Fermat's Last Theorem - Inside Your Calculator

[ThinkTank255]

I think I am very close to a proof of FLT, and something actually much greater based on a new type of mathematics I have discovered (which I found by examining the works of Fermat closely), but 500 characters is too short to contain it.

[madbowler6]

where is part 2?

[SNUGandSESOR]

n≥3 ≠ n>2 FYI

[AllHailZeppelin]

@SNUGandSESOR for integers they're the same

[LovelyYTRocks]

...And where is the second part to this video?

[972trev]

I ran across this, don't understand it or understand the significance? I was searching for the number 3 which is the first fermat's number, he just said fermat's made it as a (what they thought) joke. After so long, he just goes on and on and on and on and on.....please someone tell me why would this be necessary to learn?

[TLucretiusCarus]

Python is a wonderful computer language that works really well with really large numbers.

[insrisg]

Where can I get more information about python? G Rising

[coooldude777]

@TLucretiusCarus

are you stupid? A programming language itself has nothing to do with "how well" it works with numbers

nice try n00bie

[TLucretiusCarus]

@coooldude777 - I was merely stating that it handles large values and will calculate the large products in his sample. With a good library, BASIC could do the same thing. e.g. In Python, you could print 5034958902834239048902384**292­9348 without an issue. - So it was merely a statement. - BTW, calling others stupid wasn't out of line but calling me a noob is outright slanderous! :)

[orion285]

what's with you people? I liked this response

[NiteAngel]

could someone show just a simple proof just to understand this stuff

[subatomicparticlephy]

@NiteAngel That's the point, it's a seemingly simple question and yet there is no simple proof. In fact it takes some of the most complex mathematics in the world to prove this, and was only proven very recently

[pyramidiot1]

I solved Fermat's Next To Last theorem on my calculator: 69 * 69 / 69 + 69 - 69 = 69

[972trev]

@pyramidiot1 simple terms.1*1/1+1-1=1......let switch it backwards 1-1+1/1*1=1........lets mix it.....1+1*1-1/1=1 why is this significant? 

[jahdallah]

The following is a correction to my other comment below:

If n is a positive irrational number, then there are three unique positive rational numbers x, y and z such that gcd(denominator of x, denominator of y, denominator of z) = 1, gcd(numerator of x, numerator of y, numerator of z) = r for some positive integer r and x^n + y^n = z^n.

[jahdallah]

The following is a correction to my other comment below:

If n is a positive irrational number, then there are three unique positive rational numbers x, y and z such that gcd(denominator of x, denominator of y, denominator of z) = 1 and x^n + y^n = z^n.

[jahdallah]

If n is a positive irrational number, then the equation x^n + y^n = z^n has solutions in rational numbers x, y and z. (This means that the set of irrational numbers is countable).

If a positive real number n is different from 2 and 1, then if (x1, y1, z1) and (x2, y2, z2) are any two solutions of the equation x^n + y^n = z^n, then x1/y1 = x2/y2, x1/z1 = x2/z2 and y1/z1 = y2/z2.

[jahdallah]

1- If n is a positive irrational number, then there are three unique positive rational numbers x, y and z such that gcd(x, y, z) = 1 and x^n + y^n = z^n.

2- For every three positive rational numbers x, y and z, if x + y is not equal to z, z is greater than both x and y, and x, y, and z are not the lengths of the sides of a right triangle, then there is a positive irrational number n such that x^n + y^n = z^n.

If 1- and 2- are true, then the set of irrational numbers is countable.

[jahdallah]

1- Let n be a positive irrational number. There are three unique positive rational numbers x, y and z such that gcd(x, y, z) = 1 and x^n + y^n = z^n.

2- Let x, y and z be positive rational numbers such that x+y is not equal to z, z is greater than both x and y, and x,y, and z are not the lengths of the sides of a right triangle. There is a positive irrational number n such that x^n + y^n = z^n.

If 1- and 2- are true, then the set of irrational numbers is countable.

[hypeblast]

whats the answer???

x^2+y^2=z^2

[Entertainmentwf]

any pythagorean triple can satisfy that solution.