manudyent7

2022-09-12

Suppose we are given the system of equations

$${\alpha}_{1}A+{\beta}_{1}B+{\gamma}_{1}C=x$$

$${\alpha}_{2}A+{\beta}_{2}B+{\gamma}_{2}C+{\theta}_{2}D=y$$

$${\alpha}_{3}A+{\beta}_{3}B+{\gamma}_{3}C+{\theta}_{3}D=z$$

where ${\alpha}_{i},{\beta}_{i},{\gamma}_{i},{\theta}_{i}$ are chosen from finite field ${\mathbb{F}}_{q}$ where $q$ is prime. Note that the variables here are $A,B,C,D$.

Is it possible to use the equations above to have a unique solution for $A,B,C$?

$${\alpha}_{1}A+{\beta}_{1}B+{\gamma}_{1}C=x$$

$${\alpha}_{2}A+{\beta}_{2}B+{\gamma}_{2}C+{\theta}_{2}D=y$$

$${\alpha}_{3}A+{\beta}_{3}B+{\gamma}_{3}C+{\theta}_{3}D=z$$

where ${\alpha}_{i},{\beta}_{i},{\gamma}_{i},{\theta}_{i}$ are chosen from finite field ${\mathbb{F}}_{q}$ where $q$ is prime. Note that the variables here are $A,B,C,D$.

Is it possible to use the equations above to have a unique solution for $A,B,C$?

Karson French

Beginner2022-09-13Added 15 answers

Temporarily set $D=0$. Call the resulting equations the "reduced" equations. If the determinant for (of the $\alpha ,\beta ,\gamma $) is non-zero, Cramer's rule will give you a unique solution for the reduced system. So you will get more than one solution to the original equations.

So the interesting case is where $\mathrm{\Delta}=0$. That means there are either no or many solutions to the reduced equations. If there are many, then there are clearly many to the original equations.

If there are none, then we have to consider what happens in the original set. If ${\theta}_{1}={\theta}_{2}=0$, there are still none. If ${\theta}_{1}={\theta}_{2}\ne 0$, then the original equations will have a solution provided that we do not have the last two original equations contradicting each other. but we will have more than one solution unless $q$ is small. If ${\theta}_{1}\ne {\theta}_{2}$ then we will have more than one solution.

But some care is probably needed if $q$ is really small, eg 2. Maybe in that case some of the "many"s become 1.

So the interesting case is where $\mathrm{\Delta}=0$. That means there are either no or many solutions to the reduced equations. If there are many, then there are clearly many to the original equations.

If there are none, then we have to consider what happens in the original set. If ${\theta}_{1}={\theta}_{2}=0$, there are still none. If ${\theta}_{1}={\theta}_{2}\ne 0$, then the original equations will have a solution provided that we do not have the last two original equations contradicting each other. but we will have more than one solution unless $q$ is small. If ${\theta}_{1}\ne {\theta}_{2}$ then we will have more than one solution.

But some care is probably needed if $q$ is really small, eg 2. Maybe in that case some of the "many"s become 1.

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