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Simple Number Theory Proof, Again!

  1. Mar 21, 2010 #1
    Alright, having problems with this question too. It seems to be the same type of number theory problem, which is the problem.

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

    2. Relevant equations

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

    3. 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!
  2. jcsd
  3. Mar 21, 2010 #2


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    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).
  4. Mar 22, 2010 #3
    You've expanded it yet you factorised it? it will be taking you back to what you got initially. you should have changed your [tex]4m^{2}+4m+1[/tex] into [tex]4(m^{2}+m)+1[/tex] and explain that it is similar to the form 4k+1 .
  5. Mar 5, 2012 #4
    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.
  6. Mar 5, 2012 #5


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    The only difference is that you have "hidden" the complications under the name "mod". That's fine if the person knows about modular arithmetic but the proof originally given is much simpler in that it does not use modular aritymetic.

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