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Author
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Topic: Fermat
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chimp
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Member # 333
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posted 12. April 2004 14:21
5^1 = 1*0 + 5
5^2 = 2*10 + 5
5^3 = 3*40 + 5
5^p = p*a + 5
x^p = p*a + x
x^p = p*a + x
y^p = p*b + y
z^p = p*c + z
...x^p + y^p = z^p
p*a + x + p*b + y = p*c + z
p*[a + b - c] = z - [x + y]
p = [z - (x + y)]/[a + b - c]
[ 12. April 2004, 14:22: Message edited by: Russell E. Rierson ]
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chimp
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posted 13. April 2004 02:06
p, x, y, a,b,c, are always integers > 0, when p is a prime number. When p is 1, a = 0, b = 0, and c = 0.
z and c cannot be a +integer for p > 2 ...?
[x^p - x]/p = a
[y^p - y]/p = b
[z^p - z]/p = c
for example:
[3^2 - 3]/2 = 3
[3^3 - 3]/3 = 8
[3^5 - 3]/5 = 48
[+integer^p - +integer]/p = another +integer.
It works for all prime numbers.
3^2 + 4^2 = 5^2
2*3 + 3 = 3^2
2*6 + 4 = 4^2
2*3 + 4*3 + 3 + 4 = 6*3 + 7 = 5^2
--------------------------------------------------------------------------------
[ 13. April 2004, 02:49: Message edited by: Russell E. Rierson ]
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Evan
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posted 13. April 2004 13:06
In what way does the opening post meet the guidelines for Brainstorms? Just wondering?
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Evan
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posted 13. April 2004 23:25
Russell writes,
quote:
5^1 = 1*0 + 5
5^2 = 2*10 + 5
5^3 = 3*40 + 5
5^p = p*a + 5
x^p = p*a + x
x^p = p*a + x
Two questions:
1) In respect for the sequence for 5^p, what is the formula or rule for a, or is it just whatever number works to make the equation 5^p = p*a + 5 true?
And is there something special about p = a prime? For instance, the next terms in the sequence look like they would be 5^4 = 4*155 + 5 and then 5^5 = 5*624 +5, but so what. Is there anything special about the seuence 0, 10, 40, 155, 624? other than the fact that they are the numbers that make things work?
2) In the formula x^p = p*a + x, the number a is not a constant for different values of x. Examples
5^3 = 3*40 + 5
but
4^3 = 3*20 + 4
So the formula x^p = p*a + x is not really valid because the number a is different for all possible cases.
So what is the point of all this? It doesn’t even seem like valid number theory?
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Rex Kerr
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posted 14. April 2004 00:33
I'm also not sure what the point of this is.
It's a pretty simple to see why n^p - n is divisible by p if you know the binomial distribution. C(p,k) = p!/((p-k)! * k!) = 1 if k=0 or k=p, and has a leading term of p otherwise--the divisor has no factor as large as p, and p is prime. This is why p has to be prime.
As the simplest example for a non-prime, note that 2^4 - 2 = 16 - 2 = 14 = 2*7, not divisible by 4!
Anyway, we have n^p = (1+(n-1))^p = 1^p + p*(stuff) + (n-1)^p = 1 + p*(stuff) + (n-1)^p. Repeat on the (n-1)^p to get n^p = 1 + p*(stuff) + 1 + p*(other stuff) + (n-2)^p = 2 + p*(junk) + (n-2)^p. If we repeat k times, p^n = k + p*(blah) + (n-k)^p. Let k = n-1 and we get n^p = (n-1) + p*(whatever) + 1 = n + p*(whatever).
We could write out a hideous set of sums for "whatever", but it's just some number--call it A, and thus for any prime p and number n there exists an A such that n^p - n = p*A. A is therefore a function of n and p.
Fermat's last theorem states that x^p + y^p = z^p cannot be satisified by integers if p is greater than 2. Plug in p*A(x) + x for x^p and FLQ becomes p*A(x,p) + x + p*A(y,p) + y = z + p*A(z,p). The horrible mess which is A(.) cannot satisfy this equation either for p greater than 2 regardless of the positive integer values of x,y,z. So we have learned a tiny bit about a horrible mess that we didn't even want to write down.
If you're really a stickler for punishment, the formula for A(n,p) is
code:
sum( k = 2 to n , sum( i = 1 to p-1 , k^i*(p-1)!/(i!*(p-i)!) ) )
Yay, I suppose.
