Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

3-D Pythagorean Theorem?

  1. Jun 30, 2004 #1
    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?
     
  2. jcsd
  3. Jun 30, 2004 #2

    jcsd

    User Avatar
    Science Advisor
    Gold Member

    The logical extension of the Pythagorean theorum in 3 dimensions is

    s^2 = x^2 + y^2 + z^2
     
  4. Jun 30, 2004 #3

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  5. Jun 30, 2004 #4

    uart

    User Avatar
    Science Advisor

    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?
     
  6. Jun 30, 2004 #5
    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.
     
    Last edited: Jun 30, 2004
  7. Jun 30, 2004 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  8. Jun 30, 2004 #7

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  9. Jun 30, 2004 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    "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
     
  10. Jun 30, 2004 #9
    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.
     
  11. Jun 30, 2004 #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
     
  12. Jun 30, 2004 #11

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Is that the shortest known "proof" of Fermat's last theorem?
     
  13. Jun 30, 2004 #12

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
    Last edited: Jun 30, 2004
  14. Jun 30, 2004 #13
    x^2 + y^2 = s^2

    s^2 + z^2 = r^2

    x^2 + y^2 + z^2 = r^2
     
  15. Jul 1, 2004 #14
    I don't know. I am sure there is a short proof but I don't know how to do it.
     
  16. Jul 1, 2004 #15

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Your above post claims you do though.
     
  17. Jul 1, 2004 #16

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This (the first 3 lines are okay) doesn't make any sense to me. Can someone (Digit?) please explain ?
     
  18. Jul 1, 2004 #17

    selfAdjoint

    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    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.
     
  19. Jul 2, 2004 #18
    How was the counter-example discovered? By the use of computers?
     
  20. Jul 2, 2004 #19
    Probably. See Noam Elkies' article "On A^4 + B^4 + C^4 = D^4, Math. of Comp. 51 (Oct. 1988), 825-835". ;)
     
  21. Jul 3, 2004 #20

    uart

    User Avatar
    Science Advisor

    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 ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: 3-D Pythagorean Theorem?
  1. Pythagorean Theorem. (Replies: 9)

Loading...