de Fermat's last theorem

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	Author	Topic: de Fermat's last theorem (Read 677 times)
		Description: A trigonometric approach
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jehovajah

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de Fermat's last theorem

« on: February 11, 2011, 10:55:54 AM »

$x^n+y^n \neq z^n$

Reduces to $\frac{x^n}{ z^n}$ + $\frac{y^n}{ z^n}$ $\neq$ $1^n$ in the unit circle.

This is the same as $cos^n\theta +sin^n\theta\neq 1^n$

Which can be written as $cos^n\theta +sin^n\theta\neq (cos^2\theta +sin^2\theta)^n$

Which by de Moivre reduces to $cos^n\theta +sin^n\theta\neq (cos n\theta +i*sin n\theta)(cos n\theta - i*sin n\theta)$

You can explore these relations graphically here.

Post what you find out, think, render, or create.

Have fun dancing banana


« Last Edit: April 06, 2012, 06:57:38 AM by jehovajah »	Logged

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jehovajah

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Re: de Fermat's last theorem

« Reply #1 on: February 15, 2011, 11:02:21 PM »

$x^n+y^n \neq z^n$

This is the same as $cos^n\theta +sin^n\theta\neq 1^n$

Which can be written as $cos^n\theta +sin^n\theta\neq (cos^2\theta +sin^2\theta)^n$

Which can be expressed $(e^{i\theta}-isin\theta)^n +(-ie^{i\theta}+icos\theta)^n\neq (e^{in\theta})(e^{-in\theta})$
which can be arranged as

$(e^{i\theta}-isin\theta)^n +i^n(cos\theta-e^{i\theta})^n\neq (e^{in\theta})(e^{-in\theta})$

$(e^{i\theta}-isin\theta)^n +i^n(cos\theta+e^{-i\theta})^n\neq (e^{in\theta})(e^{-in\theta})$

You can explore these relations graphically using any fractalgenerator.

Indulge me, someone?

Post what you find out, think, render, or create.

Have fun dancing banana


« Last Edit: November 12, 2011, 10:13:39 AM by jehovajah »	Logged

jehovajah

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Re: de Fermat's last theorem

« Reply #2 on: February 17, 2014, 09:23:20 AM »

I think I have made a mistake in one of the conversions.

Anyone care to find it? embarrass


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jehovajah

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Re: de Fermat's last theorem

« Reply #3 on: March 01, 2014, 10:36:54 AM »

z = x + y

Can be understood as z is a binomial form.
Then zⁿ can be expanded by the binomial theorem.

We find then that zⁿ can never equal xⁿ + yⁿ.

However we do have a special case, proved by Pythagoras

z² = x² + y² + 2xy by the binomial expansion
= c² + 2xy by the Pythagoras theorem , using the proof of embedding in a larger square.

What this means is, by Pythagoras we can always find some c² that sums the 2 squares . We have no such identity for n>2.


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thanks and see you perhaps in 10 years again