Yet another problem

Mar 2018
18
0
California
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.
 

Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
(7,15,20) and (15,7,20)
 

Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
Wrote short looper program.
 

Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
It's simply hit and miss...brute force.
No "solving" involved.
 
Mar 2018
18
0
California
It's simply hit and miss...brute force.
No "solving" involved.
Then how did you bash out and get those specific solutions? What kinds of numbers were you looking for?
 

topsquark

Math Team
May 2013
2,519
1,049
The Astral plane
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
 
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Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
Agree. That's exactly what I did; from 1 to 999.
 
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