[ 14. April 2004, 00:35: Message edited by: Rex Kerr ]
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chimp
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posted 14. April 2004 01:15
If p = 2 :
x^2 = 2*[1 + 2 + 3+...+ x-1] + x
2^2 = 2*[1] + 2
3^2 = 2*[1 + 2] + 3
4^2 = 2*[1 + 2 + 3] + 4
5^2 = 2*[1 + 2 + 3 + 4] + 5
etc...
This congruence does not hold for p > 2
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chimp
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posted 14. April 2004 03:34
5^2 = 2*[1 + 2 + 3 + 4] + 5
5^2 = 2*[1 + 2] + 2*[3 + 4] + 5
5^2 = 2*[1 + 2] + 2*[3 + (4 - 1)] + 5 + 2*1
5^2 = 2*[1 + 2] + 2*[3 + 3] + 7
5^2 = 2*[1 + 2] + 3 + 2*[ 1 + 2 + 3] + 4
5^2 = 3^2 + 4^2
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Evan
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posted 14. April 2004 07:32
Both Rex and I have pointed out that there is nothing special about "a" in your original post. The pattern exists only because you write a for a certain coefficient that really doesn't tell us anything. (Rex gave the formula for it, but the formula is really just a backwards consequence of using a to stand for a whole bunch of missing information.)
So I wonder why you are posting: no connection to ID, not a "brainstorm", and not really even very meaningful mathematically.
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chimp
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posted 15. April 2004 01:38
Trying to figure out the nature of space-time DOES require brainstorming...
Fermat's theorem must obey relativity in real space, because measured lengths are not absolute. Space-time is a non-Euclidean geometry.
Rex did not provide any real counter argument with his "A = 'stuff' " demonstration. Since we are discussing prime numbers, we are also discussing the Riemann hypothesis, and consequently, the "Hilbert-Polya conjecture" :
http://en.wikipedia.org/wiki/Riemann_hypothesis
quote:
Dyson saw that the statistical distribution found by Montgomery was exactly the same as the pair correlation distribution for the eigenvalues of a random Hermitian matrix. Subsequent work has strongly borne out this discovery, and the distribution of the zeros of the Riemann zeta function is now believed to satisfy the same statistics as the eigenvalues of a random Hermitian matrix, the statistics of the so-called Gaussian Unitary Ensemble. Thus the conjecture of Polya and Hilbert now has a more solid basis, though it has not yet led to a proof of the Riemann hypothesis.
Rex has argued in the past, that mathematics does NOT correspond to physical existence.
Yet There appears to be a mathematical correspondence of abstract and concrete, even via the properties of randomness?... ![[Wink]](wink.gif)
Some mathematicians even think it is possible that randomness exists at the foundations of even mathematics.
Now it appears that while two integers squared can be combined to form a third integer squared, two higher powers > 2, cannot be combined to form a third higher power.
I explained what "a" was for p = 2, to Evan, but here it is again:
x^p = p*a + x
3^2 = 2*[1+2] + 3
4^2 = 2*[1+2+3] + 4
etc...
For higher powers 2*n
for example:
3^4 = 9^2 = 2*[1+2+3+4+5+6+7+8] + 9
3^6 = [3^3]^2 = 27^2 = 2*[1+2+...+26] + 27
x^[2*n] =
[x^n]^2 = 2*[1+2+...+(x^n) - 1] + x^n
The metric space has distance function r(x,y), definitively characterized by involvement with the real numbers, R, such that the metric space and R are embedded simultaneously in the full structure of manifold M. A topological space consists of sets of points which are defined[in this case] to be the intersections of cotangent bundles.
If f(x) is a homeomorphism from T onto S, and, for every point p in T, f(U(p)) = U(f(p)), the relations can be invariant under standard topological transformations.
If the world is actually "2 dimensional", how could we tell the difference?
1 + 2 + 3
2 + 3 + 4
3 + 4 + 5
____________
6 + 9 + 12 = 3^3 = 27
A "square" representation of a cube
There is no way to divide the above representation into two separate cubes.
Here is a "square" 6^3
1+2+3+4+5+ 6
2+3+4+5+6+ 7
3+4+5+6+7+ 8
4+5+6+7+8+ 9
5+6+7+8+9+10
6+7+8+9+10+11
The sum:
1+2+3
2+3+4
3+4+5
+
1+2+3+4
2+3+4+5
3+4+5+6
4+5+6+7
+
1+2+3+4+5
2+3+4+5+6
3+4+5+6+7
4+5+6+7+8
5+6+7+8+9
equals 6^3
[ 15. April 2004, 04:29: Message edited by: Russell E. Rierson ]
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Rex Kerr
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posted 16. April 2004 03:49
I wasn't posting a counterargument. I intended to (1) demonstrate that (n^p - n)/p is an integer (and that this is nearly trivial as far as number theory proofs go), and (2) demonstrate that this number is not a particularly elegant quantity in general.
