# Simple Number Theory Proof, Again!

Alright, having problems with this question too. It seems to be the same type of number theory problem, which is the problem.

## Homework Statement

Prove "The square of any integer has the form 4k or 4k+1 for some integer k.

## Homework Equations

definition of even= 2k
definition of odd= 2k+1

## The Attempt at a Solution

Basically I have: Case 1. 2k(2k) = 4k
Case 2. 2k+1(2k+1) = 4k2+4k+1 = 4k+1(k+1)

Not sure if it's correct. Do I need to use different indices? I feel I am missing something. Thanks for any help!

Mark44
Mentor
Alright, having problems with this question too. It seems to be the same type of number theory problem, which is the problem.

## Homework Statement

Prove "The square of any integer has the form 4k or 4k+1 for some integer k.

## Homework Equations

definition of even= 2k
definition of odd= 2k+1

## The Attempt at a Solution

Basically I have: Case 1. 2k(2k) = 4k
Case 2. 2k+1(2k+1) = 4k2+4k+1 = 4k+1(k+1)

Not sure if it's correct. Do I need to use different indices? I feel I am missing something. Thanks for any help!
For case 1, (2k)(2k) != 4k
For case 2, you need parentheses.
(2k + 1)(2k + 1) = 4k2 + 4k + 1 != 4k + 1(k + 1)

Your last expression above is equal to 5k + 1, which is different from (2k + 1)(2k + 1).

You've expanded it yet you factorised it? it will be taking you back to what you got initially. you should have changed your $$4m^{2}+4m+1$$ into $$4(m^{2}+m)+1$$ and explain that it is similar to the form 4k+1 .

Even though this is my first post on Physics Forums and this was done a year ago, I'm going to tell everyone you've made it way too complicated. I'm a pretty new mathematician, and I feel this isn't in the spirit of a proof, but it still works.

Take the case n^2.
if n==0 (mod 4), n^2 will also be congruent to 0 modulo 4. Check.
if n==1, 1^2 will also be congruent to 1 modulo 4. Check.
if n==2, 2^2=4 and 4 modulo 4 == 0. Check.
if n==3, 3^3=9, 9==1 modulo 4. Check.

That's all the cases.

HallsofIvy