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john !!
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#1
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#1
why can the sum of two cubes never leave a remainder of 3 or 4 when divided by 7 or 9?
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J.F.N
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What kind of explanation are you looking for? Consider the group Z/7Z under multiplication. The cube of every element in this group is either 0,1,3,6 (a subgroup of order 4). Therefore, the sum of two cubes would be either 0,2,5,6. Similarly, for Z/9Z, all cubes are either 0,1,8. So the sum of two cubes is 0,2,7.
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dvs
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Another way to look at it would be to consider cubes modulo 7:
0³ = 0, 1³ = 1, 2³ = 1, 3³ = -1, 4³ = 1, 5³ = -1, 6³ = -1, 7³ = 0

In other words, cubes are either in the form 7k, 7k-1 or 7k+1. So two cubes can be of the form:
7k+7k-1 = 7(2k)-1 = -1 = 6 (mod 7)
7k+7k+1=7(2k)+1 = 1 (mod 7)
7k-1+7k+1=7(2k) = 0 (mod 7)

You can do the same for 9.
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