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X^2 + 1 = 0 (mod 5^3).

  1. Dec 20, 2004 #1
    How would i start to solve this problem?

    x^2 + 1 == 0 (mod 5^3).

    Find all solutions.

    How do i know how many solutions there are? If i reduce it to
    x^2 + 1 == 0 (mod 5), i get that x= 2,3,7,8,12, etc.
     
  2. jcsd
  3. Dec 20, 2004 #2

    Gokul43201

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    All solutions of the latter will not be solutions of the former.

    If 0<=x<125 is a root of [itex]x^2 +1 \equiv 0 (mod~ 5^3) [/itex] then x satisfies [itex]x^2 +1 \equiv 0 (mod~ 5^2) [/itex] and is of the form y+25n, 0<=n<5, 0<=y<25.

    Clearly y=7, 18 works.

    Also since these do not satisfy[itex]f'(y) = 2y \equiv 0 (mod~ p) [/itex] , there is only one n, which will give you the principal roots 57 and 68.

    I've left some gaps for you to figure out and fill.
     
    Last edited: Dec 20, 2004
  4. Dec 21, 2004 #3
    How about this method?
    [itex] x^2+1=125t (t:+ integer) [/itex]
    [itex]x=1/2* root 4(125t-1) [/itex]
     
    Last edited: Dec 21, 2004
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