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November 1st, 2010, 05:21 PM   #1
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Pythagorean Triples II

Find all the Pythagorean triples that form an arithmetic sequence.
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November 1st, 2010, 05:46 PM   #2
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Re: Pythagorean Triples II

where

What must n be? Then what must n - k and n + k be?
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November 12th, 2010, 11:52 AM   #3
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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
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November 12th, 2010, 11:58 AM   #4
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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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