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BdMO Online Forum • View topic - Easy FE

## Easy FE

For discussing Olympiad Level Algebra (and Inequality) problems

### Easy FE

Find all function $f: N \rightarrow N$ such that ,
$f^3(1) + f^3(2) + ... + f^3(n) = (\frac{f(n)f(n+1)}{2})^2$ and $f(1) = 1$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Absur Khan Siam

Posts: 52
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Re: Easy FE

$f^3(1) + f^3(2) + ... + f^3(n) = (\frac{f(n)f(n+1)}{2})^2...(i)$
$f^3(1) + f^3(2) + ... + f^3(n) + f^3(n+1)= (\frac{f(n+1)f(n+2)}{2})^2 ...(ii)$

$(ii) - (i) \Rightarrow f(n+1) = \frac{f^2(n+2)}{4} - \frac{f^2(n)}{4} = (\frac{f(n+2) + f(n)}{2})(\frac{f(n+2) - f(n)}{2})$

Since, $f(n+1) > 0 \Rightarrow f(n+2) > f(n)$

This , implies that $f(1) < f(2) < ... < f(n)$

Let, $f(n+2) = f(n) + k$

if $k < 2 , f(n+1) < f(n)$ ; contradiction
if $k > 2 , f(n) > f(n+2)$ ; contradiction

Thus , $k = 2 , f(n+1) = f(n) + 1$

$f(1) = 1$.By induction , we can say that $f(n) = n$.

$\therefore f(n) = n$.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Absur Khan Siam

Posts: 52
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka