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

[dgamma3]

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

 

[Mark44]

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.

 

[dgamma3]

sorry mate.

Prove that the square of any integer, when divided by 3, leaves remainder 0 or 1 but never 2.
thanks

 

[haruspex]

What possibilities do you need to consider for the integer?