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# 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
2 years, 11 months ago

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

$$f(1) =2$$.

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

$$f(f(2))=f(3)= 2^2 +2 =6$$. · 2 years, 11 months ago

I don't understand with your solving.. f(1)=f(2)=1/2/3=3/2=3.. am i wrong to understand?? · 2 years, 11 months ago

Sorry. I was saying $$f(1)=$$ 1 or 2 or 3. · 2 years, 11 months ago

And how you conclude that f(1)=1or2or3?? and than continue to f(1)=2? sorry i just learned this.. · 2 years, 11 months ago

I think you ought to prove $$f(n)>n$$ for this · 2 years, 11 months ago

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, $$f(1)=2$$. · 2 years, 11 months ago