Prove that the square of any integer, when divided by 3. only by odd and even.

Click For Summary

Homework Help Overview

The discussion revolves around proving that the square of any integer, when divided by 3, leaves a remainder of 0 or 1, but never 2. Participants are exploring this proof specifically through the lens of even and odd integers rather than the typical modular approach.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • One participant attempts to expand the square of an odd integer, represented as (2x+1)^2, and groups terms to analyze the result. They express uncertainty about the next steps in their reasoning. Another participant questions the clarity of the problem statement and suggests that the original poster clarify what they are trying to prove.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the proof. Some guidance has been offered regarding the clarity of the problem statement, but there is no explicit consensus on the method to be used or the next steps to take.

Contextual Notes

Participants are focusing on proving the statement using only even and odd integers, which may limit the approaches available to them. There is also a note of confusion regarding the exact phrasing of the problem statement, which may affect the direction of the discussion.

dgamma3
Messages
12
Reaction score
0

Homework Statement



I know you could prove this by stating every integer is either 3m, 3m+1 or 3m+2. However I am trying to prove this just using either even numbers or odd numbers.

so for example, when I try:
(2x+1)^2
= 4x^2 + 4x + 1 - expand
= 3x^2 + x^2 + 3x + x + 1 - group like terms
= 3x^2 + 3x + x^2 + x + 1
= 3*(x^2 + x) + x(x+1) + 1

x(x+1) + 1 is an odd number

so
= 3*(x^2 + x) + 2p + 1

thats as far as I can go.

thanks
 
Physics news on Phys.org
dgamma3 said:

Homework Statement



I know you could prove this by stating every integer is either 3m, 3m+1 or 3m+2. However I am trying to prove this just using either even numbers or odd numbers.

so for example, when I try:
(2x+1)^2
= 4x^2 + 4x + 1 - expand
= 3x^2 + x^2 + 3x + x + 1 - group like terms
= 3x^2 + 3x + x^2 + x + 1
= 3*(x^2 + x) + x(x+1) + 1

x(x+1) + 1 is an odd number

so
= 3*(x^2 + x) + 2p + 1

thats as far as I can go.
What are you trying to prove?
Your thread title seems to be missing a few words
Prove that the square of any integer, when divided by 3. only by odd and even.
Prove that the square of any integer, when divided by 3, does what?

The problem template has a problem statement section titled, and this is where the statement of the problem should go. Use it.
 
sorry mate.

Prove that the square of any integer, when divided by 3, leaves remainder 0 or 1 but never 2.
thanks
 
What possibilities do you need to consider for the integer?
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K