Is There a Fifth Order 3D Pythagorean Theorem?

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Discussion Overview

The discussion revolves around the exploration of a potential fifth order extension of the Pythagorean theorem in three dimensions, specifically examining the equation a^3 + b^3 + c^3 = d^3. Participants consider whether this relationship has been previously established in mathematics or if it represents a new discovery. The conversation includes references to higher-order equations and conjectures related to sums of powers.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a relationship a^3 + b^3 + c^3 = d^3, using specific values (3, 4, 5) to illustrate the concept.
  • Another participant describes the three-dimensional Pythagorean theorem as s^2 = x^2 + y^2 + z^2, questioning the original claim's validity.
  • Some participants reference Fermat's conjecture regarding the absence of integer solutions for a^3 + b^3 = c^3 and discuss the implications for higher powers.
  • Euler's conjecture about the impossibility of integer solutions for x^4 + y^4 + z^4 = w^4 is mentioned, along with a counterexample discovered by Noam Elkies.
  • There is a discussion about the generalization of the Pythagorean theorem to n dimensions, with varying opinions on its implications and definitions.
  • Several participants express uncertainty about the validity of the claims and seek clarification on the definitions and terms used in the proposed theorems.
  • Questions arise regarding the discovery process of counterexamples in higher-order equations, particularly the role of computational methods.

Areas of Agreement / Disagreement

Participants express a mix of agreement and disagreement regarding the validity of the proposed relationships and the existence of higher-order equivalents to the Pythagorean theorem. The discussion remains unresolved, with multiple competing views on the topic.

Contextual Notes

Some participants note limitations in the definitions and assumptions underlying the proposed theorems, as well as the need for clarity in mathematical terms. There are references to unresolved mathematical steps and the implications of various conjectures.

Who May Find This Useful

This discussion may be of interest to those exploring mathematical conjectures, the properties of numbers in relation to powers, and the generalizations of the Pythagorean theorem in various dimensions.

dperez3894
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Has anybody else tried this?

a^3 + b^3 + c^3 = d^3

3^3 + 4^3 + 5^3 = 6^3

27 + 64 + 125 = 216

This seems to be a logical extension of the Pythagorean Theorem and it works if the values of 3, 4 and 5 are used for a, b and c.

Has this already been discovered in mathematics or is this something new?
 
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The logical extension of the Pythagorean theorum in 3 dimensions is

s^2 = x^2 + y^2 + z^2
 
The term 3d pythagoras is usually reserved to mean that the square of the length of a diagonal of a cube is the sum of the squares of the sides.

What is your theorem anyway? I only see an example that you've found some numbers whose cubes are related in a certain way.
 
Yeah that's kind of interesting, Fermat's famous conjecture was that there exist no equivalent of Pythagorean Triads for powers higher than two, eg no chance for integers a^3 + b^3 = c^3.

So what you're saying is that although there is no direct cubic "triad" equivalent there are indeed integer "cubic quartets". Interesting idea, perhaps there are also 4th power "quintets" and fifth power "sextet" etc. Does anyone know if there are existing theorems or conjectures about this?
 
Yes, Euler conjectured that there were no integers x, y, z, w such that x^4 + y^4 + z^4 = w^4 (not exactly what you were asking for, but close enough). Noam Elkies of Harvard discovered this counterexample in 1988:

2682440^4 + 15365639^4 + 18796760^4 = 20615673^4.
 
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And there is the famous example in this vein that every integer (and hence every square, cube 4th power etc) is the sum of 4 squares.
 
"This seems to be a logical extension of the Pythagorean Theorem and it works if the values of 3, 4 and 5 are used for a, b and c."

but doesn't if a=b=c=1
 
The more general question is: which sums are products?
A^3 = A^2 + A^2 + A^2 or A^3 = 3*A^2
From that A must equal 3, or for general:
N^n = n*N^n-1

So for any two sums like Z^n = X^n + Y^n
n can only be two.
Just started messing with this. Don't know where it goes.
 
  • #10
There's an n-dimensional Pythagorean theorem too isn't there? I don't see why not. How about a_1^2 + a_2^2 + ... + a_n^2 = a^2
 
  • #11
Digit said:
The more general question is: which sums are products?
A^3 = A^2 + A^2 + A^2 or A^3 = 3*A^2
From that A must equal 3, or for general:
N^n = n*N^n-1

So for any two sums like Z^n = X^n + Y^n
n can only be two.
Just started messing with this. Don't know where it goes.

Is that the shortest known "proof" of Fermat's last theorem?
 
  • #12
fourier jr said:
There's an n-dimensional Pythagorean theorem too isn't there? I don't see why not. How about a_1^2 + a_2^2 + ... + a_n^2 = a^2

Yes and no. The n dimensional version is a direct consequence of the 2d version; it is provable directly from it. Of course one might argue that this is just a formal result from making the definitions of inner products such as they are, though I must ask, is no one else actually going to say what any of the terms in their 'theorems' actually are? Pythagoras DOES NOT say that x**2+y**2=z**2, since 1,1,3 for x,y,z resp disproves that (even if we assume x,y,z must be real numbers in the first place!) it states something geometrical. Is the OP going to state what they might actually mean?
 
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  • #13
x^2 + y^2 = s^2

s^2 + z^2 = r^2

x^2 + y^2 + z^2 = r^2
 
  • #14
matt grime said:
Is that the shortest known "proof" of Fermat's last theorem?

I don't know. I am sure there is a short proof but I don't know how to do it.
 
  • #15
Digit said:
I don't know. I am sure there is a short proof but I don't know how to do it.

Your above post claims you do though.
 
  • #16
Digit said:
The more general question is: which sums are products?
A^3 = A^2 + A^2 + A^2 or A^3 = 3*A^2
From that A must equal 3, or for general:
N^n = n*N^n-1

So for any two sums like Z^n = X^n + Y^n
n can only be two.
Just started messing with this. Don't know where it goes.

This (the first 3 lines are okay) doesn't make any sense to me. Can someone (Digit?) please explain ?
 
  • #17
Euler showed that the product of two sums of four squares is again a sum of four squares. This was part of his proof that every integer is the sum of four squares (including squares of zero where necessary). He did it by working out all the partial products and collecting terms.
 
  • #18
Muzza said:
Yes, Euler conjectured that there were no integers x, y, z, w such that x^4 + y^4 + z^4 = w^4 (not exactly what you were asking for, but close enough). Noam Elkies of Harvard discovered this counterexample in 1988:

2682440^4 + 15365639^4 + 18796760^4 = 20615673^4.
How was the counter-example discovered? By the use of computers?
 
  • #19
Probably. See Noam Elkies' article "On A^4 + B^4 + C^4 = D^4, Math. of Comp. 51 (Oct. 1988), 825-835". ;)
 
  • #20
Ethereal said:
How was the counter-example discovered? By the use of computers?
My guess is it was a computer search, it's quite large seach to find those numbers, of the order of the largest LHS number to the forth if you do it by "brute force".

I wonder if anyone has tried searching for a fifth order example, a^5 + b^5 + c^5 + d^5 + e^5 = f^5 ?
 

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