Since $24 = 3 + 5 + 7 +9$, the number $24$ can be written as the sum of at least two consecutive odd positive integers.
(a) Can $2005$ be written as the sum of at least two consecutive odd positive integers? If yes, give an example of how it can be done. If no, provide aproof why not.
(b) Can $2006$ be written as the sum of at least two consecutive odd positive integers? If yes, give an example of how it can be done. If no, provide a proof why not.
$$\text{Sum of first $n$ consecutive odd positive integers}=n^2$$ Let a number $x$ be written as the sum of consecutive odd positive integers starting from the $(m+1)^{th}$ odd positive integer to the $n^{th}$ odd positive integer. Then, $$x=n^2-m^2=(n+m)(n-m)$$
(a) $x=2005$
$$2005=5\times401=(n+m)(n-m)$$ where $n,m\in\mathbb N$.
The factorization can be done as $1\times2005$ and $5\times401$. Moreover, $n+m\gt n-m$. Thus, in the first case, $$n+m=2005$$ $$n-m=1$$ $$n=1003,m=1002$$ But we need at least two consecutive odd numbers, i.e., $n-m\gt1$, so this solution is not valid.
In the second case, $$n+m=401$$ $$n-m=5$$ $$n=203,m=198$$ Thus, $2005$ can be written as the sum of odd consecutive numbers starting from the $199^{th}$ odd number to the $203^{th}$ odd number. Hence, $$2005=397+399+401+403+405$$
Similarly, solve for $x=2006$.