Cubes as Sums of Consecutive Odd Integers

David W. Hansen

Nichomachus, a mathematician from the early Christian era, is credited with being the first to state that cubical numbers are always equal to the sum of consecutive odd integers [1]. For example,

1³ = 1 = 1

2³ = 8 = 3 + 5

3³ = 27 = 7 + 9 + 11

4³ = 64 = 13 + 15 + 17 + 19

5³ = 125 = 21 + 23 + 25 + 27 + 29

Note that in each case above, the number of consecutive odd integers needed to represent the cube is equal to the base of the cube. For example, 2³ is represented by 2 consecutive odd integers; 5³ is repre-sented by 5 consecutive odd integers. In general, we will find that any cube n³ may be written as a sum of n consecutive odd integers.

Now, how can we write any cube as a sum of consecutive odd integers? It’s easy! From the article on Representing Natural Numbers by Sums of Consecutive Odd Integers [2], we find that a natural number N can be represented as a sum of consecutive odd integers if and only if it can be written as a product of two factors, each greater than 1 and of the same parity. Using the ordering method from this article, we get

2³ = 2(2²) = 2(4) = 4 + 4 = (4-1) + (4+1) = 3 + 5, using the + sign as pivot.

3³= 3(3²) = 3(9) = 9 + 9 + 9 = (9-2) + 9 + (9+2) = 7 + 9 + 11, using the middle 9 as pivot.

6³ = 6(6²) = 6(36) = 36 + 36 + 36 + 36 + 36 + 36

= (36-5) + (36-3) + (36-1) + (36+1) + (36+3) + (36+5)

= 31 + 33 + 35 + 37 + 39 + 41.

In general, for any n³, where n is a natural number, we have n³ = n(n²) = n² + . . . + n², which is a sum of n addends of n², which gives us n consecutive odd integers after the ordering method is used.

If n is even, then so is n², and if n is odd, then n² is also, so the two factors of n³ = n(n²) are of the same parity, and we can write n³ as a sum of consecutive odd integers, where, using Theorem 1 of [2] for

n³ = n(n²) = ka, we have the first term F = a - (k-1) = n² - (n-1), and the last term L = a + (k-1)

= n² + (n-1) = n² – (n-1) + 2(n-1) = F + 2(n-1). This gives us

n³ = [n² - (n-1)] + . . . + [n² + (n-1)]. (1)

Now, n³ can also be written as n²(n). Will this factorization give us a sum of consecutive odd integers different from the one we got using n(n²)? Let’s see.

Using Theorem 1 of [2] for n³ = n²(n) = ka, we have the first term F = a - (k-1) = n - (n²- 1), and the last term L = a + (k-1) = n + (n² - 1). Now, n - (n² - 1) is negative for n > 1 (see proof below), so we get

n³ = [n - (n²- 1)] + . . . + [n + (n² - 1)]

= - [(n² - 1) - n] + . . . + [n + (n² - 1)]

= - [(n² - 1) - n] + . . . + [(n² - 1) - n] + [(n² - 1) - n + 2] + . . . + [n + (n² - 1)]

= 0 + [(n² - 1) - n + 2] + . . . + [n + (n² - 1)]

= [(n² - 1) - n + 2] + . . . + [n + (n² - 1)]

= [n² - n + 1] + . . . + [n² + n - 1)]

= [n² - (n - 1)] + . . . + [n²+ (n - 1)] (2)

Since (2) is identical to (1), we see that n²(n) does not give us a different sum of consecutive odd integers for n³ for n > 1. Of course, if n = 1, we get F = n - (n²- 1) = 1 - (1² - 1) = 1, and L = n + (n²- 1) = 1 + (1² - 1) = 1, which, since F = L gives us 1³ = 1, which, strictly speaking, is not a sum of consecutive odd integers, but we shall make an exception here once!

Proof that if n is a natural number > 1, then n - (n² - 1) < 0.

If n is a natural number > 1, then n ≥ 2 > (1 + √5) / 2 ≈ 1.618.

So, n > (1 + √5) / 2

n > 1/2 + √5/2

n – 1/2 > √5/2

(n – 1/2 )² > (√5/2)²

n² – n + 1/4 > 5/4

n² – n – 1 > 0

n² – 1 > n,

or n < n² – 1,

and n - (n² - 1) < 0 for n > 1.

All of this gives us

Theorem 1

If n is a natural number, then n³ can be written as a sum of n consecutive odd integers

whose first term is F = n² – (n-1), and whose last term L = n² + (n-1) = F + 2(n-1).

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Example Write 7³ as a sum of 7 consecutive odd integers. Since n = 7, the first term is F = 7² – (7-1) = 49 – 6 = 43, and the last term is L = 7² + (7-1) = 49 + 6 = 55, or L = 43 + 2(7-1) = 43 + 12 = 55. Thus,

7³ = 43 + 45 + 47 + 49 + 51 + 53 + 55.

Now, take a look at the beginning of this section where various cubes were written as sums of consecutive odd integers. If we form the sum of the first 4 cubes and their consecutive odd integer representations, we have

1³ + 2³ + 3³ + 4³ = (1) + (3+5) + (7+9+11) + (13+15+17+19),

or 1³ + 2³ + 3³ + 4³ = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19.

