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 November 3rd, 2008, 10:07 PM #1 Senior Member   Joined: Oct 2008 Posts: 215 Thanks: 0 A Diophantine equation Please find a non-zero integer solution for equation $x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2=x_1x_2x_3x_4x_ 5x_6$
 November 5th, 2008, 06:54 PM #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: A Diophantine equation (1, -1, -2, -2, -2, 4 - sqrt(2))
November 5th, 2008, 11:09 PM   #3
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Re: A Diophantine equation

Quote:
 Originally Posted by CRGreathouse (1, -1, -2, -2, -2, 4 - sqrt(2))
4-$\sqrt{2}$ is not integer

 November 6th, 2008, 04:39 AM #4 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: A Diophantine equation Ah, missed that, sorry. I wasn't able to find any nonzero integer solutions.
 November 6th, 2008, 04:43 AM #5 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: A Diophantine equation Mathematica 6.0 fails: Code: In[1]:= FindInstance[ a^2 + b^2 + c^2 + d^2 + e^2 + f^2 == a b c d e f && a > 0, {a, b, c, d, e, f}, Integers] During evaluation of In[1]:= FindInstance::nsmet: The methods \ available to FindInstance are insufficient to find the requested \ instances or prove they do not exist. >> Out[1]= FindInstance[ a^2 + b^2 + c^2 + d^2 + e^2 + f^2 == a b c d e f && a > 0, {a, b, c, d, e, f}, Integers]
 November 6th, 2008, 04:07 PM #6 Senior Member   Joined: Oct 2008 Posts: 215 Thanks: 0 Re: A Diophantine equation Hehe, How if there're 5 or 7 variables instead of 6 variables?
 November 7th, 2008, 11:17 AM #7 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: A Diophantine equation It's can't find anything for 5. For 7, it gives (3, 2, 2, 2, 1, 1, 1).
 November 7th, 2008, 02:25 PM #8 Senior Member   Joined: Oct 2008 Posts: 215 Thanks: 0 Re: A Diophantine equation So it seems it is not so good. For five numbers, below is a solution 1 1 3 3 4
 November 7th, 2008, 02:29 PM #9 Senior Member   Joined: Oct 2008 Posts: 215 Thanks: 0 Re: A Diophantine equation In fact, I have proved that there're no non-zero solutions for 6 variables with help of computer: See more details at http://zdu.spaces.live.com/blog/cns!...2037!164.entry

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