Fermat's "Last Theorem"?

  • #1
Since it is known than the number N of primitive Pythagorean triples up to a given hypotenuse length A is given on average by N = Int(A/(2.pi)) and according to my calculations with primitive triples and A = B + C, I get on average N = Int(0.152 A^2) (for A = 10^3 I get N = 152,095 compare with N = 319 with pythagoras, and for 10^6 I get N = 151,981,776,195), would it be logical to conclude that for powers p then N =Int( K.A^(3-p))? Not sure how K would vary with p, probably not much variation seeing that 0.152 in not much different from 1/(2.pi). Anyway K must be less than one, which leads to N = 0 for all powers , higher than 2, as per Fermat.
 

Answers and Replies

  • #2
phyzguy
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An interesting observation, but "it seems logical to conclude" is not a proof.
 
  • #3
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I'm not sure where your statements come from (reference for ##N = O(A^2)## please!) and they don't seem to be correct either. I have found an empirical relation of
$$
N \approx 15.915\,\%\,A \approx \dfrac{1}{2\pi} A
$$
with data up to ##A=10^{10}##. But there wasn't a further explanation, and it isn't what you said. It is neither ##0.152## nor is it quadratic in ##A##.

On the other hand we have:
"Four consecutive Fibonacci numbers yield in ##(f_k \cdot f_{k+3}, 2f_{k+1}\cdot f_{k+2}, f_{2k+3})## a Pythagorean triple which is up to a factor ##2## always primitive." so there could as well be a connection to ##\pi##, the more as ##\left( \dfrac{x}{z} \right)^2+\left( \dfrac{y}{z} \right)^2=1## are certain points on the unit circle, and ##\pi## can be defined via Buffon's needle problem which connects the unit circle with integers. So the above relation seems at least plausible. As there are myriads of papers about Pythagorean triples, some have certainly investigated this relation further.

Your extrapolation to powers ##p##, however, is out of the blue and definitely no proof or whatsoever of FLT.
 
  • #4
phyzguy
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I'm not sure where your statements come from (reference for ##N = O(A^2)## please!) and they don't seem to be correct either. I have found an empirical relation of
$$
N \approx 15.915\,\%\,A \approx \dfrac{1}{2\pi} A
$$
with data up to ##A=10^{10}##. But there wasn't a further explanation, and it isn't what you said. It is neither ##0.152## nor is it quadratic in ##A##.

On the other hand we have:
"Four consecutive Fibonacci numbers yield in ##(f_k \cdot f_{k+3}, 2f_{k+1}\cdot f_{k+2}, f_{2k+3})## a Pythagorean triple which is up to a factor ##2## always primitive." so there could as well be a connection to ##\pi##, the more as ##\left( \dfrac{x}{z} \right)^2+\left( \dfrac{y}{z} \right)^2=1## are certain points on the unit circle, and ##\pi## can be defined via Buffon's needle problem which connects the unit circle with integers. So the above relation seems at least plausible. As there are myriads of papers about Pythagorean triples, some have certainly investigated this relation further.

Your extrapolation to powers ##p##, however, is out of the blue and definitely no proof or whatsoever of FLT.
I think the OP is saying that for Fermat's theorem it is linear in A, but for a "first order Fermat's theorem" (A=B+C) it is quadratic in A. So by logical extension, for higher orders it "might" go as A^(N-3).
 
  • #5
pbuk
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I think the OP is saying that for Fermat's theorem it is linear in A, but for a "first order Fermat's theorem" (A=B+C) it is quadratic in A.
Oh, is that what he was trying to say? And not only linear, but with a constant that may be tantalizingly close to the constant for pythagorean triples? That in itself would be a surprising result given that there is no apparent connection to pi for the linear equation.
 
  • #6
I think the OP is saying that for Fermat's theorem it is linear in A, but for a "first order Fermat's theorem" (A=B+C) it is quadratic in A. So by logical extension, for higher orders it "might" go as A^(N-3).
I think the OP is saying that for Fermat's theorem it is linear in A, but for a "first order Fermat's theorem" (A=B+C) it is quadratic in A. So by logical extension, for higher orders it "might" go as A^(N-3).
Thanks. I made an error in N = 318 at A = 1000, it should be 158 (actually from 975 to 1008)
Not intended to be a proof, just a way of describing the theorem.
My counts of N are based on b > c, so would be doubled if they are reversed.
 
  • #7
Thanks. I made an error in N = 318 at A = 1000, it should be 158 (actually from 975 to 1008)
Not intended to be a proof, just a way of describing the theorem.
My counts of N are based on b > c, so would be doubled if they are reversed.
I wonder what K would be in the linear case and much higher value of A? It took me about 4 days to count N with A = 10^ 6 so I'm not going any further.
Apparently Euler's totient function is related to my work.
An interesting calculation of 1/2.pi *3/pi =1.5/pi^2 = 0.151982 is very close to my value for K
probably just a coincidence.
 
  • #8
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I wonder what K would be in the linear case and much higher value of A?
What do you mean? In the linear case of ##A^1=B^1+C^1## we have all decompositions ##A=1+(A-1)## until ##A=\frac{A}{2}+\frac{A}{2}## before they repeat due to symmetry, so ##N(A,1)=\frac{1}{2}A##.
The quadratic case is ##N(A,2) = \frac{1}{2\pi}A## and ##N(A,p)=0## for ##p>2##.
 
