Nice and hard problem!

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marfak
Posts: 1
Joined: Sun May 07, 2017 6:15 pm

Nice and hard problem!

Unread post by marfak » Sun May 07, 2017 6:36 pm

Let $f(n)$ polynomial (not costant) with integer coefficients and:
If n is an odd number $2f(n)^{4}=f(f(n)^{2}-458)+1832f(n)^{2}$

If m is an even number
$f(m)-f(1)$ is multiple of $m+1$.
Find $f(457)+f(459)+f(461)$

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asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
Location: Sylhet,Bangladesh

Re: Nice and hard problem!

Unread post by asif e elahi » Mon May 15, 2017 11:24 am

From the first equation if $y=f(n)^2-458$, then $f(y)=2y^2-2.458^2$
We can rewrite it as $f(y)-2y^2+2.458^2=0$. Suppose $g(y)=f(y)-2y^2+2.458^2$. So $g(y)$ is a polynomial and it has infinite zeros (obviously $f(n)^2-458$ can take infinite values). So $g(y)$ must be zero for all $y$. The rest is easy.

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