You have noticed that A(n,2) = sum(i=1..(n-1),i) = n(n-1)/2. However, this is not generally true; for example, A(n,3) is not sum(i=1..(n-1),i^2):
sum(i=1..(n-1),i^2) = n(2n-1)(n-1)/6 = n^3/3 - n^2/2 +n/6 which is not n^3/3 - n/3.
Writing n^3 as a square is pretty neat. I hadn't noticed that before. Sadly, it doesn't seem to generalize. I don't think it has much to do with the nature of space-time, though. The Riemann hypothesis, Hilbert-Polya conjecture, and the like, are the result of deep thought from talented mathematicians studying advanced mathematics--and such is well beyond the capability of HTML to adequately express, even in the event that someone is able to understand what is being written.
[ 16. April 2004, 03:52: Message edited by: Rex Kerr ]
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chimp
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posted 17. April 2004 01:18
quote:
Rex Kerr wrote:
Writing n^3 as a square is pretty neat. I hadn't noticed that before. Sadly, it doesn't seem to generalize. I don't think it has much to do with the nature of space-time, though.
This is not a counter argument, ...again.
The volume of a 3 dimensional space, "n^3" , is the sum of the elements in a 2 dimensional[square] array, which is the scalar product of two n+k dimensional vectors.
1+2+3 = 6
2+3+4 = 9
3+4+5 = 12
6+9+12 = 27 = 3^3
< 1, 2, 3, 4, 5 >*< 1, 2, 3, 2, 1> =
1*1 + 2*2 + 3*3 + 4*2 + 5*1 = 27 = 3^3
1+2+3+4 = 10
2+3+4+5 = 14
3+4+5+6 = 18
4+5+6+7 = 22
10 + 14 + 18 + 22 = 64 = 4^3
<1,2,3,4,5,6,7>*<1,2,3,4,3,2,1> =
1*1+2*2+3*3+4*4+5*3+6*2+7*1 = 64 = 4^3
[ 17. April 2004, 01:43: Message edited by: Russell E. Rierson ]
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Rex Kerr
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posted 17. April 2004 02:36
It's not supposed to be a counterargument, again. It only works for p=3, not for arbitrary p, and not for nonprime powers.
Are the vectors (1,2,3,4,5) and (1,2,3,2,1) supposed to have something more to do with spacetime than, say, (1,0,0,0,0) and (0,1,0,0,0)?
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chimp
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posted 17. April 2004 05:00
I sent the "idea" to Dr. Math. It is very old news
quote:
Dr. Math:
Hi,
Is there something you are trying to prove?
Neither of the statements you have made are novel.
They are different
statements of very well known results in number
theory. And they have
very little use in actual physics.
- Doctor Roy, The Math Forum
1 2 3
2 3 4
3 4 5
gives 3^3
25 26 27
26 27 28
27 28 29
gives 3^5
79 80 81
80 81 82
81 82 83
gives 3^6
0 0 0
0 1 2
1 2 3
<1,2,3>*<2,2,1> = 1*2 + 2*2 + 3*1 = 3^2
0 0 0 0
0 0 0 0
0 1 2 3
1 2 3 4
1*2 + 2*2 +3*2 + 4*1 = 4^2
0 0 0 0 0
0 0 0 0 0
0 1 2 3 4
1 2 3 4 5
1*2 +2*2 +3*2 +4*2 +5*1 = 5^2
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Rex Kerr
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posted 17. April 2004 12:21
Yes, that's what I've been trying to say.
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chimp
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posted 18. April 2004 01:23
Three equidistant[comoving] points form an equilateral triangle ABC
Rotate the equilateral triangle to BCA, CAB, it is invariant to ABC
A B C
B C A
C A B
the invariance of rotation for comoving points A,B,C appears to correspond to an array of elements in a 2D[square] matrix. Information is encoded on the surface of space.
According to Hawking, the maximum entropy of a closed region of space cannot exceed 1/4 of the area of the circumscribing surface A/4 .
So information is stored on the 2 dimensional boundary of space analogously to the way a 3D holgram can be encoded on a 2D surface.
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