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January 8th, 2017, 05:42 PM  #1 
Newbie Joined: Jan 2017 From: Mangalore, Karnataka, India Posts: 11 Thanks: 0  Integer solutions to (x^2+x) + (y^2+y) = (z^2+z)
Is there any integer solution to the equation, (x^2+x) + (y^2+y) = (z^2+z) ?

January 8th, 2017, 08:12 PM  #2 
Senior Member Joined: Feb 2010 Posts: 590 Thanks: 82 
$\displaystyle (3^2+3)+(5^2+5)=(6^2+6)$

January 9th, 2017, 12:26 AM  #3 
Newbie Joined: Jul 2015 From: tbilisi Posts: 26 Thanks: 7  
January 9th, 2017, 11:24 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 16,577 Thanks: 1198 
(x, y, z) = (0, t, t), where t is an integer is the easiest solution. A better question would be to find all the solutions in positive integers. Let t be a natural number, then some (not necessarily distinct) solutions in positive integers are given below. (x, y, z) = (3t, 4t + 1, 5t + 1) (x, y, z) = (3t + 2, 4t + 2, 5t + 3) (x, y, z) = (t + 2, (t + 1)(t + 4)/2, (t + 2)(t + 3)/2) (x, y, z) = (3t + 5, (3t + 2)(t + 3)/2, (3tÂ² + 11t + 12)/2) One might speculate that every solution with x < y can be given in a similar way. 
January 9th, 2017, 12:03 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,424 Thanks: 569 
z = {1 + SQRT[1 + 4(x^2 + x + y^2 + y)]} / 2 Kinda interesting to "look at"!! 

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