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Author
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Topic: Supplementary Fermat Conjecture
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chimp
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Member # 333
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posted 12. June 2003 05:38
While at work, I scribbled this equation in the margins of my log book.
x does not equal y
[(x+y)^n + (x-y)^n ]/(x^n + y^n) = n
This equation holds true only when n = 2
True or False ?
Is this equation related to Fermat's last theorem?
Russ
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Evan
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Member # 164
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posted 12. June 2003 22:20
No to your last question, I think.
I imagine that if someone wanted to write out the general binomial expansions for (x+y)^n and (x-y)^n and combine terms, you could prove the rule fairly directly.
And, of course, this has nothing to do with ID, complexity, etc.
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chimp
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posted 13. June 2003 00:54
If there exists a mathematical structure isomorphic to our universe, exactly how do we go about discovering what the mathematical structure is? Mathematics would apparently have everything to do with ID and complexity.
What would a universe be like that didn't abide by the rule, a^2 + b^2 = c^2 ???
Some theories postulate that reality is a mathematical language based on 2-valued logic.
The question becomes, which is more basic? the syntax of the language or the invariance principles, i.e. the symmetry of the syntax?
There is no way? to prove that the 2-valued logic syntax is the most basic. Symmetry is the most basic, and the most "fundamental" ![[Wink]](wink.gif)
Russ
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Rex Kerr
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posted 13. June 2003 01:41
Diophantine equations are hard to solve, despite being deceptively easy to write down.
I wouldn't expect the above to be any easier to solve than Fermat's Last Theorem, and I'd expect any number of similarly simple-to-write equations to be equally hard to solve. There is no reason to assume that the problem is solvable, as it has been proven that there is no algorithm that will yield the solution to an arbitrary diophantine equation.
Added in edit: unless you mean that the formula holds for all x,y (not just that there exists an integer solution), in which case it is trivial because if we rewrite as
(x+y)^n + (x-y)^n = n(x^n+y^n)
then for sufficiently large x the lhs goes as 2x^n and the rhs goes as nx^n, and thus there exist values of x for every y,n such that the two are not equal when n is not 2.
[ 13. June 2003, 01:51: Message edited by: Rex Kerr ]
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Evan
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posted 13. June 2003 17:36
Assuming that n is a positive integer, it's easy to show that n = 2 is the only value for which Russel's theorem works.
Typing it all out algebraically would be way too cumbersome. Let me first show why n = 2 works, and then why n = 3 doesn't. From there, the result should be clear.
First, rewrite as (x+y)^n + (x-y)^n = n (x^n + y^n), as Rex did.
For n = 2, we have
x^2 + 2xy + y^2 + x^2 - 2xy + y^2 = 2 (x^2 + y^2)
=> 2x^2 + 2y^2 = 2(x^2 + y^2), which is true.
For n = 3, we have on the left side
x^3 + 3x^2y + 3xy^2 + y^3 + x^3 - 3x^2y + 3xy^2 - y^3
=> 2x^3 + 6xy^2 != 3(x^3 + y^3)
From this example, it is easy to see why no value other than n = 2 works. Two reasons:
1) The first term of the left hand side will always be 2x^n, never nx^n (except when n = 2)
2) Secondly, there will always be various "middle terms" in the binomial expansions that will not cancel out, so it will not be possible (except for n = 2) to get just terms involving x^n and y^n.
So Russel's theorem is not a mystery, and not related to Fermat's Last Theorem.
That's my take on it, anyway.
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Evan
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posted 13. June 2003 17:45
Russell writes,
"What would a universe be like that didn't abide by the rule, a^2 + b^2 = c^2 ???"
The Pythgorean Theorem only holds on a flat surface. It doesn't apply to triangles on a positively curved surface such as a sphere nor on a negatively curved surface.
So the answer to Russel's question would be "a non-flat universe," which might very well be how the universe is.
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chimp
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posted 14. June 2003 03:29
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. So, the good ol' Pythagorean theorem cannot be dismissed so easily.
Some truths are invented and some truths are discovered. The best truths are discovered truths. The Frey equation used in Dr. Wiles proof appears to be an invented truth. Still true, but invented nonetheless
Fermat Last Theorem:
x^n + y^n = z^n
The natural Fermat equation:
z > y > x
[ (x!)/(x-1)! ]^n! + [ (y!)/(y-1)! ]^n! - [(z!)/(z-1)! ]^n! = 0
n = 2
Russ
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Evan
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posted 14. June 2003 08:25
I'm curious, Russell - do you accept my demonstration that the formula in your opening post is easily shown to be true for n = 2 only?
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chimp
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posted 14. June 2003 20:31
Yes, thanks for the demonstration Evan.
x^n + y^n = z^n = [(x+y)^n + (x-y)^n ]/n ,
when x, y, z, n, are positive integers, z > y > x , the equation is true, only when n is 2 .
[ 14. June 2003, 20:41: Message edited by: Russell E. Rierson ]
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Rex Kerr
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posted 15. June 2003 04:39
But x^2 + y^2 = z^2 is true for only a 1D manifold in R^3. How is this related to x^2 + y^2 = ((x+y)^2+(x-y)^2)/2 for all x,y?
And it seems even less related to the existence of a nonzero intersection of the manifold of solutions to x^n + y^n = z^n and N^3 only when n<3, which is Fermat's Last Theorem.
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chimp
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posted 15. June 2003 14:16
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
Russ
[ 15. June 2003, 15:00: Message edited by: Russell E. Rierson ]
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Rex Kerr
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posted 16. June 2003 00:53
Despite clearly understanding the math, I still don't see the connection. I suspect it is a difference in how our brains perceive analogies, and won't worry about it.
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chimp
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posted 16. June 2003 14:12
x^2 + (x+1)^2 = (x+2)^2
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.
x^3 + (x+1)^3 + (x+2)^3 = (x+3)^3
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chimp
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posted 19. June 2003 12:59
3^2 + 4^2 = 5^2
x^2 + (x+1)^2 = (x+2)^2
x^2 - 2x - 3 = 0
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 physical law.
Elementary geometry:
If a polyhedron has V vertices, F faces, and E edges and is topologically equivalent to the sphere, this equation is true:
V + F - E = 2
Thanks to Rene Descartes and Leonhard Euler.
A polyhedron with the least possible number of faces is a tetrahedron. Topologically, the tetrahedron is equivalent to a sphere.
4_V + 4_F - 6_E = 2
There exists invariance principles for circle and straight edge.
Russ
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chimp
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posted 25. June 2003 02:37
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
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