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Vol. 2, No. 1, March 2013 | At Right Angles 11
feature
The following identity is very well known: for all positive integers n,
(1) 13
+ 23
+ 33
+ ... + n3
= (1 + 2 + 3 + ... + n)
2
For example, when n = 2 each side equals 9, and when n = 3 each
side equals 36. The result is seen sufficiently often that one may not
quite realize its strangeness. Just imagine: a sum of cubes equal to
the square of a sum!
Identity (1) is generally proved using the method of mathemati- cal induction (indeed, this is one of the standard examples used to
illustrate the method of induction). The proof does what it sets out to
do, but at the end we are left with no sense of why the result is true.
In this article we give a sense of the ‘why’ by means of a simple figure
(so this is a ‘proof without words’; see page 85 of [1]; see also [2]). Then
we mention a result of Liouville’s which extends this identity in a
highly unexpected way.
Slicing a cube
Sum of Cubes and
Square of a Sum
Understanding your identity
Memorisation is often the primary skill exercised when learning algebraic
identities. Small wonder that students tend to forget them well before their use-by
date! Here, the sum of cubes identity is unpacked using a series of pictures more
powerful than symbols. It doesn’t stop there — the article then investigates other
sets of numbers for which ‘the sum of the cubes is equal to the square of the sum’.
Giri Kodur
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12 At Right Angles | Vol. 2, No. 1, March 2013
1. A visual proof
We represent n3
using a cube measuring n × n × n,
made up of n3
unit cubes each of which measures
1 × 1 × 1. We now divide this cube into n slabs of
equal thickness (1 unit each), by cuts parallel to its
base; we thus get n slabs, each measuring n × n × 1
and having n2
unit cubes.
When n is odd we retain the n slabs as they are.
When n is even we further divide one of the slabs
into two equal pieces; each of these measures
n/2 × n × 1. Figure 1 shows the dissections for
n = 1, 2, 3, 4, 5. Observe carefully the difference be- tween the cases when n is odd and when n is even.
Now we take one cube each of sizes 1 × 1 × 1,
2 × 2 × 2, 3 × 3 × 3, . . . , n × n × n, dissect each one in
the way described above, and rearrange the slabs
into a square shape as shown in Figure 2. (We have
shown a slant view to retain the 3-D effect.) Note
carefully how the slabs have been placed; in par- ticular, the difference between how the even and
odd cases have been handled.
Figure 2 makes it clear ‘why’ identity (1) is
true. For, the side of the square is simply
1 + 2 + 3 + ... + n, and hence it must be that
13
+ 23
+ 33
+ ... +n3
= (1 + 2 + 3+... + n)
2
.
It is common to imagine after solving a problem
that the matter has now been ‘closed’. But math- ematics is not just about ‘closing’ problems! Often,
it is more about showing linkages or building
bridges. We build one such ‘extension-bridge’ here:
a link between the above identity and divisors of
integers.
2. A generalization of the identity
First we restate identity (1) in a verbal way: The
list of numbers 1, 2, 3, . . . , n has the property that
the sum of the cubes of the numbers equals the
square of the sum of the numbers. The wording
immediately prompts us to ask the following:
Query. Are there other lists of numbers with the
property that “the sum of the cubes equals the square
of the sum”?
It turns out that there are lists with the SCSS (short
for ‘sum of cubes equals square of sum’) property.
Here is a recipé to find them. It is due to the great
French mathematician Joseph Liouville (1809–
1882), so ‘L’ stands for Liouville.
L1: Select any positive integer, N.
L2: List all the divisors d of N, starting with 1 and
ending with N.
L3: For each such divisor d, compute the number of
divisors that d has.
L4: This gives a new list of numbers which has the
SCSS property!
