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Math Help - [SOLVED] Quadratic Congruences

  1. #1
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    [SOLVED] Quadratic Congruences

    Regarding quadratic congruences in the form,  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:  y^2 \equiv 13 (\bmod 35) , then  y \equiv 15 (\bmod 35) and  y \equiv 38 (\bmod 35) . I do not understand the jump from the first step to the next.

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

    *How do you solve for a linear congruence in this form (for example):
     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
     y^2 \equiv 13 (\bmod 35) , then  y \equiv 15 (\bmod 35) and  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
     y^2 \equiv 13 (\bmod 35) , then  y \equiv 15 (\bmod 35) and  y \equiv 38 (\bmod 35)
    is not true. Check it.
    Ah, yes you right. Dyslexia :S It should be

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