I Fermat's "Last Theorem"?

Terry Coates

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.

phyzguy

An interesting observation, but "it seems logical to conclude" is not a proof.

fresh_42

Mentor
2018 Award
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.

phyzguy

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).

pbuk

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.

Terry Coates

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.

Terry Coates

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.

fresh_42

Mentor
2018 Award
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 $p>2$.

Terry Coates

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

fresh_42

Mentor
2018 Award
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 $N=2$.

Terry Coates

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 $N=2$. add
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.

landica

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]

"Fermat's "Last Theorem"?"

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