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Say we have a system of $m$ equations of the form: $$a_{11} x_1 + a_{12} x_2 + ... + a_{1n} x_n = p_1$$ $$...$$ $$a_{m1} x_1 + a_{m2} x_2 + ... + a_{mn} x_n = p_m$$ Where the $p_i,x_j \in \mathbb{R}$, but the $a_{ij}$'s are boolean values, i.e $a_{ij}\in\{0,1\}$.

So, if I want to solve for $\{a_{ij}\}$ and $\vec{x}$ for a given $\vec{p}$, a total of $(m+1)n$ variables,

  1. What are the conditions on $m$ ensuring the existence of at most one solution?
  2. Does it change anything if $p_i,x_j \in \mathbb{Z}$ instead of $\mathbb{R}$?
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[Ignore this] Seemed like the total is $m\,(n+1)$ unknowns, but I mistook $p_i$ for the unknowns rather than $x_i$. –  ccorn Jul 10 '13 at 21:30
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up vote 1 down vote accepted

Here is a generic construction for this problem. Let $\vec{x} = \vec{1}$ and set all $a_{ij}=0$ except $a_{ii} = p_i$. That is a solution.

Note that this is an integer solution as well, as long as $\vec{p}$ is an integer vector.

I am surprised how you set up the problem though - typically, don't we have $a_{ij}$ fixed?

EDIT Even more than that. For any non-zero vector $\vec{x}$ there is a solution which works for any $\vec{p}$. Let $a_{ij} = 0$ when $i \neq j$ and let $a_{ii} = p_i/x_i$.

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$p$ is real.. but I think what you are saying is that there will always be a trivial solution.. –  nbubis Jul 10 '13 at 21:35
    
Actually, I'm claiming there will be an uncountably multiplicity of solutions. About the integer solution: you must at least supply integer boundary conditions and then you are guaranteed that an integer solution will also exist... –  gt6989b Jul 10 '13 at 21:40
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