Finding weights that remove residual dot product I have three time-series $m_i$, $k_i$, $s_i$, for $i \in [1,N]$.
The problem is to find constants $A_m$, $A_k$, $A_s$, 
such that 
for $p_i = ( A_m \cdot m_i ) + ( A_k \cdot k_i ) + ( A_s \cdot s_i )$ ,
and a given integer constant $t$
 \[ \sum_{i=1}^{N-t} { (m_i - p_i)  (p_{t+i} - p_i) } = 0  \]
 \[\sum_{i=1}^{N-t} { (k_i - p_i)  (p_{t+i} - p_i) } = 0 \]
 \[\sum_{i=1}^{N-t} { (s_i - p_i)  (p_{t+i} - p_i) } = 0 \]
Not sure if such constants would always exist.
Any pointers ?
 A: Such constants often exist.  We can readily characterize when they do and provide a method to find them.
I presume that nonzero solutions are sought, because $(A_m, A_k, A_s) = (0,0,0)$ always works.
To simplify the notation and reveal what's going on, let $x_1 = A_m$, $x_2 = A_k$, $x_3 = A_s$, and $\mathbf{x} = (x_1,x_2,x_3)'$.  Let $U$ be the $3$ by $n$ matrix with rows $\mathbf{m}$, $\mathbf{k}$, and $\mathbf{s}$ and let $U_1$ be the first $N-t$ columns of $U$ and $U_t$ be the last $N-t$ columns of $U$.  Set $V = U_t - U_1$.  In these terms,
$$\mathbf{p} = \mathbf{x}' U.$$
Similarly, the $t^\text{th}$ difference of $\mathbf{p}$ equals $\mathbf{x}' V$.
Look at the terms in the first equation, for example, and write them in terms of $\mathbf{x}$:
$$\eqalign{
(m_i - p_i)  (p_{t+i} - p_i) &= (u_{1i} - (\mathbf{x}' U_1)_i) (\mathbf{x}' V)_i \\
&= u_{1i}(\mathbf{x}' V)_i - (\mathbf{x}' U_1)_i (\mathbf{x}' V)_i.
}$$
Summing over $i$ and equating to $0$ for the three equations (for $\mathbf{m}$, $\mathbf{k}$, and $\mathbf{s}$) yields the conditions
$$\mathbf{x}' U_1 V' \mathbf{x} = (\mathbf{x}' V U_1')_1 = (\mathbf{x}' V U_1')_2 = (\mathbf{x}' V U_1')_3.$$
Note that the left hand side is a quadratic form in $\mathbf{x}$, $Q(\mathbf{x})$.  
Provided $V U_1'$ is nonsingular, we can find an $\mathbf{x}_0$ for which each of the three right hand expressions has the common value of $1$.  Assuming $Q(\mathbf{x}_0) \ne 0$, which in general will be the case, set $t = 1/Q(\mathbf{x}_0)$, $\mathbf{x} = t \mathbf{x}_0$, and compute
$$Q(\mathbf{x}) = Q(t \mathbf{x}_0) = t^2 Q(\mathbf{x}_0) = 1/Q(\mathbf{x}_0) = t = t(1) = t(\mathbf{x}_0' V U_1')_i = (\mathbf{x}' V U_1')_i,$$
$i = 1, 2, 3.$ This proves $\mathbf{x}$ is a solution.
We encountered two conditions to assure a nonzero solution: first, $V U_1'$ must be a nonsingular ($3$ by $3$) matrix.  (Actually, it suffices that $(1,1,1)$ be in the span of its rowspace.)  Second, $\mathbf{x}_0$ must not be in the null space of the quadratic form $Q$ with matrix $U_1 V'$.

Example
Let the three time series be $\mathbf{m}=(2,4,6,4,2)$, $\mathbf{k}=(2,4,6,8,10)$, and $\mathbf{s}=(8,2,0,2,8)$.  Thus
$$U = \left(
\begin{array}{ccccc}
 2 & 4 & 6 & 4 & 2 \\
 2 & 4 & 6 & 8 & 10 \\
 8 & 2 & 0 & 2 & 8
\end{array}
\right).$$
Suppose $t=1$.  Then the first differences of the time series (of the columns of $U$, that is) are
$$V = \left(
\begin{array}{cccc}
 2 & 2 & -2 & -2 \\
 2 & 2 & 2 & 2 \\
 -6 & -2 & 2 & 6
\end{array}
\right).$$
We compute
$$V'U_1 = \left(
\begin{array}{ccc}
 -8 & -16 & 16 \\
 32 & 40 & 24 \\
 16 & 40 & -40
\end{array}
\right).$$
The solution to $\mathbf{x}_0 V U_1' = (1,1,1)$ is
$$\mathbf{x}_0 = \left(-\frac{1}{16},\frac{1}{32},-\frac{1}{32}\right).$$
Then, because $Q(\mathbf{x}_0) = -\frac{1}{16}$,
$$(A_m, A_k, A_s) = \mathbf{x} = (1, -\frac{1}{2}, \frac{1}{2}).$$
As a check, for this solution we have
$$\mathbf{p} = (5,3,3,1,1).$$
Subtracting $\mathbf{p}$ from the rows of $U_1$ gives
$$\left(
\begin{array}{cccc}
 -3 & 1 & 3 & 3 \\
 -3 & 1 & 3 & 7 \\
 3 & -1 & -3 & 1
\end{array}
\right).$$
The inner products of each of these rows with the first ($t^\text{th}$) differences of $\mathbf{p}$, equal to $(-2,0,-2,0)$, are indeed all $0$, as required.
If you work through the case $t=2$ you will find that this time $VU_1'$ is singular.  Nevertheless, there exists a solution $\mathbf{x}_0 = (\frac{1}{32},0,\frac{1}{32})$ to the equation $\mathbf{x}_0 V U_1' = (1,1,1)$, leading to $(A_m, A_k, A_s) = (\frac{1}{2}, 0, \frac{1}{2})$.  This shows that nonsingularity of $V U_1'$ is not a necessary condition for a solution to exist: we merely need that $(1,1,1)$ be in the span of its rowspace.
When $t=3$, once more $VU_1'$ is singular, but now there exists no $\mathbf{x}_0$.  Thus there cannot be a solution to the original problem in this case.  (The three equations are inconsistent.)
