November 1st, 2010, 06:21 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, 06:46 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs  Re: Pythagorean Triples II where What must n be? Then what must n  k and n + k be? 
November 12th, 2010, 12:52 PM  #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^22nk=2nk now by substraction we get n^24nk=0 then we have n(n4k)=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, 12:58 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs  Re: Pythagorean Triples II
Yes, you wind up with (3k,4k,5k) or integral multiples of the wellknown (3,4,5) triple. 

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