Modular Congruences of Integer Squares

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Discussion Overview

The discussion revolves around proving modular congruences related to the squares of integers, specifically that for any integer n, \( n^{2} \cong 0 \) or \( 1 \) (mod 3), and \( n^{2} \cong 0, 1, 4 \) (mod 5). The focus is on theoretical aspects of modular arithmetic.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant suggests that proving \( n \cong m \) (mod 3) implies \( n^{2} \cong m^{2} \) (mod 3), but struggles to prove \( n \cong 0 \) (mod 3).
  • Another participant points out that it is sufficient to consider \( n^{2} \) (mod 3) for the cases when \( n = 0, 1, 2 \), and suggests a similar approach for mod 5.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the proof methods, and multiple approaches are being discussed without resolution.

Contextual Notes

Participants have not fully explored the implications of their assumptions or the completeness of their proofs, particularly regarding the cases for mod 3 and mod 5.

phyguy321
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prove that for any integer n, n[tex]^{2}[/tex] [tex]\cong[/tex] 0 or 1 (mod 3), and n[tex]^{2}[/tex] [tex]\cong[/tex] 0,1,4(mod 5)
 
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And what have you tried...?
 
The only thing i found was that if you can prove n[tex]\cong[/tex]m mod 3 than n[tex]^{2}[/tex] [tex]\cong[/tex] m[tex]^{2}[/tex] mod 3

but i couldn't prove n [tex]\cong[/tex] 0 mod 3 so i gave up
 
You only have to consider n^2 (mod 3) when n=0,1,2. A same type of comment applies mod 5. (Why?)
 

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