March 29th, 2018, 10:45 AM  #1 
Newbie Joined: Mar 2018 From: California Posts: 18 Thanks: 0  Yet another problem
Solve in positive integers $x^3 + y^3 +3xyz = z^3 + 2018$ This problem doesn't seem to be solvable because there are three variables in one equation. Could anyone provide some assistance? Full solutions are appreciated with justifications. 
March 29th, 2018, 12:20 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
(7,15,20) and (15,7,20)

March 29th, 2018, 01:13 PM  #3 
Newbie Joined: Mar 2018 From: California Posts: 18 Thanks: 0  
March 29th, 2018, 02:12 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
Wrote short looper program.

March 29th, 2018, 03:04 PM  #5 
Newbie Joined: Mar 2018 From: California Posts: 18 Thanks: 0  
March 29th, 2018, 03:23 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
It's simply hit and miss...brute force. No "solving" involved. 
March 29th, 2018, 04:31 PM  #7 
Newbie Joined: Mar 2018 From: California Posts: 18 Thanks: 0  
March 29th, 2018, 05:03 PM  #8 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,299 Thanks: 960 Math Focus: Wibbly wobbly timeywimey stuff. 
I would say that you simply do three loops: x, y, z from 1 to 50 and check to see if the x,y,z values work. Nothing too complicated. (You can also be crafty and note that if x, y, z is a solution, then so is y, x, z.) Dan 
March 29th, 2018, 05:32 PM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
Agree. That's exactly what I did; from 1 to 999.

March 29th, 2018, 05:37 PM  #10 
Newbie Joined: Mar 2018 From: California Posts: 18 Thanks: 0  