Consecutive Integer Summation

All sums of two or more consecutive positive integers can be visualized as the number of objects that are in the shape of a trapezoid. For example, 2 + 3 + 4 can be visualized as having 2 objects in the first row, 3 objects in the second row, and 4 objects in the bottom row, as pictured below.

If we let $x$ be the number of objects in the top row (and the first number in the consecutive series), and $y$ be the number of objects in the bottom row (and the last number in the consecutive series), then the number of rows can be expressed as $h = y - x + 1$ . Following the area formula for a trapezoid, the total number of objects (and the sum of consecutive positive integers from $x$ to $y$ ) can be expressed as $S = \frac{1}{2}(x + y)(y - x + 1)$ .

A positive integer $S$ that is not a power of 2 must have at least one odd factor. There are three cases to examine:

Case 1 : $S$ is odd.
Then set $S = x + y$ and $1 = \frac{1}{2}(y - x + 1)$ , which solves to $x = \frac{S - 1}{2}$ and $y = \frac{S + 1}{2}$ . Both these equations give positive integer values for $x$ and $y$ such that $0 < x < y$ and the sum of the consecutive integers from $x$ to $y$ is $S$ . (For example, if $S = 11$ , then $x = 5$ and $y = 6$ , and 5 + 6 = 11.)

Case 2 : $S$ is even with an odd factor $o > 1$ and an even factor $e$ such that $S = eo$ and $2e > o$ .
Then set $e = \frac{1}{2}(x + y)$ and $o = y - x + 1$ , which solves to $x = e - \frac{1}{2}(o - 1)$ and $y = e + \frac{1}{2}(o - 1)$ . Both these equations give positive integer values for $x$ and $y$ such that $0 < x < y$ and the sum of the consecutive integers from $x$ to $y$ is $S$ . (For example, if $S = 12$ , then $e = 4$ and $o = 3$ , and $x = 3$ and $y = 5$ , and 3 + 4 + 5 = 12.)

Case 3 : $S$ is even with an odd factor $o > 1$ and an even factor $e$ such that $S = eo$ and $2e < o$ .
Then set $e = \frac{1}{2}(y - x + 1)$ and $o = x + y$ , which solves to $x = \frac{1}{2}(o + 1) - e$ and $y = \frac{1}{2}(o - 1) + e$ . Both these equations give positive integer values for $x$ and $y$ such that $0 < x < y$ and the sum of the consecutive integers from $x$ to $y$ is $S$ . (For example, if $S = 10$ , then $e = 2$ and $o = 5$ , and $x = 1$ and $y = 4$ , and 1 + 2 + 3 + 4 = 10.)

In all three cases, we can find positive integer values for $x$ and $y$ such that $0 < x < y$ and the sum of the consecutive integers from $x$ to $y$ is $S$ . Therefore, every positive integer that is not a power of 2 can be written as a sum of two or more consecutive positive integers.

BONUS: If $x$ and $y$ have the same parity (that is, if they are both odd or if they are both even), then $y - x + 1$ is odd. If $x$ and $y$ have a different parity (that is, if one is odd and the other is even), then $x + y$ is odd. Therefore, the sum $S = \frac{1}{2}(x + y)(y - x + 1)$ must necessarily have at least one odd factor, so no positive integer that is a power of 2 can be written as a sum of two or more consecutive positive integers.

Start by considering the numbers that can be written as the sum of two consecutive integers, $n + (n+1) = 2n + 1$ . This means that every odd integer can be written as the sum of two consecutive integers. Let's now suppose that we have a number expressible as $x = 2^k c$ where c is an odd number. Then $c = 2n+1$ for some integer n so $x = 2^{k+1}n + 2^k = \sum_{i= (1 - 2^k)}^{2^k} (n+i)$ which proves that any positive integer may be written as a sum of consecutive integers.

Now, the attentive reader may notice that this is weird. Even powers of 2 can be written this way. For example, 4 = -3 + -2 + -1 + 0 + 1 + 2 + 3 + 4. This means that the problem is trivial, and the question is ill-posed. Well, maybe not. I believe that the question means "as a sum of two or more consecutive positive integers". This would disqualify my trivial sum for 4. And more than that, it would add the difficulty of having to show the sum I propose is actually not trivial for numbers that are not perfect powers of 2:

So let's suppose that $x = 2^k c$ where $c>1$ . This would imply that $2n + 1 > 1$ and therefore $n>0$ . This, in turn, would ensure that at least all the terms from $n$ to $n + 2^k$ would be positive. Meanwhile, at most only the terms from $n + (1 - 2^k)$ to $n - 2$ would be negative, leaving at least 3 positive consecutive integers.

Example: $12 = 2^2 \times 3 = 4(2\times1 + 1) = -2 - 1 + 0 + 1 + 2 + 3 +4 + 5 = 3 + 4 + 5.$