How to solve modular equivalence with polynomial \(\displaystyle{x}^{{2}}-{3}{x}+{2}\equiv{0}\pm{o}{d}{\left\lbrace{14}\right\rbrace}\)

peggyleuwpodg

peggyleuwpodg

Answered question

2022-03-29

How to solve modular equivalence with polynomial
x23x+20±mod14

Answer & Explanation

Cason Singleton

Cason Singleton

Beginner2022-03-30Added 13 answers

Step 1
If 14(x1)(x2), we cannot conclude 14x1 or 14x2, because 14 is not '.
But
14=2×7,
So 14(x1)(x2)
means 2(x1)(x2) and 7(x1)(x2)
Since 2 and 7 are ', this means (2x1 or 2x2) and (7x1 or 7x2)
Thus, there are four solutions (mod 14).

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