What Are the Solutions to the Equation x² + 1 ≡ 0 (mod 5³)?

  • Thread starter Thread starter brute26
  • Start date Start date
Click For Summary
SUMMARY

The equation x² + 1 ≡ 0 (mod 5³) has specific solutions derived from its reduction to x² + 1 ≡ 0 (mod 5). The roots of the latter are x = 2, 3, 7, 8, 12, etc., but not all are valid for the original equation. Valid solutions for 0 ≤ x < 125 are of the form y + 25n, where y = 7 or 18, and n ranges from 0 to 4. The principal roots identified are 57 and 68, as they satisfy the necessary conditions.

PREREQUISITES
  • Understanding of modular arithmetic, specifically modulo operations.
  • Familiarity with quadratic equations and their solutions in modular contexts.
  • Knowledge of the Chinese Remainder Theorem for solving systems of congruences.
  • Basic calculus concepts, particularly derivatives in relation to roots.
NEXT STEPS
  • Study the Chinese Remainder Theorem for solving modular equations.
  • Explore quadratic residues and non-residues in modular arithmetic.
  • Learn about the implications of derivatives in modular equations.
  • Investigate higher powers of primes in modular equations, particularly mod 5ⁿ.
USEFUL FOR

Mathematicians, students studying number theory, and anyone interested in solving modular equations and understanding their properties.

brute26
Messages
6
Reaction score
0
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.
 
Physics news on Phys.org
All solutions of the latter will not be solutions of the former.

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

Clearly y=7, 18 works.

Also since these do not satisfyf&#039;(y) = 2y \equiv 0 (mod~ p) , 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:
How about this method?
x^2+1=125t (t:+ integer)
x=1/2* root 4(125t-1)
 
Last edited:

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
7
Views
1K
Replies
8
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
32
Views
2K
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K