Results 1 to 2 of 2

Math Help - How does this go?

  1. #1
    Newbie
    Joined
    Apr 2006
    Posts
    7

    How does this go?

    Prove that there are no solutions in integers x and y to the equation
    2x^2 + 5y^2 = 14

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by jzon
    Prove that there are no solutions in integers x and y to the equation
    2x^2 + 5y^2 = 14

    Thanks in advance
    Here is one way.

    y can only be 0 or +,-1, because if y were |2| or greater, then 5y^2 is already 20 or more---which is greater than 14.
    If y=0, then x^2 = 7, and x is not an integer.
    If y= |1|, then x^2 = 9/2, and x is not an integer.

    x can only be 0, |1|, or |2|. If x were |3| or more, then 2x^2 is already 18 or more.
    If x=0, y^2 = 14/5 and y is not an integer.
    If x=|1|, y^2 = 12/5 and y is not an integer.
    If x=|2|, y^2 = 16/5 and y is not an integer.

    Therefore, there is no solution in x and y as integers.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum