USAMO 2008: Problem 1

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sourav das
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USAMO 2008: Problem 1

Unread post by sourav das » Wed Sep 28, 2011 7:15 pm

(Titu Andreescu) Prove that for each positive integer $n$, there are pairwise relatively prime integers $k_0, k_1 ... k_n$, all strictly greater than 1, such that $k_0 k_1 .... k_n -1$ is the product of two consecutive integers.
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Re: USAMO 2008: Problem 1

Unread post by *Mahi* » Thu Sep 29, 2011 10:32 pm

We will use induction on $k_n$. Let there be $k_0,k_1,k_2..k_{n-1}$ satisfying the given condition. So we can write $\prod k_i = a^2+a+1$. Now , some $k_n$ satisfying the condition $k_n \prod k_i = (a+m)^2+(a+m)+1$ for some $m \in \mathbb N$ and $(\prod k_i , k_n)=1$ will prove the fact.
Now we can write $(a+m)^2+(a+m)+1=a^2+a+1 + m(2a+ m +1)$ . So it is enough to set $m= a^2+a+1$, which will make $k_n$ an integer. Setting $m=a^2+a+1$ , we will get $k_n= a^2+a+1 +2a+2=a^2+3a+3$.Now as $(a^2+a+1,a^2+3a+3)=1$, $k_n=a^2+3a+3$ is a solution.
Indeed, $(a^2+a+1)(a^2+3a+3)= (a+1)^4+(a+1)^2+1$,so the proof is complete.
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zadid xcalibured
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Re: USAMO 2008: Problem 1

Unread post by zadid xcalibured » Sun Apr 07, 2013 5:18 pm

$\prod k_i = a^2+a+1$ now we can take $k_n=a^2-a+1$ satisfying $(\prod k_i,k_n)=1$ and $(a^2+a+1)(a^2-a+1) =a^4+a^2+1$

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