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 March 12th, 2014, 02:18 AM #1 Senior Member     Joined: Oct 2013 From: Far far away Posts: 427 Thanks: 18 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 $(3a+1)^{2}+(3b+1)^{2}\,=\,9a^{2}+6a+1+9b^2+6b+1\,= \,3(3a^2+3b^2+2a+2b)+2$ SO $\frac{x^{2}+y^{2}}{3}$ has remainder = 2 2) x = 3a + 1 and y = 3b + 2 $(3a+1)^{2}+(3b+2)^{2}\,=\,9a^2+6a+1+9b^2+12b+4\,=\ ,3(3a^2+3b^2+2a+4b+1)+2$ SO $\frac{x^2+y^2}{3}$ 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 $\frac{x^2+y^2}{3}$ 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 $z^2=(3n)^{2}=9n^2$ SO $\frac{z^2}{3}$ has a remainder 0 2) z = 3n + 1 $(3n+1)^2\,=\,9n^2+6n+1\,=\,3(3n^2+2n)+1$ SO $\frac{z^2}{3}$ has a remainder 1 3) z = 3n + 2 $(3n+2)^2\,=\,9n^2+12n+4\,=\,3(3n^2+4n+1)+1$ SO $\frac{z^2}{3}$ 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?
 March 12th, 2014, 08:57 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra 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.
 March 12th, 2014, 09:31 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,758 Thanks: 2138 Re: Pythagorean triples Is the notation used below acceptable? It would allow a more succinct presentation. As 2² ? 1 (mod 3), n ? 2 (mod 3) implies n² ? 1 (mod 3).
 March 12th, 2014, 11:13 AM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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
 March 12th, 2014, 02:51 PM #5 Global Moderator   Joined: May 2007 Posts: 6,770 Thanks: 700 Re: Pythagorean triples You could use the following which applies to all P triples. m and n integers, m > n x = m² - n², y = 2mn (z = m² + n²) 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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