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

[brute26]

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.

[brute26]

ok, so now i know that it has 2 solutions, because x^2 + 1 == 0 (mod 5) has only two solutions, namely x= 4, x= -4.

however f '(4) and f '(-4) are not congruent to 0 (mod 5). So these roots are nonsingular?

[Hurkyl]

Please don't double post.

http://www.physicsforums.com/showthread.php?t=57425