  • #9
I am only counting the primitive pairs, i.e co-prime pairs. So up to A = 8 there are for a = 3,4,5,6,7,8 there are 1,1,2,1,3,2 pairs respectively making a total N = 10
 
  • #10
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I am only counting the primitive pairs, i.e co-prime pairs. So up to A = 8 there are for a = 3,4,5,6,7,8 there are 1,1,2,1,3,2 pairs respectively making a total N = 10
Sorry, I don't get it. For ##A=8## we have ##8=1+7=2+6=3+5=4+4## where ##(1,7)## and ##(3,5)## are coprime, so ##N=2##.
 
  • #11
Sorry, I don't get it. For ##A=8## we have ##8=1+7=2+6=3+5=4+4## where ##(1,7)## and ##(3,5)## are coprime, so ##N=2##.
N is the total number of pairs with a maximum sum A, not only those adding up to A.
I will see if using Euler's totient function can get my count programme to run much faster and then go to say A = 10^7.
It might be worth looking at the linear case as a flattened triangle.
 
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  • #12
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Terry, I want to help you.

Your number N of coprime primitive solutions to x+y=z for z up to n is just:

N=(φ(3)+φ(4)+...+φ(n))/2. The proof of this is 3 lines. No need to write it down here. Ask me if you wish.

You might be tempted to think that as n tends to infinity this number approaches the totient summatory Φ(n) function divided by n^2 = 3/(π^2) but this isn't so!

Further help: (φ(3)+φ(4)+...+φ(n))/n^2 is approximately equal to (φ(1)+φ(2)+...+φ(n))/n^2=Φ(n)/n^2 as n tends to infinity but Φ(n)/n^2 isn't just 3/(π^2)!

Reference: http://mathworld.wolfram.com/TotientSummatoryFunction.html

Best wishes!

Here's a Mathematica program for you for z up to 10^9. One line lots of RAM required:

n = 1000000000; N[n^2/Total[Table[EulerPhi[k], {k, 1, n}]], 20]
 
  • #13
In my original contribution I wondered how k would vary with p I have now found a disconnected curve that answers this, and it results in zero N for all p greater than 2. I'll try to insert this curve.
k = y and p = x
242315
 
  • #14
Error in above chart . should be a=e^(2/3.pi^2) b = e^2.pi()
 
  • #15
Is what follows a possible simple proof of FLT?
If you calculate the irrational values of Y = (1- X^p)(1/p). with p > 2 it is found that in a triangle with unit base and sides X and Y the angle between X and Y is a minimum with X = Y and is between 60 and 90 degrees, approaching 60 as p approaches infinity. So at infinity we get an equilateral triangle with unit sides. Compare this with power 2 with 90 degrees instead of 60 or 180 degrees with unit power. Now look at 1^infinity which is indeterminate, say X = Y = 1 and
(X +Y)^(1/infinity ) = 1
Now we can say only when p is infinite 1^p + 1^p =1^p which is represented by the equilateral triangle. Can we conclude that Fermat is correct for all powers greater than 2 and less than infinity?
 
  • #17
Is what follows a possible simple proof of FLT?
If you calculate the irrational values of Y = (1- X^p)(1/p). with p > 2 it is found that in a triangle with unit base and sides X and Y the angle between X and Y is a minimum with X = Y and is between 60 and 90 degrees, approaching 60 as p approaches infinity. So at infinity we get an equilateral triangle with unit sides. Compare this with power 2 with 90 degrees instead of 60 or 180 degrees with unit power. Now look at 1^infinity which is indeterminate, say X = Y = 1 and
(X^p+Y^p)^(1/p ) = 1 when p = infinity and X^p & Y^p are any two integers
Now we can say only when p is infinite 1^p + 1^p =1^p which is represented by the equilateral triangle. Can we conclude that Fermat is correct for all powers greater than 2 and less than infinity?
Do the sides of the equilateral triangle represent the cube roots of 1 and -1 = 0.5+/-0.866 i or perhaps the 3 x infinity root?
It might be argued that anything can happen at infinity, but here is a specific case where it is shown to be true.
 
  • #19
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It is quite easy with FLT.

a.) If there was an easy proof, we surely would have found in 350 years.
b.) If there was an easy proof, Wiles wouldn't have spent a decade to find another one.

These two arguments are heuristic, admitted, but nevertheless they are strong, because they limit the chances to a value close to zero, that there is an easier proof - and certainly none by inspection.
 
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  • #20
berkeman
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Thread is now closed. We will wait to discuss this work once it is published in a peer-reviewed journal.
 
  • #21
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Much of this thread has been discussing the possibility of an alternate simpler proof of FLT based on numerical observations over limited ranges of numbers and extrapolating to infinity.

In very few cases of mathematical research, has this approach ever led to new breakthroughs. However, in many other cases, it leads to discovering a counter example that breaks the observation and ends the line of research. We have seen this occur over and over again and it’s what makes mathematics such an endearing and challenging subject of study. It is also why proof has become so important as a means to build solid foundations.

This thread is now closed and we thank everyone who contributed to the discussion. Perhaps some day it will lead to a peer reviewed proof that we can review and discuss but until that time comes we will just have to wait for word of its publishing.

Thank you again,
Jedi
 
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