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Topic: Proof by induction  (Read 1275 times)

Offline swagmaster420x

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Proof by induction
on: September 22, 2015, 08:35:33 AM
Prove (3 + sqrt(5))^n + (3 - sqrt(5))^n is divisible by 2^n for all positive integers n.

Offline josh93248

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Re: Proof by induction
Reply #1 on: September 22, 2015, 09:47:47 AM
Uh... What is the point of this?
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Offline toryka

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Re: Proof by induction
Reply #2 on: September 25, 2015, 01:46:18 PM
What is this? Looks so scary to me  :o

Offline swagmaster420x

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Re: Proof by induction
Reply #3 on: September 25, 2015, 03:28:18 PM
Lol learn math noobs w t f

Offline ted

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Re: Proof by induction
Reply #4 on: September 27, 2015, 03:53:36 AM


x^n+y^n=(x+y){x^(n-1)+y^(n-1)}-xy{x^(n-2)+y^(n-2)}

Substitute 3+sqrt5 for x and 3-sqrt5 for y, assume it is true for the first two n, and the result follows because x+y=6 and xy=4.
"Mistakes are the portals of discovery." - James Joyce

Offline swagmaster420x

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Re: Proof by induction
Reply #5 on: September 27, 2015, 08:35:02 AM

x^n+y^n=(x+y){x^(n-1)+y^(n-1)}-xy{x^(n-2)+y^(n-2)}

Substitute 3+sqrt5 for x and 3-sqrt5 for y, assume it is true for the first two n, and the result follows because x+y=6 and xy=4.
Clever , bravo !
Is that expansion something you just knew, or played around with to get?
Another solution which results in the same form is observing that 3 +- sqrt5 are the roots of
x^2 - 6x + 4.

Offline ted

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Re: Proof by induction
Reply #6 on: September 27, 2015, 09:07:04 AM
Is that expansion something you just knew, or played around with to get?

I don't know, just a guess really, although it is a standard technique to express a symmetric function in terms of the elementary symmetric functions. I might have remembered that from fifty years ago I suppose !
"Mistakes are the portals of discovery." - James Joyce
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