A natural number written as an arithmetic progression 
Let $a$ and $n>1$ be positive integers and $$x = a+(a+1)+\cdots+(a+n-1) = \dfrac{n(2a+n-1)}{2}$$ where $x$ is also a positive integer. Prove that there exist $a,n$ if and only if $x$ has an odd divisor.

Since one of $n$ and $2a+n-1$ is odd and the other is even and both are greater than $1$, $x$ has an odd factor greater than or equal to $3$. On the other hand, for every $x$ with an odd divisor $p > 3$, there must exist $a,n$ since if $x$ doesn't contain a factor of $2$ then let $n = 2$ and we can make $2a+1$ be any odd number. 
What do we do in the case that $x$ contains a factor of $2$? 

$$4\cdot 3=3+4+5$$
$$4\cdot 5=2+3+4+5+6$$
$$4\cdot 7=1+2+3+4+5+6+7$$
We are now at a roadblock. But we can continue by changing the method.
$$4\cdot 9=1+2+3+4+5+6+7+8$$
$$4\cdot 11=2+3+4+5+6+7+8+9$$
$$4\cdot 13=3+4+5+6+7+8+9+10$$
$$4\cdot 15=4+5+6+7+8+9+10+11$$
etc., etc.
Do you see how this method could be used to analyze other powers of 2?
