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 June 7th, 2012, 09:31 AM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory diophantine equation Hello, solve: x+y+z=43 x³+y³+z³=17299 Thanks in advance.
 June 7th, 2012, 09:34 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: diophantine equation Are the variables integers, positive integers, or nonnegative integers?
 June 7th, 2012, 10:06 AM #3 Member   Joined: May 2012 From: Chennai,India Posts: 67 Thanks: 0 Re: diophantine equation (25,11,7) is a solution.. i am still working on proving it. cube root of 17299 = 25 (approx) cube root of (17299-25^3) = cube root of 1674 = 11 (approx) cube root of (1674 - 11^3) = 343 = 7^3
June 7th, 2012, 10:13 AM   #4
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Re: diophantine equation

Quote:
 Originally Posted by CRGreathouse Are the variables integers, positive integers, or nonnegative integers?
If x+y+z=43
x³+y³+z³=17299
And x,y,z belong to $\mathbb{+Z}$
Then x,y,z=?
Quote:
 Originally Posted by karthikeyan.jp (25,11,7) is a solution.. i am still working on proving it. cube root of 17299 = 25 (approx) cube root of (17299-25^3) = cube root of 1674 = 11 (approx) cube root of (1674 - 11^3) = 343 = 7^3
I have done this too, but we both are iterating here, is there a proper diophantic non iterating method to solve it? However, thank you for the help

June 7th, 2012, 12:45 PM   #5
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Re: diophantine equation

Quote:
 Originally Posted by karthikeyan.jp (25,11,7) is a solution.
It's the only one.

 June 7th, 2012, 03:41 PM #6 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Re: diophantine equation x³ = 17299-y³-z³ x³ < 17299 x < 26 So all x,y,z satisfy n < 26. Also, 3n³ = 17299 n = 17.93 .. So at least one satisfies n>17. x³=17299-18³-z³ So, at least one other satisfies n<22. To sum up, we have: x>17,y<22,z<26 Hope this helps! :)
 June 8th, 2012, 04:50 AM #7 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: diophantine equation Maybe I got the solution: Consider the triplet 1,5,7 1+5+7=13(the last digit is 3) and 1³+5³+7³=469(the last digit is 9) so, we can assume that x=10a+1, y=10b+5, z=10c+7 It implies that a+b+c=3, and we are done
June 8th, 2012, 06:15 AM   #8
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Re: diophantine equation

Quote:
 Originally Posted by mathbalarka Maybe I got the solution:
I did post it above...

June 8th, 2012, 11:53 AM   #9
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Re: diophantine equation

Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by mathbalarka Maybe I got the solution:
I did post it above...
I got the method to find the solution

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