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?
JD2718 
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
interesting..
thanx for this post
—
zaytuun | Fractal Art WebGallery
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.
Fox…
that brings us to 6…
For 5,
= -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.
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!
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.
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.
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?
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).
How do we know it generates all the primitive triples? I think this is a really good question.
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.)
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…