Soo huge?
x 1 3 + x 2 3 + x 3 3 + x 4 3 + x 5 3 + x 6 3 + x 7 3 + x 8 3 + x 9 3 + x 1 0 3 = y 1 3 + y 2 3 + y 3 3 + y 4 3
Consider the equation above where all the variables are non-zero integers.
What can we say about the number of solutions?
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3 solutions
Here is one subset of possible solutions for the given equation:
( 3 x ) 3 + ( 4 x ) 3 + ( 5 x ) 3 + ( 3 y ) 3 + ( 4 y ) 3 + ( 5 y ) 3 + ( 3 z ) 3 + ( 4 z ) 3 + ( 5 z ) 3 + w 3 = ( 6 x ) 3 + ( 6 y ) 3 + ( 6 z ) 3 + w 3 for any integers x , y , z , w .
Note that 3 3 + 4 3 + 5 3 = 6 3 .
Boy, that was elegant.
I am not posting a solution for it, I am working on a variety of problems to publish so I can't disclose the actual solution. But just to see it has solutions consider this example,
( 1 1 3 ) 3 + ( 1 8 1 ) 3 + ( 1 7 9 ) 3 + ( − 2 3 3 ) 3 = ( 5 7 4 2 ) 3 + ( 4 4 5 8 ) 3 + ( − 6 3 7 7 ) 3 + ( − 2 5 4 7 ) 3 + ( − 1 2 7 3 ) 3 + ( − 1 ) 3 + ( − 2 ) 3 + ( − 2 ) 3 + ( − 3 ) 3 + ( − 7 ) 3
All natural numbers can be the sum of at most 9 cubes, so since the sum of four arbitrary cubes is a natural number, there will always exist 9 cubes that add up to that number, if some can be 0 . If fact, if we're allowed both positive and negative cubes, only at most 5 are needed.
See Waring's Problem