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Thread: [SOLVED] Quadratic Congruences

  1. #1
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    [SOLVED] Quadratic Congruences

    Regarding quadratic congruences in the form, $\displaystyle y^2 \equiv d (\bmod p) $ where p i s a prime

    1. How exactly do you solve for quadratic congruences in this form?

    This is an example from online: $\displaystyle y^2 \equiv 13 (\bmod 35) $ , then $\displaystyle y \equiv 15 (\bmod 35) $ and $\displaystyle y \equiv 38 (\bmod 35) $. I do not understand the jump from the first step to the next.

    2. In this form, $\displaystyle ax^2 + bx +c \equiv 0 ( \bmod p) $:
    From one website, it said to:
    i.Let $\displaystyle y = 2ax + b $ and $\displaystyle d = b^2-4ac $
    ii. Then get $\displaystyle y^2 \equiv d (\bmod p) $ which is the I have above
    ii. Then substitute y = 2ax + b back into the solutions of $\displaystyle y^2 \equiv d (\bmod p) $ and solve for x.

    *How do you solve for a linear congruence in this form (for example):
    $\displaystyle 10x - 3 \equiv 15 ( \bmod 53) $

    Thank you for reading. Any help is greatly appreciated.
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  2. #2
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    Firstly
    $\displaystyle y^2 \equiv 13 (\bmod 35) $ , then $\displaystyle y \equiv 15 (\bmod 35) $ and $\displaystyle y \equiv 38 (\bmod 35) $
    is not true. Check it.
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  3. #3
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    Quote Originally Posted by whipflip15 View Post
    Firstly
    $\displaystyle y^2 \equiv 13 (\bmod 35) $ , then $\displaystyle y \equiv 15 (\bmod 35) $ and $\displaystyle y \equiv 38 (\bmod 35) $
    is not true. Check it.
    Ah, yes you right. Dyslexia :S It should be

    $\displaystyle y^2 \equiv 13 (\bmod 53) $, then $\displaystyle y \equiv 15 ( \bmod 53) $ and $\displaystyle y \equiv 38 (\bmod 53) $
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