Solving a simple quadratic in modular arithmetic x^2 \equiv 9\pmod{13}

dificultehcm

dificultehcm

Answered question

2022-04-19

Solving a simple quadratic in modular arithmetic
x29±mod13

Answer & Explanation

RormFrure6h1

RormFrure6h1

Beginner2022-04-20Added 13 answers

Step 1
One way to see this is solving x2=9 over the integers yields x=±3.
Clearly, 3 is one solution, and similarly, 310(±mod13) so 10 is another solution.
But if 3 is a solution, so must be 3+13=16, etc.

Assainato9gf

Assainato9gf

Beginner2022-04-21Added 15 answers

Step 1
Because 13 is a ' number. Then Z13 is a finite field.
That means that
x20±mod13
has exactly two solutions
x{3,3}={3,10}

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