1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook