Mary Reyes

2021-12-19

Does the equation ${x}^{2}\equiv x\cdot x\equiv 2x\cdot 4xb\text{mod}7$ show that factorization of polynomials $b\text{mod}7$ is not unique? Why or why not?

Cassandra Ramirez

Step 1
${x}^{2}\equiv x\cdot x\equiv 2x\cdot 4xb\text{mod}7$
To show: factorization of polynomials $b\text{mod}7$ is not unique.
Other examples can be taken as
${x}^{2}\equiv 3x\cdot 5xb\text{mod}7$
${x}^{2}\equiv 6x\cdot 6xb\text{mod}7$
Step 2
From the above example it is clear that factorization of polynomials mod 7 is not unique as there are other factorization also.

Orlando Paz

Step 1
Given that, the equation is ${x}^{2}\equiv x\cdot x\equiv 2x\cdot 4x\left(b\text{mod}7\right)$.
${x}^{2}\equiv x\cdot x\left(b\text{mod}7\right)$
${x}^{2}\equiv 2x\cdot 4x\left(b\text{mod}7\right)$
From above equation, it is observed that the factorization of the polynomials $\left(b\text{mod}7\right)$ is not unique.
Step 2
From above equation, it is observed that the factorization of the polynomials $\left(b\text{mod}7\right)$ is not unique.
For example:
${x}^{2}\equiv 2x\cdot 4x\left(b\text{mod}7\right)$
${x}^{2}\equiv \left(2\cdot 4\right)\left(x\cdot x\right)\left(b\text{mod}7\right)$
${x}^{2}\equiv \left(3\cdot 5\right)\left(x\cdot x\right)\left(b\text{mod}7\right)$
From above example, it is observed that the factorization of the polynomials $\left(b\text{mod}7\right)$ is not unique.
Because, ${x}^{2}\equiv 1\left(b\text{mod}7\right)$ can have both solutions $x\equiv ±1\left(b\text{mod}7\right)$. That is, .

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