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 June 2nd, 2017, 12:37 PM #1 Newbie   Joined: Jun 2017 From: United Kingdom Posts: 1 Thanks: 0 Primitive Pythagoras nth term Hi new to this forum. I was messing about with generating pythagorean triples using: x^2-y^2,2xy,x^2+y^2 and the difference between b and c and noticed this. I knew that if x-1=y b would be 1 less than c. I realised this meant you could organise these triples using x=n+1 and y=n but realised that for triples where b+2=c the nth term would look like this: x= [(2n+3)2^0.5]/2 and y=[(2n+1)2^0.5]/2 these always make primitive triples and I was wondering whether anyone knew if it could be done for places where there is different gaps between b and c. Thanks. Last edited by greg1313; June 2nd, 2017 at 12:47 PM.
 June 3rd, 2017, 08:42 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,768 Thanks: 651 These 7 Pythagorean triples all have shorter leg = 660: 660,779,1021 660,989,1189 660,2989,3061 660,4331,4381 660,12091,12109 660,27221,27229 660,108899,108901 Also, that is the only case where short leg < 1000. Perhaps that'll help you in "checking different gaps".
 June 3rd, 2017, 09:41 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,854 Thanks: 2228 Math Focus: Mainly analysis and algebra If you use $a=mn, \, b=\frac12(m^2-n^2), \, c=\frac12(m^2+n^2)$, then every coprime pair $(m,n)$ of odd natural numbers gives you a primitive Pythagorean Triple. Furthermore, this scheme gives you every primitive Pythagorean Triple.

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