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Assume that the $a$'s in the following equations must be limited to a binary values of $\{0,1\}$ and the $c$'s are integer constants: \begin{align} 2a_{34}a_{23}a_{24} &= c_1\\ a_{23} + a_{24} + a_{34} &= c_2\\ 2 a_{34}a_{13}a_{14} &= c_3\\ a_{13} + a_{14} + a_{34} &= c_4\\ 2 a_{24}a_{12}a_{14} &= c_5\\
a_{12} + a_{14} + a_{24} &= c_6 \end{align} I recognize that these are highly restricted form of Diophantine equations, but I've been unable to find what they are called (once I know their names I can lookup proofs on any properties they have). If they don't have a specific name, how could one show that a] a solution exists, b] it is unique c] if not unique, how many solutions there are?

My CS background forces to acknowledge that I could enumerate through all $2^6$ possibilities, but I don't think I'd understand what is going if I did that. Also, I'd like to know how this would scale when there are more then 6 free variables.

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You have to be more specific, especially when you talk of scaling. – Aryabhata Jan 6 '11 at 19:48
@Hooked: I assumed that when you said "binary values" you meant that each $a$ has to be either $0$ or $1$, so I changed the notation ($[0,1]$ is the interval of all real numbers $x$ with $0\leq x\leq 1$; the set whose only elements are $0$ and $1$ is denoted with curly brackets). If this is not the case, then you'll need to say what you really meant by "binary values". – Arturo Magidin Jan 6 '11 at 19:53
@ Moron - I realize this is still vague, but the upper bound is clearly 2^n, where n is the number of c's. Given some structure of these "binary Diophantine", can this number be reduced? – Hooked Jan 6 '11 at 19:53
@ AM, that is correct. I intended to say the set containing only the elements zero and one. Thanks! – Hooked Jan 6 '11 at 19:54
@Hooked: I was actually talking about the vagueness of the structure. Just an example is not enough to deduce what you had in mind... – Aryabhata Jan 6 '11 at 19:58
up vote 1 down vote accepted

I think this could easily be translated into a boolean problem. We will interpret $x=0$ with $x$ is false and $x=1$ with $x$ is true. Now Consider quoations of the form


where $z$ is an integer and $a, b, c \in \{0,1\}$.

Now z gives you the amount of variables that you can set to "true". So we can observe because of the interval of $a,b,c$ that $z \in [0,3]$ it will give you basically 4 cases.

If $z=0$ you get the equoation $a = 0 \land b = 0 \land c = 0$, as all variables are $0$ and therefore "false".

If $z=1$, exactly one variable is "true", so you have multiple cases here, you get the equoations

$(a = 1 \land b = 0 \land c = 0) \lor (a = 0 \land b = 1 \land c = 0) \lor (a = 0 \land b = 0 \land c = 1)$

If $z=2$ you get

$(a = 1 \land b = 1 \land c = 0) \lor (a = 1 \land b = 0 \land c = 1) \lor (a = 0 \land b = 1 \land c = 1)$

If $z=3$ you get

$a = 1 \land b = 1 \land c = 1$

The $2abc=y$ equations are quite simple, you will either have $y = 0$ or $y = 1$. If you get $y = 0$ the equivalent logical equation is

$a = 0 \lor b = 0 \lor c = 0$

As at least one of the variables is 0 in that case.

Else if $y = 1$ you have $a = 1 \land b = 1 \land c = 1$ as all variables are "true" in this case.

Now you are left with a boolean "SAT"-Problem ( ) where many algorithms exist that are clever and will work for a huge number of quoations although the problem is NP-complete.

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I don't think that this kind of restriction occurs in diophantine equations to begin with, so I doubt they carry a special name.

There are a lot of very simple observations that you can make, though: since each $a$ can only take the value $0$ or $1$, then the only possible values of $c_1$, $c_3$, and $c_5$ are $0$ or $2$; and $c_2$, $c_4$, and $c_6$ just tell you how many of the $a$s are equal to $1$ and how many are equal to $0$, so the only possible values are $0$, $1$, $2$, or $3$. In particular, $c_1 = 2$ if and only if $c_2=3$; and similarly with $c_3$ and $c_4$, and with $c_5$ and $c_6$. Your variables are divided into three non-disjoint subsets: $\{a_{23}, a_{24}, a_{34}\}$, $\{a_{13}, a_{14}, a_{34}\}$, and $\{a_{12}, a_{14}, a_{24}\}$; any two of the sets have exactly one element in common, and the triple intersection is trivial. Since you will know how many in each set are equal to $1$ and how many are equal to $0$, it will be easy to determine any solutions using these facts.

You can also delete the $2$s from the first, third, and fifth equation (suitably modifying $c_1$, $c_3$, and $c_5$ if necessary), and reduce $c_2$, $c_4$, and $c_6$ modulo $2$ so that you have a system of equations over $\mathbb{F}_2$, the field with two elements; solve it over that field and then check which (if any) of the solutions still works over $\mathbb{Z}$ (which is just a matter of whether $c_2$, $c_4$, and $c_6$ were originally $2$ instead of $0$, or $3$ instead of $1$).

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