), \(f(a)>f(b)\) and \(f(f(x))=x^2+2\). What is the value of \(f(3)\)?

This also happens to be my first note!

Thanks!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest\(f(f(1))=3\) => \(f(1)=\) \(1\) or \(2\) or \(3\). Easy investigation gives

\(f(1) =2\).

\(f(f(1))= f(2)= 3\).

\(f(f(2))=f(3)= 2^2 +2 =6\).

Log in to reply

I don't understand with your solving.. f(1)=f(2)=1/2/3=3/2=3.. am i wrong to understand??

Log in to reply

Sorry. I was saying \(f(1)=\) 1 or 2 or 3.

Log in to reply

Log in to reply

I think you ought to prove \(f(n)>n\) for this

Log in to reply

If \(f(1)= 1\) then \(f(f(1))= f(1) = 1\). But \(f(f(1))=3\) If \(f(1)=3\) then \(f(f(1))=f(3) = 3\) So, \(f(f(3)) = f(3)= 3\) But \(f(f(3))=11\)

So, \(f(1)=2\).

Log in to reply

Congrats on your first note Siam! It was a good one.

Log in to reply