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 November 1st, 2010, 05:21 PM #1 Senior Member   Joined: Apr 2009 Posts: 106 Thanks: 0 Pythagorean Triples II Find all the Pythagorean triples that form an arithmetic sequence.
 November 1st, 2010, 05:46 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: Pythagorean Triples II $(n-k)^2+n^2=(n+k)^2$ where $k\in\mathbb Z$ What must n be? Then what must n - k and n + k be?
 November 12th, 2010, 11:52 AM #3 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 Re: Pythagorean Triples II all u should do is to open the brackets u will get n^2 -2nk +k^2 +n^2= n^2+2nk +k^2 now eliminate the identical elements from both side u will get n^2-2nk=2nk now by substraction we get n^2-4nk=0 then we have n(n-4k)=0 we have too solutions n=0 the other is n=4k actually we have infinite solutions try for example n=4 k=1 u get 3^2+4^2=5^2
 November 12th, 2010, 11:58 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: Pythagorean Triples II Yes, you wind up with (3k,4k,5k) or integral multiples of the well-known (3,4,5) triple.

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