# Sum of $k$ consecutive integers

I'm trying to prove or disprove:

Characterize $k \in\Bbb N$ that makes the following statement true: “$a$ is divisible by $k$ if and only if $a$ can be written as a sum of $k$ consecutive integers”.

I tried the converse:

Proof. If $k\mid a$ and $k=2q+1$, $q \in\Bbb Z$, then $a=c(2q+1)$ for $c \in\Bbb Z$.

Notice: \begin{align*}ck &= (c - q) + \big(c - (q-1)\big) + \ldots+ (c -1) + c\\&+\, (c +1) +\ldots+ \big(c + (q-1)\big) + (c + q)\;,\end{align*} so $ck$ is the sum of the $2q+1$ consecutive integers.

Hence $a$ is expressable as a sum of $k$ consecutive integers

Is this okay? Other methods?

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For the first part where you say $k|a \implies k=2q+1...$ is wrong. It should be $kq=a$ for some $q\in\mathbb{Z}$. And maybe have two cases one for when $k$ is odd and then when $k$ is even. – user60887 Oct 28 '13 at 1:05
I guess you meant "I tried the forward direction." – Lord Soth Oct 28 '13 at 1:05

Since the statement you are trying to prove contains "if and only if", you also need to prove

"if $a$ can be written as the sum of $k$ consecutive integers, $a$ is divisible by $k$."

Also, I'm not sure if it's just me being tired, but I'm not following your 'Notice:' line of reasoning.

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The reasoning is correct; the OP omitted a lot of parentheses that would have made it much clearer. I’ve taken the liberty of inserting them. – Brian M. Scott Oct 28 '13 at 5:26

You were asked to find a characterization of those positive integers $k$ that have a certain property; this means that you’re to find and prove a statement of the form

a positive integer $k$ has the specified property if and only if $k$ is something.

What you’ve proved is that $k$ has the desired property if $k$ is odd; to complete the characterization you must show that if $k$ is even, it does not have the specified property. That means that you must show that if $k$ is an even positive integer, then there is some integer $a$ such that either

• $k\mid a$, but $a$ cannot be written as the sum of $k$ consecutive integers, or
• $k\nmid a$, but $a$ can be written as the sum of $k$ consecutive integers.

HINT: What if $a=k$?

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