
Author 
Message 
coolcreep
Joined: 18 Feb 2006 Posts: 588

Posted: Wed Nov 19, 2008 9:33 pm Post subject: A Math Problem! 


Ok here is the math problem:
An infinite sequence a0, a1, a2, a3... satisfies
a(mn) + a(m+n) = a(2m) + a(2n)
for all integers with m>=n>=0
Prove that all integers in the sequence are equal. 

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Onyx
Joined: 16 Oct 2004 Posts: 11


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Amaterasu
Joined: 14 Sep 2006 Posts: 74

Posted: Wed Nov 19, 2008 9:49 pm Post subject: 


...if n can = 0...then, damn it...it's 0...which makes the problem rather easy :/ 

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aldaryn
Joined: 02 Sep 2004 Posts: 501

Posted: Wed Nov 19, 2008 10:02 pm Post subject: 


The original problem isn't clear... (mn), (m+n), (2m), and (2n) are subscripts, right? I think that is where Amaterasu got confused, thinking it was normal multiplication.
Anyway, are you trying to prove that the actual integers IN the sequence are equal, i.e. a0=a1=a2=....? Your wording is not very clear. 

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blamin8or
Joined: 31 Aug 2008 Posts: 17

Posted: Thu Nov 20, 2008 2:48 am Post subject: 


aldaryn wrote:  are you trying to prove that the actual integers IN the sequence are equal, i.e. a0=a1=a2=....? Your wording is not very clear. 
coolcreep wrote:  Prove that all integers in the sequence are equal. 
lol
i think the key is that all integers m>=n. Setting m=n if the series is divergent it breaks and if it is convergent it breaks so it must be neither. 

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coolcreep
Joined: 18 Feb 2006 Posts: 588

Posted: Thu Nov 20, 2008 12:30 pm Post subject: 


I used a proof that was basically the same one that Evenkael used, first showing that a_0=a_2m, then showing that a_0=a_(2m1). The problem with the proof by contradiction is that you do not justify changing a_(m+n), a_2m and a_2n all to k. By doing so you are assuming that all terms excluding a_i are equal, which is too narrow an assumption. 

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