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?

xx, aa, bb are positive numbers such that if a>ba>b, f(a)>f(b)f(a)>f(b) and f(f(x))=x2+2f(f(x))=x^2+2. What is the value of f(3)f(3)?

This also happens to be my first note!

Thanks!

Note by Siam Habib
5 years, 8 months ago

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f(f(1))=3f(f(1))=3 => f(1)=f(1)= 11 or 22 or 33. Easy investigation gives

f(1)=2f(1) =2.

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

f(f(2))=f(3)=22+2=6f(f(2))=f(3)= 2^2 +2 =6.

Fatin Farhan - 5 years, 8 months ago

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

Alessio Di Lorenzo - 5 years, 8 months ago

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

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

Fatin Farhan - 5 years, 8 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??

Hafizh Ahsan Permana - 5 years, 8 months ago

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

Fatin Farhan - 5 years, 8 months ago

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

Hafizh Ahsan Permana - 5 years, 8 months ago

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

Peter Taylor Staff - 5 years, 8 months ago

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