Math Is Fun Forum

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Pythagoras Triple

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Numbers are the essence of the Universe

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Pythagoras Triple

I am so stupid , I have no clue,help me out

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Numbers are the essence of the Universe

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Pythagoras Triple

Edit: The following is horribly and utterly wrong
Edit #2: Upon further review, it's wrong, but not horribly wrong.

For any Pythagorean triple (x, y, z), 3 divides at least one of them, 4 divides at least one of them, and 5 divides at least one of them.  Of course, this simply means that 3 divides exactly one, 4 divides exactly one, and 5 divides exactly one.  Once you prove this, your proof becomes easy.

But I recall that the proof of this is fairly long (a page or two), and so if you aren't allowed to assume it, then I would recommend finding a better way.  I'll try to think of a better way, but if no one comes up with one, I'll give an outline of the above fact.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Pythagoras Triple

yeah , I know I need to prove this one , but I can't crack it. I think maybe using the Pythagoras number generating forumla will work?

,oh , it won't work either...gotta find a way around

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Numbers are the essence of the Universe

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Pythagoras Triple

My above post is wrong.  But it's close.  And its exactly what you want.

For any triple (x, y, z), one entry is divisible by 3, one entry is divisible by 4, and one entry is divisible by 5.  Note that these need not be unique.  For example, 5, 12, 13:

3 | 12
4 | 12
5 | 5

It should be fairly straight forward to prove that xyz = 0 (mod 60) using this as xyz = 3*4*5k.

But now how to prove this?  By your post, I assume you're allowed to assume all pythagorean triples have the form (2mn, m^2 - n^2, m^2 + n^2).

If 3 | m or 3 | n (or both), then 3 | 2mn, so 3 divides one of numbers in the triplet.
Assume 3 does not divide m and 3 does not divide n.  Then it must be that m^2 = 1 (mod 3) and n^2 = 1 (mod 3).  So m^2 - n^2 = 0 (mod 3), and 3 | m^2 - n^2.

The proofs for 4 and 5 are remarkably similar.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Pythagoras Triple

Yeah, but can't it be solved by Modulo method?

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Numbers are the essence of the Universe