A Math Problem!

coolcreep · Joined: 18 Feb 2006 Posts: 410

Ok here is the math problem:

An infinite sequence a0, a1, a2, a3... satisfies

a(m-n) + a(m+n) = a(2m) + a(2n)

for all integers with m>=n>=0

Prove that all integers in the sequence are equal.

Onyx · Joined: 16 Oct 2004 Posts: 11

www.dumb.com

Amaterasu · Joined: 14 Sep 2006 Posts: 74

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

aldaryn · Joined: 02 Sep 2004 Posts: 475

The original problem isn't clear... (m-n), (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.

Evenkeal · Joined: 20 Nov 2006 Posts: 1

Call * = "a(m-n) + a(m+n) = a(2m) + a(2n) "

Try m=n ; then a_0 = a_{2m} for all m. So all the even terms are the same. So the right side of * is constant. So a_{m-n} + a_{m+n} is constant.

So plug in m,n = m,1 ; m+1,2.
You get a_{m-1} + a_{m+1} = a_{m-1} + a_{m+3}. So a_{m+1} = a_{m+3} for all m. So all the odd terms are the same. But * also implies that two odd terms added up equal two even terms, so all the terms are the same.

blamin8or · Joined: 31 Aug 2008 Posts: 2

ck3gds · Joined: 21 Jun 2005 Posts: 255

Here I would actually use a proof by contradiction (reductio ad absurdum).

Let us assume that the statement you want to prove is false - i.e., "Not all integers in the sequence are equal".

This implies that there exists a value i ,where i >= 0, such that a_0 = a_1 = ... = a_(i-1) = a_(i+1) = a_(i+2) = ... = k for any arbitrary constant k, and a_i != k.

Without loss of generality we can let any term in the equation be a_i. Let's say we let i = m - n. Then we have

a_i + a_(m+n) = a_2m + a_2n
=> a_i + k = k + k
=> a_i = k + k - k = k

But this contradicts what we assumed, that a_i != k. Therefore, our assumption must have been wrong, and so the negation of our assumption must be true.

NOT("Not all integers in the sequence are equal.") = "All integers in the sequence are equal."

(QED)

coolcreep · Joined: 18 Feb 2006 Posts: 410

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_(2m-1). 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.