Results 1 to 8 of 8

Math Help - Elementary Number Theory

  1. #1
    Member cassiopeia1289's Avatar
    Joined
    Aug 2007
    From
    chicago
    Posts
    101

    Elementary Number Theory

    Prove the equation x^2 = 4y + 3 has no integer solutions.

    So, I was thinking, prove by contradiction that statement is not true. So:
    Assume x^2 = 4y+ 3 has integer solutions.

    This is where I get stuck. Do you let x and y be some integers k and j? How do you prove (or rather, disprove) integer solutions??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    If a number x is even, then it is of the form 2n and so its square is 4n^2 and so can not be in the form 4y + 3.

    If a number x is odd, then it is of the form 2n+1 and so its square is ...?

    Can that form of x be of the form 4y + 3?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member cassiopeia1289's Avatar
    Joined
    Aug 2007
    From
    chicago
    Posts
    101
    OK - you lost me with the not written in 4y+3 form.
    Why would 4k^2 have to look like 4y+3 : y and k are not the same variables ... ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    y is an integer. k^2 is an integer. 4k^2 is not in the same form as 4y+3 because it's 4y + 3 is 3 more than a multiple of 4. 4k^2 is zero more than a multiple of 4.

    Have you covered modulo arithmetic?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, cassiopeia1289!

    Prove the equation x^2 \:= \:4y + 3 has no integer solutions.
    Assume there exist integers x and y that satisfy the equation.

    We have: . x^2 \:=\:4y+3 \quad\Rightarrow\quad y \:=\:\frac{x^2-3}{4}



    x must either even or odd



    If x is even, then x \:= \:2k for some integer k.

    . . Then: . y \;=\;\frac{(2k)^2-3}{4} \;=\;\frac{4k^2-3}{4}\;=\;k^2 -\frac{3}{4}

    Hence, y cannot be an integer.



    If x is odd, then x \:= \:2k+1, for some integer k.

    . . Then: . y \;=\;\frac{(2k+1)^2 - 3}{4} \;=\;\frac{4k^2 + 4k - 2}{4} \;=\;k^2+k - \frac{1}{2}

    Hence, y cannot be an integer.



    Therefore, no integer solution exists.

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member cassiopeia1289's Avatar
    Joined
    Aug 2007
    From
    chicago
    Posts
    101
    Quote Originally Posted by Soroban View Post
    Hello, cassiopeia1289!

    Assume there exist integers x and y that satisfy the equation.

    We have: . x^2 \:=\:4y+3 \quad\Rightarrow\quad y \:=\:\frac{x^2-3}{4}



    x must either even or odd



    If x is even, then x \:= \:2k for some integer k.

    . . Then: . y \;=\;\frac{(2k)^2-3}{4} \;=\;\frac{4k^2-3}{4}\;=\;k^2 -\frac{3}{4}

    Hence, y cannot be an integer.



    If x is odd, then x \:= \:2k+1, for some integer k.

    . . Then: . y \;=\;\frac{(2k+1)^2 - 3}{4} \;=\;\frac{4k^2 + 4k - 2}{4} \;=\;k^2+k - \frac{1}{2}

    Hence, y cannot be an integer.



    Therefore, no integer solution exists.

    Thank you - thank you - thank you!
    I get that.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member cassiopeia1289's Avatar
    Joined
    Aug 2007
    From
    chicago
    Posts
    101
    ok sorry - same part of homework: new question(s)

    "If you get at least a C on the final exam, then you will pass the course." So if you got a D on the final - then what can you conclude?

    He didn't pass OR there's still hope

    OK: and second one, similar in nature:

    "If you get at least a B on the final, then you will pass the course" You pass the course. What can you conclude?

    He got at least a B on the final OR you can't conclude anything
    (On this one, I'm inclined to think you can't conclude anything, because P does not cause Q, so he could still pass some other way.)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    A new question needs a new thread. Read the rules.

    There's still hope, and you can conclude nothing.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. elementary number theory question
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 9th 2011, 10:23 AM
  2. elementary number theory: sum of digits
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: October 17th 2009, 11:40 AM
  3. Elementary Number Theory - when to use converse?
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: September 5th 2008, 09:45 AM
  4. Elementary Number Theory - beginnings
    Posted in the Number Theory Forum
    Replies: 10
    Last Post: August 27th 2008, 08:20 PM
  5. elementary Number Theory
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 25th 2008, 08:45 PM

Search Tags


/mathhelpforum @mathhelpforum