Interchanging mathematical operations proof

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The discussion revolves around two mathematical proofs concerning the conditions under which certain polynomial identities hold true. The first proof states that (x+y)^(2n+1) equals x^(2n+1) + y^(2n+1) only if x and y are both zero or y equals -x. The second proof claims that (x+y)^(2n) equals x^(2n) + y^(2n) if either x or y is zero. The original poster has attempted induction and summation techniques but has encountered difficulties, while also discovering a related equality that may assist in proving the first statement. A suggestion to explore differentiation as a potential method for proving these identities was made, highlighting the need for further exploration of this approach.
homegrown898
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I'm working on the following two proofs:

1.) (x+y)2n+1 = x2n+1 + y2n+1 if and only if x=0, y=0 or y=-x

and

2.) (x+y)2n = x2n + y2n if and only if x=0 or y=0

I've tried using induction and get stuck at a certain point. I've also tried playing around with summations since there is a binomial expansion in there. I haven't had any luck. Earlier today however I found that this equality is true:

x2n+1 + y2n+1 = (x+y)(x2n - x2n-1y + x2n-2y2 - ... - xy2n-1 + y2n)

I haven't played around with this yes, but I think setting this equal to (1) might help prove this.

I'm just wondering if anyone has any ideas on this.
 
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hi homegrown898! :smile:

have you tried differentiating?
 
I haven't tried that no and I'm not exactly sure what you mean
 
if eg (1 + x)6 - 1 - x6 = 0 at x = 0 and at x = something else,

then its derivative must be zero somewhere in between :wink:
 
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