We saw this wonderful identity in
Sum of Cubes:
13 + 23
+ ... + n3 =
(1 + 2 + ... + n)2.
Hence the set of numbers {1,2,...,n} has the property that
the sum of its cubes is the square of its sum.
Are there any other collections of numbers with this
property? Yes, and the following method is guaranteed to
generate such a set.
Pick a number, any number.
Did I hear you say 63? Fine.
List the divisors of 63, and for each divisor of 63,
count the number of divisors it has:
63 has 6 divisors (63, 21, 9, 7, 3, 1)
21 has 4 divisors (21, 7, 3, 1)
9 has 3 divisors (9, 3, 1)
7 has 2 divisors (7, 1)
3 has 2 divisors (3, 1)
1 has 1 divisor (1).
The resulting collection of numbers has the same cool
property. Namely
63 + 43 + 33 +
23 + 23 + 13
= 324 = (6+4+3+2+2+1)2.
Neat, huh?
Presentation Suggestions:
If you are short on time, you can just present the sum
of cubes fact.
The Math Behind the Fact:
From number theory, multiplicative functions
are functions f defined over the positive integers that
satisfy f(xy)=f(x)f(y) whenever integers x,y have no common factors.
Observation.
Once you know the value of f for all prime powers you can
determine f(N) for all integers N.
Easy examples of multiplicative functions are
f(x)=xn for any fixed n. Note also that
if f(x) is multiplicative, so is [f(x)]2 and
[f(x)]2, etc.
It is harder to prove (reference NZM) that
Theorem (*). If f is multiplicative, so is F(n) defined by
F(n)=SUMm|nf(m), where the sum is over all
divisors of n.
Using (*) we see that
d(n)=SUMm|n1 must be multiplicative since f(x)=1
is. But then [d]3 is multiplicative, and
(*) shows that SUMm|nd3(m) is.
Also, from (*) and squaring,
[SUMm|nd(m)]2 is
multiplicative. We wish to show that
SUMm|nd3(m) =
[SUMm|nd(m)]2.
By the Observation, it is enough to show this equality holds
for n any prime power. But this reduces to the
simpler sum of cubes Fun Fact, which is easy to verify!
How to Cite this Page:
Su, Francis E., et al. "Sum of Cubes and Beyond."
Math Fun Facts.
<http://www.math.hmc.edu/funfacts>.
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