# Am I Missing Something?

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! Note by Siam Habib
7 years, 4 months ago

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## Comments

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

$f(1) =2$.

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

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

- 7 years, 4 months ago

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

- 7 years, 4 months ago

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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, $f(1)=2$.

- 7 years, 4 months ago

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

- 7 years, 4 months ago

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

- 7 years, 4 months ago

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And how you conclude that f(1)=1or2or3?? and than continue to f(1)=2? sorry i just learned this..

- 7 years, 4 months ago

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

Staff - 7 years, 4 months ago

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