FOM: Interesting set of integers

Joe Shipman shipman at
Wed Nov 3 14:43:57 EST 1999

Matthew Frank writes:

>>>I think it would be very interesting to find an elegant formula A(x),

number-theoretic at least in the sense of not involving the above
for which we could prove that {x: A(x)} is not computable.  The best
suggestion I've seen was Barry Mazur's outlandish (though surprisingly
well-justified) one that "x is the sum of three cubes of integers" might

work.  (Note that these integers are allowed to be either positive or
negative, and that such a simple question as whether thirty is the sum
three cubes is still unknown after much computer work.)<<<

Where did Mazur suggest this, and what is the surprising justification
you refer to?  I agree, this is the best example I have ever seen of a
simply defined enumerable set of integers which we don't know (or even
have any reason to suspect) is recursive.

What IS known about the set of sums of 3 cubes?  What is the smallest
integer known not to be in this set?

I know it is conjectured that every integer is the sum of  4 cubes, and
there is strong numerical evidence supporting the proposition that this
is true for almost all integers (even if no changes of sign are
allowed).  So we have strong reason to believe that the set of sums of 4
cubes is cofinite and therefore recursive.

5 cubes is enough to get any integer:   (n+1)^3 + (-n)^3 + (-n)^3 +
(n-1)^3 = 6n , so we can get any multiple of 6 with 4 cubes, and thus
can get any integer with 5 cubes because the cubes exhaust the residue
classes mod 6.

-- JS

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