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March 12th, 2014, 02:18 AM   #1
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Pythagorean triples

Prove or disprove: If x^2 + y^2 = z^2 for whole numbers x, y and z then, either x or y is a multiple of 3

My attempt:
Assume that x and y are not divisible by 3
So x and y must be of the form 3m + 1 or 3m + 2 (leaving a remainder 1 or 2 when divided by 3)
We have 3 possibilities:
1) x = 3a + 1 and y = 3b + 1
SO has remainder = 2

2) x = 3a + 1 and y = 3b + 2
SO has a remainder 2

3) x = 3a + 2 and y = 3b + 2
[latex](3a+2)^{2}+(3b+2)^{2}\,=\,9a^2+12a+4+9b^2+12b+4\,= \,3(3a^2+3b^2+4a+4b+2)+2[latex] SO has a remainder of 2

Now consider z. It can be only one of the following forms 3n, 3n + 1 and 3n + 2

1) z = 3n
SO has a remainder 0

2) z = 3n + 1
SO has a remainder 1

3) z = 3n + 2
SO has a remainder 1

If x and y are not multiples of 3 then the remainder of (x^2 + y^2)/3 is 2 BUT the remainder of (x^2+y^2)/3 cannot be 2 (it has to be 0 or 1).
Therefore at least one of x or y has to be a multiple of 3.

Is this proof acceptable? Is there a better proof?
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March 12th, 2014, 08:57 AM   #2
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Re: Pythagorean triples

A = There exist integers x, y, z such that x^2 + y^2 = z^2
B = Either x or y is a multiple of three

You have proved not(B) => not(A), which is equivalent to A => B.

So yes, that would be a valid proof.

Of course, the proof has nothing to say about the existence of such triplets, but neither do the conditions of the hypothesis.
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March 12th, 2014, 09:31 AM   #3
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Re: Pythagorean triples

Is the notation used below acceptable? It would allow a more succinct presentation.

As 2 ? 1 (mod 3), n ? 2 (mod 3) implies n ? 1 (mod 3).
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March 12th, 2014, 11:13 AM   #4
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Re: Pythagorean triples

Just as a note , you can reduce your work on the proof by 2/3 if you consider Primitive Pythagorean Triples.

For Primitive Pythagorean Triples:By elementary arguments ,

1) gcd(x,y) = 1

2) x and y cannot both be odd

It follows that one of the two legs must be even , the other leg must be odd and the hypotenuse must be odd.

So you only need to consider (WLOG) the case

x = 3a + 1 , y = 3b + 2

Here are some fun facts about PT's and PPT's.

http://en.m.wikipedia.org/wiki/Pythagorean_triple

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March 12th, 2014, 02:51 PM   #5
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Re: Pythagorean triples

You could use the following which applies to all P triples.
m and n integers, m > n
x = m - n, y = 2mn (z = m + n)
If m or n is a multiple of 3, then y is.
If m?n (mod 3), then m-n is a multiple of 3, therefore x is.
Otherwise m+n is a multiple of 3, and x is.
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