January 9th, 2017, 06:05 PM  #1 
Newbie Joined: Jan 2017 From: Mangalore, Karnataka, India Posts: 11 Thanks: 0  (ax^3+bx^2+cx+d) + (ay^3+by^2+cy+d) = (az^3+bz^2+cz+d)
Is there any way to find positive integer solutions to (ax^3+bx^2+cx+d) + (ay^3+by^2+cy+d) = (az^3+bz^2+cz+d), where a,b,c,d are some integers?

January 9th, 2017, 07:34 PM  #2 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 
Note that for the special case $a = 1, b = c = d = 0$, the equation becomes $x^3 + y^3 = z^3$. Although that equation was proved (by Euler, I think) to have no positive integer solutions, the proof, while elementary, is not simple. Is it possible that the same techniques could completely resolve your more general equation? Possibly, but I doubt it. In any case, are you familiar with the proof that the equation $x^3 + y^3 = z^3$ has no positive integer solutions? Consider that as prerequisite knowledge for any meaningful investigation of your question. 
January 9th, 2017, 08:33 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,208 Thanks: 556  
January 9th, 2017, 11:05 PM  #4 
Newbie Joined: Jul 2015 From: tbilisi Posts: 26 Thanks: 7 
You want to get a simple answer. It can not be. Parameterization itself looks very cumbersome  you do not want. Here the deadlock  because you want what can not be. Quite often hard to solve Diophantine equations. And besides, the solutions themselves look very bulky. For this equation, the number of solutions for given factors  if any. The final number. Formula communication solutions with coefficients can not write  because it does not exist. Sami factors still have to parameterize. 
January 10th, 2017, 05:34 AM  #5 
Newbie Joined: Jan 2017 From: Mangalore, Karnataka, India Posts: 11 Thanks: 0  Are you all suggesting that solution does not exist or method does not exist?
Are you all suggesting that solution does not exist or method does not exist?. Could you people be little more specific, I am novice in this field.

January 10th, 2017, 05:45 AM  #6 
Newbie Joined: Jan 2017 From: Mangalore, Karnataka, India Posts: 11 Thanks: 0  Yes sir, I made up this problem for curiosity. 
January 10th, 2017, 06:01 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,208 Thanks: 556 
Ok. WHAT are the givens: a, b, c and d? Do you realise that a and b are interchangeable: if as example we get a solution with a=3 and b=5, then a=5 and b=3 is also a solution. No matter how you cut it, a cubic equation needs to be solved: agree? Here's a solution for x: WolframAlpha: Computational Knowledge Engine Perhaps in the future you can go to that site to "see" what your madeup equations mean... 
January 10th, 2017, 05:16 PM  #8  
Newbie Joined: Jan 2017 From: Mangalore, Karnataka, India Posts: 11 Thanks: 0  some examples Quote:
$(x^3 + 3*x^2 + 2*x + 6) + (y^3 + 3*y^2 + 2*y + 6) = (z^3 + 3*z^2 + 2*z + 6)$, $a=1, b=3, c=2, d=6$, This has the following integer solutions, $(4,5,6), (5,7,8 ), (14,32,33), (25, 75, 76), (27, 84, 85), (8, 9, 11), (22, 43, 45), (24, 49, 51), (63, 207, 209), (17, 19, 23), (229, 707, 715), (117, 228, 238 )$. Last edited by Naveenchandra Kumar; January 10th, 2017 at 05:54 PM.  
January 10th, 2017, 05:30 PM  #9  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,208 Thanks: 556  Quote:
 
January 10th, 2017, 05:37 PM  #10 
Newbie Joined: Jan 2017 From: Mangalore, Karnataka, India Posts: 11 Thanks: 0  Sir, are you serious? 