User Name Remember Me? Password

 Geometry Geometry Math Forum

 March 12th, 2014, 02:18 AM #1 Senior Member   Joined: Oct 2013 From: Far far away Posts: 431 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 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? March 12th, 2014, 08:57 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 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: 21,035 Thanks: 2271 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). March 12th, 2014, 11:13 AM #4 Math Team   Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 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: 6,835 Thanks: 733 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. Tags pythagorean, triples Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shunya Geometry 3 March 20th, 2014 07:12 AM elim Number Theory 8 August 21st, 2011 11:29 AM Pell's fish Number Theory 3 August 5th, 2011 10:38 AM julian21 Number Theory 2 November 13th, 2010 12:01 PM julian21 Number Theory 3 November 12th, 2010 11:58 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top       