# Is zero a cube?

March 2, 2005 5:16 PM Subscribe

This page mentions, among other things, that "100 is the smallest square which is also the sum of 4 consecutive cubes." Obviously, this refers to the sum of 1^3 (1), 2^3 (8), 3^3 (27) and 4^3 (64). But it seems to me that the sequence can be pushed back to start with 0^3 (0), in which case you can get four consecutive cubes adding up to 36, which is a square. Is zero then not considered a cube?

I've been thinking about this for far too long, and I can see it going either way. For example: zero should not be considered a cube, because the logical implication of this is that it can be divided by its cube root. But its cube root is zero, and division by zero is an undefined operation.

But at the same time, there is a number

I've been thinking about this for far too long, and I can see it going either way. For example: zero should not be considered a cube, because the logical implication of this is that it can be divided by its cube root. But its cube root is zero, and division by zero is an undefined operation.

But at the same time, there is a number

*n*such that

*n*^3 = 0, which seems to me to be the very definition of a cube.

Where x^(-y) = 0, x is undefined. Zero breaks negative exponents in the accepted laws for using exponents. Therefore zero breaks positive exponents, e.g. x^(-(-y)) = 0, therefore zero is not an accepted cube, or any other multiplier.

posted by AlexReynolds at 5:28 PM on March 2, 2005

posted by AlexReynolds at 5:28 PM on March 2, 2005

*Zero breaks negative exponents in the accepted laws for using exponents. Therefore zero breaks positive exponents...*

That's not really a good justification for excluding it, though, since one could say the same of logs of negative numbers, but that wouldn't be justification for excluding ln(-(-5)).

It's really a matter of definitions. In this case, the page you give links to Mathworld's definition of "cubic number," which gives it as the cube of a positive integer, nicely excluding zero. One reason for this is that the cubic numbers are often studied as a configuration of points (in this case, a cubic configuration), and many theorems would break down if one allowed empty configurations.

In addition, it's

*extremely*useful to look at generating functions for integer sequences, and it's difficult to "work in" zero, since a

_{0}x

^{0}is often already taken. Here's a good book on generating functions with neat applications to a variety of combinatorical problems.

posted by j.edwards at 5:49 PM on March 2, 2005

*But it seems to me that the sequence can be pushed back to start with 0^3 (0), in which case you can get four consecutive cubes adding up to 36, which is a square.*

In which case, couldn't you push it back to zero, being the sum of itself, cubed, thrice?

posted by pompomtom at 6:05 PM on March 2, 2005

*Where x^(-y) = 0, x is undefined. Zero breaks negative exponents in the accepted laws for using exponents. Therefore zero breaks positive exponents, e.g. x^(-(-y)) = 0, therefore zero is not an accepted cube, or any other multiplier.*

No. You're confusing the unary negation operator with the property of being negative.

Zero raised to any positive power is zero. In particular, 0 is an nth power for all positive n. Zero raised to a negative exponent causes the blackboard to burst into flames. Now, 0^0, that's some crazy business.

Wu and Jedwards have it pretty much right. The page you reference was a little bit sloppy in leaving out the word "nontrivially". You also occasionally see Fermat's Last Theorem referred to as the statement that x^n=y^n+z^n has no solutions -- the trivial ones (0,0,0) and (0,1,-1) (for n odd) are ruled out by the assumption that the problem is interesting.

posted by gleuschk at 6:46 PM on March 2, 2005

Honestly, you need to refine the definitions a bit more. I mean, I can cube 1.1, 1.2, 1.3, and 1.4 and then add them all up. Is this consecutive whole numbers? Consecutive positive integers? The page it links to seems not to consider 0. "A cubic number is a figurate number of the form

posted by Eideteker at 6:52 PM on March 2, 2005

*n³*with*n*a**positive integer**." [emphasis mine]posted by Eideteker at 6:52 PM on March 2, 2005

*the logical implication of this is that it can be divided by its cube root*

No, that doesn't follow; don't get hung up on that. Zero is a cube. 0^3=0. Their link stipulates that, by 'cube', they only mean positive integer cubes. That's a nonstandard use of the word, and in its more conventional sense your solution is valid. I'd be much happier if they just made the wording say '100 is the smallest square which is a sum of four consecutive positive cubes', but whatever.

posted by Wolfdog at 7:48 PM on March 2, 2005

Oddly enough that Mathworld page links to the sequence of Cubic Numbers which starts with 0.

posted by vacapinta at 10:01 PM on March 2, 2005

posted by vacapinta at 10:01 PM on March 2, 2005

I have discovered a truly marvellous proof for this question, which the textarea is too small to contain.

posted by turaho at 7:07 PM on March 3, 2005

posted by turaho at 7:07 PM on March 3, 2005

This thread is closed to new comments.

The history of the concept of zero is pretty damn fascinating. Check out this book for more.

I'm sure others who know more about math than I do (which is pretty much everyone) can add to this.

posted by Dr. Wu at 5:25 PM on March 2, 2005