Here we were able to write the sum of the first 4 cubes as the sum of the first 10 consecutive odd integers! How very nice! Could it be possible that we could write the sum of the first n cubes as a sum of consecutive odd integers starting at 1? Let’s see. Perhaps some of the work we did in earlier sections can help us here.

First, recall that 1 + 2 + . . . + n = n(n+1)/2, (3)

and 1 + 3 + 5 + . . . + (2n-1) = n². (4)

Second, in the article Cubes to a Square [3], we proved that

1³ + 2³ + . . . + n³ = (1+2+ . . . + n)². (5)

If we let S = 1 + 2 + . . . + n in (5), we have

1³ + 2³ + . . . + n³ = (1 + 2 + . . . + n)². = S². (6)

Since n can be any natural number, we may replace n by S in (4) above to get

S² = 1 + 3 + . . . + (2S-1), and (6) then becomes

1³ + 2³ + . . . + n³ = (1 + 2 + . . . + n)². = 1 + 3 + 5 + . . . + (2S-1), (7)

From (3) above, we know that S = 1 + 2 + . . . + n = n(n+1)/2. So, replacing S by n(n+1)/2 in (7), we have

1³ + 2³ + . . . + n³ = (1 + 2 + . . . + n)² = 1 + 3 + 5 + . . . + (2S-1),

= 1 + 3 + 5 + . . . + [2n(n+1)/2 – 1]

= 1 + 3 + 5 + . . . + (n² + n – 1) (8)

Now, each of the cubes on the left-hand side of this equation is represented by the sum of the number of consecutive odd integers equal to its base. It follows then that the total number of consecutive odd integers

k used to represent this sum of cubes must equal the sum of the bases of the cubes; namely,

k = S = 1 + 2 + . . . + n, giving us the following theorem.

Theorem 2

Let W be the sum of the consecutive cubes from 1 to n, and let S be the sum of the

bases of these cubes. Then, W can be written as the sum of all the consecutive odd

integers from either 1 to 2S – 1 or 1 to n² + n – 1. The total number k of these conse-

cutive odd integers is S, and W = S². Thus, we have

W = 1³ + 2³ + . . . + n³ = 1 + 3 + 5 + . . . + (2S-1)

= 1 + 3 + 5 + . . . + (n² + n –1),

= S² = (1 + 2 + 3 + . . . + n)².

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Example a) Determine how many consecutive odd integers starting at 1 are needed to write

1³ + 2³ + 3³+ 4³ + 5³ + 6³ as a sum. b) Find the last integer in this sum. c) Find all the integers.

a) The number k of consecutive odd integers needed is equal to the sum S of the bases of the cubes;

that is, S = 1 + 2 + 3 + 4 + 5 + 6 = 21.

b) The last integer is either 2S - 1 = 2(21) - 1 = 41, or, since n = 6,

n² + n - 1 = 6² + 6 - 1 = 41.

c) The consecutive odd integers are:

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41.

Example 71 is the last term in a set of consecutive odd integers starting at 1 which represents a sum of

consecutive cubes starting at 1. Find

a) the value of the sum of cubes,

b) its largest cube,

c) the number of consecutive odd integers used in the representation of the sum of cubes. Then,

d) write out the sum of cubes, and

e) write out the sum of the consecutive odd integers.

a) From the first part of Theorem 2, the last term is 2S – 1. Thus, 2S – 1 = 71, 2S = 72, and S = 36,

the sum of the bases. The value of the sum of cubes is S² = 36² = 1296.

b) Its largest cube is n³. To find n, we use the second part of Theorem 2, which tells us the last

term is n² + n – 1. Thus, n² + n –1 = 71, n² + n – 72 = 0, (n-8)(n+9) = 0, and n = 8.

The largest cube is then 8³= 512.

c) There are S = 36 integers in the representation of the sum of cubes by a sum of consecutive odd

integers.

d) Since n = 8, the sum of cubes is 1³ + 2³ + 3³ + 4³ + 5³ + 6³ + 7³ + 8³.

e) Since the first term is 1 and the last term is 71, we have

1 + 3 + 5 + 7 + 9 + . . . + 65 + 67 + 69 + 71.

In summary,

1³ + 2³ + 3³ + 4³ + 5³ + 6³ + 7³ + 8³ = 1 + 3 + 5 + 7 + 9 + . . . + 65 + 67 + 69 + 71

= (1 + 2 + 3 + . . . + 8)²

= 36²

= 1296

References

[1] Reid, Constance, From Zero To Infinity, 3rd ed., New York: Thomas Y. Crowell Company, 1966

(Apollo Edition), pp. 110-111

[2] Hansen, David W., Representing Natural Numbers by Sums of Consecutive Odd Integers.

(mathematics1.fortunecity.com/ConsecutiveOddIntegers.html)

[3] Hansen, David W., Cubes to a Square.

(mathematics1.fortunecity.com/CubesToSquare.html)

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