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March 29th, 2018, 10:45 AM   #1
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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.
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March 29th, 2018, 12:20 PM   #2
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(7,15,20) and (15,7,20)
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March 29th, 2018, 01:13 PM   #3
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Quote:
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(7,15,20) and (15,7,20)
Could you explain how you arrived at these answers?
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March 29th, 2018, 02:12 PM   #4
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Wrote short looper program.
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March 29th, 2018, 03:04 PM   #5
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Quote:
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Wrote short looper program.
Could you explain how the looper program computes the values for x y and z? Like what stops is it taking?
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March 29th, 2018, 03:23 PM   #6
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It's simply hit and miss...brute force.
No "solving" involved.
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March 29th, 2018, 04:31 PM   #7
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Quote:
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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?
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March 29th, 2018, 05:03 PM   #8
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Math Focus: Wibbly wobbly timey-wimey 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
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March 29th, 2018, 05:32 PM   #9
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Agree. That's exactly what I did; from 1 to 999.
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March 29th, 2018, 05:37 PM   #10
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Agree. That's exactly what I did; from 1 to 999.
So you created like a computer program to do so? Or bash them all out by hand?
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