Show that every integer can be expressed as a sum of five or less cubes of integers .
e.g. 9=2^{3} + 1^{3} ; 22=3^{3}+(2)^{3} +1^{3} +1^{3}+1^{3}
Binomial expansions will reveal that (a + 1)^{3} + (a  1)^{3} = 2a^{3} + 6a
so it’s possible to express any multiple of 6 in terms of four cubes as follows:
6a = (a + 1)^{3} + (a  1)^{3}  a^{3}  a^{3} (1)
Now, we need to somehow split the 6a up into one more cube together with a linear term. Since the product of any three consecutive integers will always have the factors 2 and 3, it follows that (n  1)n(n + 1) is divisible by 6, for any integer n.
In other words, for any integer n, n^{3}  n = 6a for some integer a.
Using (1) n^{3}  n = (a + 1)^{3} + (a  1)^{3}  a^{3}  a^{3}
which gives: n = n^{3} + a^{3} + a^{3}  (a + 1)^{3}  (a  1)^{3}
n = n^{3} + a^{3} + a^{3} + (a  1)^{3} + (1  a)^{3} where a = (n^{3}  n)/6
                         
So it can always be done for any n, but sadly these five cubes are rarely as small as they might be. If we try it out on the examples given in the problem.
When n = 9, a = 120, so 9 = 9^{3} + 120^{3} + 120^{3} + (121)^{3} + (119)^{3}
when n = 22, a = 1771, so
22 = 22^{3} + 1771^{3} + 1771^{3} + (1772)^{3} + (1770)^{3}

Posted by Harry
on 20100803 19:44:29 