The recipé may sound confusing (divisors of divi- sors! What next, you may ask) so we give a few ex- amples. (In the table, ‘# divisors’ is a short form for
‘number of divisors’)
FIGURE 1. Dissecting the cubes into flat slabs
(credits: Mr Rajveer Sangha)
FIGURE 2. Rearranging the slabs into a square shape;
note how the odd and even-sized cubes are handled
differently (credits: Mr Rajveer Sangha)
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Vol. 2, No. 1, March 2013 | At Right Angles 13
Example 1. Let N = 10. Its divisors are 1,2,5,10
(four divisors in all). How many divisors do these
numbers have? Here are the relevant data, exhib- ited in a table:
We get this list: 1, 2, 2, 4. Let us check whether this
has the SCSS property; it does:
The sum of the cubes is 13
+ 23
+ 23
+ 43
= 1 + 8 + 8 + 64 = 81.
The square of the sum is (1 + 2 + 2 + 4)2
=
92
= 81.
Example 2. Let N = 12. Its divisors are 1, 2, 3, 4,
6, 12 (six divisors in all). How many divisors do
these numbers have? We display the data in a
table:
This time we get the list 1, 2, 2, 3, 4, 6. And the
SCSS property holds:
The sum of the cubes is 13
+ 23
+ 23
+ 33
+ 43
+
63
= 1 + 8 + 8 + 27 + 64 + 216 = 324.
The square of the sum is
(1 + 2 + 2 + 3 + 4 + 6)2
= 182
= 324.
Example 3. Let N = 36. Its divisors are 1, 2, 3, 4, 6,
9, 12, 18, 36 (nine divisors). Counting the divisors
of these numbers (this time we have not displayed
the data in a table) we get the list 1, 2, 2, 3, 4, 3, 6,
6, 9. Yet again the SCSS property holds true:
The sum of the cubes is 13
+ 23
+ 23
+ 33
+ 43
+
33
+ 63
+ 63
+ 93
= 1296.
The square of the sum is
(1 + 2 + 2 + 3 + 4 + 3 + 6 + 6 + 9)2
= 362 = 1296.
Now we must show that equality holds for each N.
The full justification involves a fair bit of algebra;
we shall do only the initial part, leaving the rest
for you. It turns out that a critical role is played
by the prime factorization of N. We consider two
cases: (i) N is divisible by just one prime number;
(ii) N is divisible by two or more distinct prime
numbers.
A key observation which we shall use repeatedly
is the following: A divisor of a positive integer
N has for its prime factors only those primes
which divide N. For example, the divisors of a
power of 2 can only be powers of 2. If N is divisible
by only two primes p and q, then every divisor of
N must be made up of the very same two primes.
The case when N is divisible by just one prime
number. Rather conveniently, this case turns
out to reduce to the very identity with which we
started! Suppose that N = pa
where p is a prime
number and a is a positive integer. Since the
divisors of a prime power can only be powers
of that same prime number, the divisors of pa
are the following a + 1 numbers:
1, p, p2
, p3
, . . . , pa
.
How many divisors do these numbers have? 1 has
just 1 divisor; p has 2 divisors (1 and p); p2
has 3
divisors (1, p and p2
); p3
has 4 divisors (1, p, p2
and
p3
); . . . ; and pa
has a + 1 divisors. So after carrying
out Liouville’s recipé we get the following list of
numbers:
1, 2, 3, . . . , a + 1.
Does this have the SCSS property? That is, is it true
that
13
+ 23
+ 33
+ . . . +(a + 1)3
= (1 + 2 + 3 + . . . + (a + 1))2
?
Yes, of course it is true! — it is simply identity (1)
with n = a + 1. And we know that the identity is
true. So Liouville’s recipé works when N = pa
.
d
Divisors of d
# divisors of d
1
{1}
1
2
{1, 2}
2
5
{1, 5}
2
10
{1, 2, 5, 10}
4
d
Divisors of d
# divisors of d
1
{1}
1
2
{1, 2}
2
3
{1, 3}
2
4
{1, 2, 4}
3
d
Divisors of d
# divisors of d
6
{1, 2, 3, 6}
4
12
{1, 2, 3, 4, 6, 12}
6