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A Pattern of Signs

Shailesh A Shiralimath@rishivalley.org

July 25, 2011

Abstract

We describe an amusing trail involving a pattern of signs. When applied to the

sequences of squares, cubes, etc, the patterns yield unexpected results which look 

quite magical at first sight. But the underlying algebra is very simple.

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Take any four consecutive integers a,b, c,d . Then it is obvious that a + d  = b + c;

that is, the sum of the first and last number equals the sum of the middle two num-

bers Let us write this relation as +a − b − c + d  = 0. Note the sequence of signs:

+,−, −,+ . We shall write the above statement in the following form:

The sign sequence +,−,−,+ if applied to any 4 consecutive integers

 yields a sum of 0.

For example, if the 4 consecutive integers are 12, 13, 14, 15 we have:

+12 − 13 − 14 + 15 = 0.

Now reverse all the signs in this string; we get the string −,+,+,− . Concatenating

these two strings of length 4 together we get the following string:

+, −,−,+,−,+,+,− ,

of length 8. Now here is a surprising fact:

The sign sequence +,−,−,+, −,+,+,− if applied to any 8 consecutive

squares yields a sum of  0.

For example, take the 8 consecutive squares 9, 16, 25, 36, 49, 64, 81, 100. We have

now:

+9 − 16 − 25 + 36 − 49 + 64 + 81 − 100 = 0.

Please verify this statement. Then try it out on other collections of 8 consecutive

squares.

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Now take the string +,−,−,+,−,+,+,− and reverse all its signs; we get the

string

−,+,+, −,+, −,−,+ .

Concatenating these two strings of length 8 together, we get the following string:

+,−,−,+,−,+,+, −,−,+,+,−,+,−,−,+ ,

of length 16. Here is the next surprising fact:

The sign sequence

+,−,−,+,−,+,+,−,−,+,+, −,+, −,−,+

if applied to any 16 consecutive cubes yields a sum of  0.

Here is an example: take the 16 consecutive cubes 27, 64, 125, 216, 343, 512, 729,

1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832. Applying the sign sequence

 just given to these numbers we get:

+ 27 − 64 − 125 + 216 − 343 + 512 + 729 − 1000 − 1331

+ 1728 + 2197 − 2744 + 3375 − 4096 − 4913 + 5832 = 0.

Try it out for other collections of 16 consecutive cubes!

I guess you will be able to guess what comes next in this sequence of statements ....

Why do we have such a pattern?

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Explanation

Define f 1 by f 1(n) = n − (n + 1); obviously, f 1(n) = −1 for all n. Hence

 f 1(n) = f 1(n + 2),

i.e.,

 f 1(n) −  f 1(n + 2) = 0,

which yields:

n − (n + 1) − (n + 2) + (n + 3) = 0.

Note a corollary to this: If g(n) is any linear expression in n, then the quantity

g(n) − g(n + 1) − g(n + 2) + g(n + 3)

is identically 0. The reason should be clear.

The sequence of squares

Next, let

 f 2(n) = n

2

− (n + 1)2

− (n + 2)2

+ (n + 3)2.

Note that we have used the pattern of signs obtained at the end of the preceding step.

It is obvious what will be the outcome of this formulation, even without doing any

computation: the n2 terms will cancel out, while the linear terms will yield 0 by virtue

of the comment made just above; so f 2(n) will reduce to a constant, whose value will

be −12 − 22 + 32 = 4. Hence we have this identity:

n2 − (n + 1)2 − (n + 2)2 + (n + 3)2 = 4.

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From this relation we get, for any n,

n2 − (n + 1)2 − (n + 2)2 + (n + 3)2 = (n + 4)2 − (n + 5)2 − (n + 6)2 + (n + 7)2,

implying that the following relation is true for all n:

n2 − (n + 1)2 − (n + 2)2 + (n + 3)2 − (n + 4)2 + (n + 5)2 + (n + 6)2 − (n + 7)2 = 0.

As earlier, we have a corollary which follows naturally: If h(n) is any quadratic expres-

sion in n, then the quantity

h(n) − h(n + 1) − h(n + 2) + h(n + 3) − h(n + 4) + h(n + 5) + h(n + 6) − h(n + 7)

is identically 0.

The sequence of cubes

Next we move the cubic case; let

 f 3(n) = n3 − (n + 1)3 − (n + 2)3 + (n + 3)3 − (n + 4)3 + (n + 5)3 + (n + 6)3 − (n + 7)3.

Note that we have used the pattern of signs obtained at the end of the preceding step. As

earlier, it is obvious what will be the outcome of this formulation, even without doing

any computation: the n3 terms will cancel out; the quadratic terms will yield 0 by virtue

of the comment made just above; and so will the linear terms. So f 3(n) will reduce to a

constant, whose value will be −13 − 23 + 33 − 43 + 53 + 63 − 73 = −48. Hence we have

this identity:

 f 3(n) = f 3(n + 8),

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and so:

n3 − (n + 1)3 − (n + 2)3 + (n + 3)3 − (n + 4)3 + (n + 5)3 + (n + 6)3

− (n + 7)3 − (n + 8)3 + (n + 9)3 + (n + 10)3 − (n + 11)3

+ (n + 12)3 − (n + 13)3 − (n + 14)3 + (n + 15)3 = 0.

It is obvious from this derivation that the process can be continued indefinitely. K

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