Results 1 to 8 of 8

Math Help - Show that sys. of eq. has no integer solutions

  1. #1
    Junior Member
    Joined
    Jul 2012
    From
    Earth
    Posts
    56

    Show that sys. of eq. has no integer solutions

    I'm to show that the system of equations below has no integer solutions.

    \begin{cases} 11x - 5y = 7 \: & (\text {I})\\9x + 10y = -3 \: & (\text {II}) \end{cases}

    Reducing to modulo 2, we get

    \begin{cases} 11x - 5y \underset{2}{\equiv} 7 &    \Leftrightarrow \ \  x + y \underset{2}{\equiv} 1 \\9x + 10y \underset{2}{\equiv} -3  &  \Leftrightarrow \ \  x  \underset{2}{\equiv} 1  \end{cases}

    This means that x must be odd and the sum x + y odd as well. A sum can only be odd if one number is odd and the other odd, thus y must be an even number. I haven't seemingly arrived at a condraction, I am still however doing something wrong because the equation in fact doesn't have any integer solutions. Help anybody?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Show that sys. of eq. has no integer solutions

    You may as well solve the system. Multiply first equation by -2:

    22x - 10y = 14

    9x + 10y = -3

    Add the two equations to get 31x = 11. Clearly, x is not an integer so we're done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2012
    From
    Earth
    Posts
    56

    Re: Show that sys. of eq. has no integer solutions

    Quote Originally Posted by richard1234 View Post
    You may as well solve the system. Multiply first equation by -2:

    22x - 10y = 14

    9x + 10y = -3

    Add the two equations to get 31x = 11. Clearly, x is not an integer so we're done.
    Yes I might as well do that but that's not the assignment.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,719
    Thanks
    635

    Re: Show that sys. of eq. has no integer solutions

    Hello, MathCrusader!

    Yes, I might as well do that, but that's not the assignment.
    Well, that's just great!

    Could you possibly tell us: what is the assignment?

    And do you have any more surprises for us?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Show that sys. of eq. has no integer solutions

    Quote Originally Posted by MathCrusader View Post
    Yes I might as well do that but that's not the assignment.
    I second Soroban, what is ​the assignment? The assignment was to show that the system has no integer solution (x,y), right? It shouldn't matter ​how you do it, as long as it's valid.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Show that sys. of eq. has no integer solutions

    Quote Originally Posted by Soroban View Post
    Hello, MathCrusader!


    And do you have any more surprises for us?
    Lol...not allowed to use mods or Bezout's identity. You must use the solution I show you. I hate when teachers do that >.<
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jul 2012
    From
    Earth
    Posts
    56

    Re: Show that sys. of eq. has no integer solutions

    My bad I forgot to write that it needs to be done using modular arithmetic. Considering that I was attempting to do it using modular arithmetic and I posted it under 'Number Theory', it's surprises that it wasn't evident how it's meant to be solved, but I digress.

    So, how would we move on from:

    \begin{cases} 11x - 5y \underset{2}{\equiv} 7 &    \Leftrightarrow \ \  x + y \underset{2}{\equiv} 1 \\9x + 10y \underset{2}{\equiv} -3  &  \Leftrightarrow \ \  x  \underset{2}{\equiv} 1  \end{cases}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Show that sys. of eq. has no integer solutions

    Quote Originally Posted by MathCrusader View Post
    My bad I forgot to write that it needs to be done using modular arithmetic. Considering that I was attempting to do it using modular arithmetic and I posted it under 'Number Theory', it's surprises that it wasn't evident how it's meant to be solved, but I digress.

    So, how would we move on from:

    \begin{cases} 11x - 5y \underset{2}{\equiv} 7 &    \Leftrightarrow \ \  x + y \underset{2}{\equiv} 1 \\9x + 10y \underset{2}{\equiv} -3  &  \Leftrightarrow \ \  x  \underset{2}{\equiv} 1  \end{cases}
    You obtain y \equiv 0 (\mod 2) but that doesn't lead much to a contradiction.

    If we look at it mod 5, we obtain

    \begin{cases} 11x \equiv 7 (\mod 5) \\ 9x \equiv -3 (\mod 5)   \end{cases}

    Adding, we get 20x \equiv 4 (\mod 5), contradiction, x is not an integer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integer solutions to a^2+b^2=c^3
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 6th 2014, 08:22 AM
  2. # of integer solutions
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 2nd 2011, 09:43 PM
  3. Replies: 2
    Last Post: May 8th 2010, 10:59 PM
  4. No integer solutions to x^4 + y^4=100
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 17th 2009, 10:04 AM
  5. Integer Solutions
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: July 31st 2009, 09:18 AM

Search Tags


/mathhelpforum @mathhelpforum