Representing Natural Numbers by Sums
of Consecutive EVEN Integers
David W. Hansen
© 2004
In other articles, we have determined which natural numbers can be represented by a sum of either
represented as a sum of consecutive even integers.
Neither 1, 2, 3, 4, or 5 can be written as a sum of consecutive even integers, as one can easily verify. In
fact, since the sum of any two even integers is an even integer, then any sum of consecutive even integers
must be even. Thus, only even natural numbers can be represented as a sum of consecutive even
integers.
In some sense, even integers are the “opposites” of odd integers. Let’s see if Theorem 1 of the article
using the “opposite” word where appropriate. (The words in bold-face have been changed.)
Theorem 1
A natural number N can be represented as a sum of consecutive even integers
if and only if it can be written as a product of two factors, k and a, each greater
than 1 and of opposite parity. Then the first term F and the last term L of this sum
will be, respectively, F = a - (k -1) and L = a + (k -1), where N = ka and k is the
number of terms. The number of ways that N can be written as a sum of
consecutive even integers equals the number of ways that N can be written as
the products of two factors, each greater than 1 and of opposite parity.
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Using Theorem 1 above (as modified for consecutive even integers), we have
6 = 2(3) = 2 3’s, where k = 2 and a = 3. So, F = 3 - (2-1) = 2 and L = 3 + (2-1) = 4,
and 6 = 2 + 4, a sum of consecutive even integers. Also,
= 3(2) = 3 2’s, where k = 3 and a = 2. So, F = 2 – (3-1) = 0 and L = 2 + (3-1) = 4
and 6 = 0 + 2 + 4 = 2 + 4, as above.
8 = 4(2) and these two factors are not of opposite parity. Therefore, 8 cannot be written as a
sum of consecutive even integers.
10 = 2(5) = 5 + 5 = (5-1) + 5 +1) = 4 + 6. Also,
= 5(2) = 2 + 2 + 2 + 2 + 2 = (2-4) + (2-2) + 2 + (2+2) + (2+4) = (-2) + 0 + 2 + 4 + 6
= 4 + 6.
12 = 4(3) = 3 + 3 + 3 + 3 = (3-3) + (3-1) + (3 + 1) + (3 + 3) = 0 + 2 + 4 + 6 = 2 + 4 + 6, or
= 3(4) = 4 + 4 + 4 = (4-2) + 4 + (4 +2) = 2 + 4 + 6.
As these three examples show, commuting the factors does NOT give us a different sum of consecutive
even integers.
No natural number which is a power of 2 can be written as a sum of consecutive even
integers since all of its factors greater than 1 are even, and thus are of the same parity.
18 = 2(9) = 9 + 9 = (9-1) + (9+1) = 8 + 10
= 3(6) = 6 + 6 + 6 = (6-2) + 6 + (6+2) = 4 + 6 + 8
90 = 2(45) = 45 + 45 = 44 + 46
= 3(30) = 30 + 30 + 30 = 28 + 30 + 32
= 5(18) = 18 + 18 + 18 + 18 + 18 = 14 + 16 + 18 + 20 + 22
= 6(15) =15 + 15 + 15 + 15 + 15 + 15 = 10 + 12 + 14 + 16 + 18 + 20
= 9(10) = 9 10’s, where k = 9 and a = 10. So, F = 10 - (9-1) = 2 and
L = 10 + (9-1) = 18, and 90 = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18.
As these examples suggest, Theorem 1 is valid. It may be proved in a manner similar to the proof of
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