Chimp

For z > y > x,

z,x,y,n are positive integers.


x^n + y^n = z^n = [(x+y)^n + (x-y)^n ]/n ,

only for n > 0 , and, n, =< 2 .


Yours truly,

Simian Mathematician

Chimp

Jinto

One problem. It should be n > 0. There is no nonzero integral solution for x^0 + y^0 = z^0.

X^0 = 1

1 + 1 != 1

Chimp

Thank you for the helpful advice Jinto.

Chimp

x^n + y^n = z^n

z > y > x

x = x

y > x

z > y

y = {(x+1), (x+2) , (x+3), ...(x+k) }

z = {(y+1), (y+2) , (y+3), ...(y+L) } = (x+k+L)

k and L are the required constants.

x^n + (x+k)^n = (x+k+L)^n

Fermat's last theorem is really a binomial expansion, when the equation is reduced to one variable.

x^n + (x+k)^n = (x+k+L)^n

x,y,z,k,L,n are positive integers

braces_for_impact

/head explodes

SimplyAtheistic

Yes, and binomial expansion takes way too much time.
Jake

Chimp

I recall that Riemann solved the problem of a non-flat intrinsically
curved surface, with the metric tensor g_uv . A powerful
generalization. Albert Einstein built on Riemann's work *with* the
tensors of Riemann. Now, the problem seems to be the unification of the Euclidean(flat space) and non-Euclidean(curved geometry) perspectives.

Some truths are invented and some truths are discovered. The best truths are discovered truths. The Frey equation used in Dr. Wiles' proof of Fermat's last theorem appears to be an invented truth IMHO. Still true, but invented nonetheless.

Fermat Last Theorem:

x^n + y^n = z^n

The natural Fermat equation:


[(x!)/(x-1)!]^2 + [(y!)/(y-1)!]^2 = [(z!)/(z-1)!]^2



3^2 + 4^2 = 5^2

x^2 + (x+1)^2 = (x+2)^2 ...corresponds to x^2 - 2x - 3 = 0

By reducing the Fermat equation to terms of one variable, we see that certain classes of polynomials are perfect squares, perfect cubes, perfect n powers.

Symmetry groups.

(1.) 3^3 + 4^3 + 5^3 = 6^3

x^3 + (x+1)^3 + (x+2)^3 = (x+3)^3 ...corresponds to 2x^3 - 12x - 18 = 0

(2.) 95800^4 + 217519^4 + 414560^4 = 422481^4

x^4 + (x+121719)^4 + (x+318760)^4 = (x+326681)^4 ...corresponds to

2x^4 + 455192x^3 + 58217860800x^2 - 2686703548661128x -
845587479769079937600 = 0

Circle and square are unified through the symmetry of mathematical law.

Russ

Chimp

The brilliant physicist Galileo Galilei discovered that the natural numbers and the squares of natural numbers, could be put into a one to one correspondence:

1--->1^2

2--->2^2

3--->3^2

n--->n^2

The two sets are equal.

Sets that are equivalent to the set of natural numbers are called denumerably infinite sets.

Dr. Georg Cantor defined an infinite set as one that can be put in one to one correspondence with itself. Cantor denoted the number of
elements in the denumerably infinite set of naturals, as the cardinal number aleph_0

The different cardinality of these infinite sets e.g. aleph_0, aleph_1, aleph_2 etc. are degrees of infinity called the "transfinite numbers".

According to Dr. Cantor,

aleph_0 + aleph_0 = aleph_0

Interesting...

Getting back to Fermat's last theorem:

x^n + y^n = z^n

cannot be solved in positive integers x,y,z

if n > 2

What about Cantor's transfinite numbers?

x^aleph_0 + y^aleph_0 = z^aleph_0 ?

True? or False?

I say it could be false but I am not quite sure

Look at the equation [a^x + b^x]^(1/x)

If c > b > a then if we take a limit as x--->oo , and oo means "infinity"

Limit
x--->oo [a^x + b^x]^(1/x) = b

So [a^aleph_0] + [b^aleph_0] does not equal
[c^aleph_0] according to standard analysis.

Russ

Jesse

If x, y, and z are integers larger than 1 it'll be true, but in a trivial way--any integer over 1 raised to the aleph_0 power will have the same cardinality, the cardinality of the set of real numbers (the 'continuum'). For example, if you look at sequences of the digits 0-9, there are 10^n possible sequences n digits long, and therefore 10^aleph_0 possible sequences aleph_0 digits long, and every real number can be represented in base 10 as such a sequence (there's not quite a one-to-one relationship between these sequences and the reals because of redundant cases like 3.99999... = 4.00000..., but it can be proved that the cardinality of all such sequences is no larger than the cardinality of the reals, so it must be equal). Of course the same would be true of base 2 or any other integer>1 base.

No one knows whether the cardinality of the continuum is equal to aleph_1 or to some other transfinite cardinal--it's not even clear if this question has a single true answer. See The Continuum Hypothesis.

lazcatluc

Hello, it should be "correspondence with a proper subset (a subset which is not the set itself)".