My Math Forum (ax^3+bx^2+cx+d) + (ay^3+by^2+cy+d) = (az^3+bz^2+cz+d)

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 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. Thanks from topsquark and Naveenchandra Kumar
January 9th, 2017, 08:33 PM   #3
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Quote:
 Originally Posted by Naveenchandra Kumar (ax^3+bx^2+cx+d) + (ay^3+by^2+cy+d) = (az^3+bz^2+cz+d)
Same as:
ax^3+bx^2+cx + ay^3+by^2+cy = az^3+bz^2+cz-d

Is that a TRUE problem. or did you make it up?

 January 9th, 2017, 11:05 PM #4 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 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. Thanks from Naveenchandra Kumar
 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
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Yes sir, I made up this problem for curiosity.

Quote:
 Originally Posted by Denis Same as: ax^3+bx^2+cx + ay^3+by^2+cy = az^3+bz^2+cz-d Is that a TRUE problem. or did you make it up?
Yes sir, I made up this problem for curiosity, after looking into $x^n+y^n=z^n$ and Fermat's Last Theorem.

 January 10th, 2017, 06:01 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,381 Thanks: 730 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: Wolfram|Alpha: Computational Knowledge Engine Perhaps in the future you can go to that site to "see" what your made-up equations mean...
January 10th, 2017, 05:16 PM   #8
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some examples

Quote:
 Originally Posted by Denis 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: Wolfram|Alpha: Computational Knowledge Engine Perhaps in the future you can go to that site to "see" what your made-up equations mean...
For example, given
$(x^3 + 3*x^2 + 2*x + 6) + (y^3 + 3*y^2 + 2*y + 6) = (z^3 + 3*z^2 + 2*z + 6)$, $a=1, b=3, c=2, d=6$, 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
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Quote:
 Originally Posted by Naveenchandra Kumar For example, given $(x^3 + 3*x^2 + 2*x + 6) + (y^3 + 3*y^2 + 2*y + 6) = (z^3 + 3*z^2 + 2*z + 6), a=1, b=3, c=2, d=6$, This has the following integer solutions, (4,5,6), (5,7,, (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, 23.
OK. But what does this achieve?

January 10th, 2017, 05:37 PM   #10
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Sir, are you serious?

Quote:
 Originally Posted by Denis OK. But what does this achieve?
I mean, what can mathematics achieve? What is the achievement of Fermat's Last Theorem?.

 Tags ax3, ay3, az3, bx2, by2, bz2, integer, solution

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