Looking for solution of a linear equation (which is a very important lemma in my research). Given $M\geq 2$ and $1<\beta_i<2$, $1\leq i \leq M$ and the equation:
$h_1\beta_1^{L-1}+h_2\beta_1^{L-2}+\cdots+ h_{L-M-1}\beta_1^{M+1}=h_L+\beta_{1}^{L}$
$h_1\beta_2^{L-1}+h_2\beta_2^{L-2}+\cdots+ h_{L-M-1}\beta_2^{M+1}=h_L+\beta_{2}^{L}$
$\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$
$h_1\beta_M^{L-1}+h_2\beta_M^{L-2}+\cdots +h_{L-M-1}\beta_M^{M+1}=h_L+\beta_{M}^{L}$
where $M\geq 2$ is an integer and $L$ is a lager integer that $L>2M+2$,  $h_1\,,h_2\,,h_{L-M-1}\,,h_{L}$ are variables. 
Can we find a solution $(h_1\,,h_2\, \ldots,h_{L-M-1}\,,h_{L})$ such that $$\sum\limits_{k=1}^{L-M-1}|h_k|+|h_{L}|\leq 2$$?
Thank you for your discussion.
 A: Your system is overdetermined if I understand correctly, and you wish to find a solution $\mathbf{h}$ with small norm. Your norm here is not the euclidean norm, and while the stricly minimum solutions between the two norms may differ, you may still benefit with using the euclidean norm over the norm given. I say that because least squares uses the euclidean norm.
Your system:
$$\pmatrix{\beta_1^{L-1} & \beta_1^{L-2} & \cdots & \beta_{1}^{M+1} & -1\\
  \beta_2^{L-1} & \beta_2^{L-2} & \cdots & \beta_{2}^{M+1} & -1\\
  \vdots & & \vdots & & \vdots\\
  \beta_M^{L-1} & \beta_M^{L-2} & \cdots & \beta_{M}^{M+1} & -1}\pmatrix{h_1 \\ h_2 \\ \vdots \\h_{L-M-1} \\ h_L} = \pmatrix{\beta_1^L \\ \beta_2^L \\ \vdots \\\beta_{M}^L \\ }$$
This actually is somewhat similar to some of my own previous research. One hopefully helpful key term is called ridge regression. The explanation about ridge regression that I like follows. 
Ridge regression basically extends the over-determined system to allow it to be solved with least squares. This is necessary as the least squares finds the best fit solution where there are no exact solutions. Your situation here is the opposite, there are many solutions. In matrix notation your system is
$$A\mathbf{h} = \mathbf{b}$$
If you extend it as such
$$ \left[\matrix{A & I}\right] \pmatrix{\mathbf{h} \\ \mathbf{t}} = \mathbf{b}$$
then you are "weighting" each equation equally. If you use something other than $I$ then the weighting is different. $\mathbf{t}$ is basically a dummy variable. The problem then becomes how to vary those two parameters for your desired results.
I believe this is an open problem, and I feel it is one that does have a solution. I could be wrong, maybe it is solved, but I haven't seen it unless it is written in one of the many publications that I can not understand. And if the solution does exist, it either solves some difficult lattice reduction problems, or does not discretize very well (it is much easier to find a least squares solution in a continuous field like the real numbers than it is to find a solution among integers only for example; that is the main difference between least squares and lattice reduction).
