Multiple of $X^N -1$ So I have two polynomials defined as 
$f(X) = a_{N-1}X^{N-1}+...+a_0$
$ g(X) = b_{N-1}X^{N-1}+...+b_0 $ 
as well as the operation $f*g = c_{N-1}X^{N-1} +...+ c_0$ where $c_i = \sum_{j+k \equiv i}a_jb_k$  meaning a sum of pairs of intergers $j,k$ such that $j+k \equiv i \mod{N}$
I want to show that $fg-f*g $ is a multiple of $X^N-1$. 
Now I have written $fg$ as $\sum_{i=0}^{2N-2}\sum_{j=0}^{i}a_jb_{i-j}X^i$ but I am having trouble to simplify $fg - f*g$ 
I have $fg-f*g =\sum_{i=0}^{2N-2}\sum_{j=0}^{i}a_jb_{i-j}X^i - (c_{N-1}X^{N-1} +...+ c_0) $
and now I am not sure how to simplify...
 A: First, note your written expression for $fg$ is not quite correct. You state it's
$$fg = \sum_{i=0}^{2N-2}\sum_{j=0}^{i}a_j b_{i-j}X^i \tag{1}\label{eq1A}$$
This problem can be fairly easily seen by checking the maximum value of $i$, i.e., $2N - 2$. The value of $b_{i-j}$ at $j = 0$ would be $b_{2N-2}$ while the value of $a_{j}$ at $j = i = 2N - 2$ would be $a_{2N-2}$. However, the indices for $a$ and $b$ only go up to $N - 1$. Instead, a correct way to write $fg$ would be
$$fg = \sum_{i=0}^{N-1}\sum_{j=0}^{N-1}a_{i}b_{j}X^{i+j} = \sum_{i + j = 0, \; 0 \le i,j \le N - 1}^{2N-2} a_{i}b_{j}X^{i+j} \tag{2}\label{eq2A}$$
For simpler notations below, summations involving $i + j$ will implicitly have the limits of $0 \le i,j \le N - 1$. Now, with your operation
$$f*g = c_{N-1}X^{N-1} +...+ c_0 = \sum_{i = 0}^{N-1}c_i X^i \tag{3}\label{eq3A}$$
you have
$$c_i = \sum_{j+k \equiv i \pmod N}a_{j}b_{k} \tag{4}\label{eq4A}$$
Since $0 \le j,k \le N - 1$, you have that $0 \le j + k \le 2N - 2$. Thus, $j + k \equiv i \pmod N \implies j + k = i \; \text{ or } \; j + k = N + i$. Thus, \eqref{eq4A} can be rewritten to include the limits on $j,k$ values and split it into $2$ parts as
$$c_i = \sum_{j+k = i, \; 0 \le j,k \le N - 1}a_{j}b_{k} + \sum_{j+k = N + i, \; 0 \le j,k \le N - 1}a_{j}b_{k} \tag{5}\label{eq5A}$$
Substituting \eqref{eq5A} into \eqref{eq3A}, and adjusting the indices, gives
$$f*g = \sum_{i + j = 0}^{N-1} a_{i}b_{j}X^{i+j} +
\sum_{i + j = N}^{2N-2} a_{i}b_{j}X^{i+j-N} \tag{6}\label{eq6A}$$
As can be seen, the first summation on the right of \eqref{eq6A} matches the first $N - 1$ terms in \eqref{eq2A}, so they cancel out, leaving just the difference of the remaining $N-1$ terms from, where the values in \eqref{eq2A} are just $x^N$ times the ones in \eqref{eq6A}. In particular, you get
$$\begin{equation}\begin{aligned}
fg-f*g & = \sum_{i + j = 0}^{2N-2} a_{i}b_{j}X^{i+j} - \left(\sum_{i + j = 0}^{N-1} a_{i}b_{j}X^{i+j} +
\sum_{i + j = N}^{2N-2} a_{i}b_{j}X^{i+j-N}\right) \\
& = \sum_{i + j = N}^{2N-2} a_{i}b_{j}X^{i+j} - \sum_{i + j = N}^{2N-2} a_{i}b_{j}X^{i+j-N} \\
& = X^N\left(\sum_{i + j = N}^{2N-2} a_{i}b_{j}X^{i+j-N}\right) - \sum_{i + j = N}^{2N-2} a_{i}b_{j}X^{i+j-N} \\
& = \left(X^N - 1\right)\left(\sum_{i + j = N}^{2N-2} a_{i}b_{j}X^{i+j-N}\right)
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
This shows $X^N - 1$ is a factor of $fg-f*g$. If you have trouble understanding any of this, I suggest manually trying $f$ and $g$ for several small values of $N$, in particular, such as $2$,$3$ and/or $4$. You should see how some of the lower power terms cancel, while the remaining lower power terms and the higher power terms match up, as I show above.
