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Puzzle: Does 60 always divide abc?

November 12, 2008 am30 6:08 am

So McRib tells me a kid told him that you can tell if three numbers are a Pythagorean triple, just check to see if their product is a multiple of 60.

Well, no. 6, 7, 10 is not a Pythagorean triple.

But, wait. Is every Pythagorean triple a multiple of 60?

3*4*5 = 60
5*12*13 = 780
6*8*10 = 480
7*24*25 = 4200
8*15*17 = 2040
9*12*15 = 1620
20*21*29 = 12180

Ok, so it’s ok for a few. But all?

Well, at least one number needs to be even  (even + even = even or odd + even = odd), but where do we get the 2nd factor of 2, and the factors of 3 and 5?

Is this true? Can you explain it?

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13 Comments leave one →
  1. Bart permalink
    November 12, 2008 am30 7:25 am 7:25 am

    After repeated attempts to solve this one, it turns out that nobody knows.

    It is an “open question” whether the ordered series of products for all Pythagorean triangles has a repeated item in it or if all the products are in fact unique:-
    Article: Unsolved Problems in Number Theory, R K Guy (2nd ed.) Springer-Verlag (1994), Problem D21 “Triangles with Integer Sides, Medians, and Area” pages 188-190. – http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/pythag.html

  2. zaytuun permalink
    November 12, 2008 am30 8:18 am 8:18 am

    interesting..
    thanx for this post


    zaytuun | Fractal Art WebGallery

  3. November 12, 2008 am30 8:27 am 8:27 am

    Consider n^2 (mod 3). The result can only be 0 or 1.

    Then if a^2 + b^2 = c^2, it must be that a^2 == b^2 == c^2 == 0 (mod 3), or a^2 == 1, b^2 == 0, c^2 == 1 (mod 3), or a^2 == 0, b^2 == 1, c^2 == 1 (mod 3).

    In either case, one of them is always divisible divisible by 3.

    Doing something similar, if I remember correctly, you can get that one must be divisible by 5 and on by 4.

    Done!

    I think, anyway.

  4. November 12, 2008 am30 9:07 am 9:07 am

    Fox…
    that brings us to 6…

    For 5, n^2 = -1, 0 or 1 mod 5.
    -1 + -1 = ??? doesn’t work
    -1 + 0 = -1
    -1 + 1 = 0
    0 + 0 = 0, but divide out the 5
    0 + 1 = 1
    1 + 1 = nope, doesn’t work

    So it looks like there must be a multiple of 5 in the mix.

    We’re up to 30.

    Almost done.

  5. November 12, 2008 am30 9:11 am 9:11 am

    Sorry, I should’ve been more complete ^^

    n^2 (mod 4) gives you 0 or 1 only, so the solutions for the (mod 3) case apply as well.

    3*4*5, done!

  6. November 12, 2008 am30 9:19 am 9:19 am

    And now it really is.

    Bart, I think you were referring to another problem, but I can’t figure out what it is.

    And finally, everyone, the next post should be interesting.

  7. November 12, 2008 pm30 2:07 pm 2:07 pm

    Another tack is to show that every primitive pythagorean triple has its abc product divisible by 60; that leads you to analyzing
    2mn(m^2-n^2)(m^2+n^2)=2mn(m^4-n^4)

    4 is clearly a factor, since either at least one of m,n is even, or they’re both odd and then m^4-n^4 is even.

    3 is a factor, since either 3 divides m or n, or if not then both m^4 and n^4 = 1 mod 3.

    Finally, 5 is a factor of abc, since if 5 isn’t a factor of either m or n, then both m^4 and n^4 = 1 mod 5.

  8. McRib permalink
    November 14, 2008 am30 2:45 am 2:45 am

    TwoPi,

    It is true that 2mn, m^2-n^2, and m^2+n^2 generate pythagorean triples, but how do we know that this algorithm generates ALL triples?

  9. November 14, 2008 am30 5:39 am 5:39 am

    It only generates all “primitive triples”, triples where a,b, and c have no factor in common.

    But every other triple is of the form ka, kb, kc where a,b,c is primitive; so if we know 60 divides abc, then it also divides (ka)(kb)(kc).

  10. November 16, 2008 am30 1:06 am 1:06 am

    How do we know it generates all the primitive triples? I think this is a really good question.

  11. TwoPi permalink
    November 16, 2008 am30 4:51 am 4:51 am

    I’ve posted a proof at the 360 blog.

    (With apologies, note that comment 9 by “Ξ” was actually written by me; I was at her computer, and forgot to check the username that popped up.)

  12. November 16, 2008 am30 4:56 am 4:56 am

    So I think I corrected the misattribution, but the e-mails are all snarled. I have eu, hl, and mc, and not much of an idea of who is who. I guess it doesn’t matter much – I’ll find all of you at 360…

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