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Math Help - Is it possible to show this equation has no solutions

  1. #1
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    Is it possible to show this equation has no solutions

    Hi,

    I'm trying to prove the following equation has no solutions, with the following conditions on a, b, c and d:

    These 4 variables are integers. They can be negative, but each of them must have a unique absolute value. That is, if one of the variables is equal to x, then none of the others may be -x. Note that this rules out the trivial solution a = b = c = d = 0.

    The equation is

    3a^2 + a + 3b^2 + b = 3c^2 + c + 3d^2 + d

    What techniques could I employ to show this has no solutions?

    Both sides of the equation must also be equal to 2e^2, which means they must be equal to a number that is 2 times a perfect square.
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  2. #2
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    Re: Is it possible to show this equation has no solutions

    Quote Originally Posted by kiwijoey View Post

    What techniques could I employ to show this has no solutions?
    You can try to rearrange the equation such that it has the sum of squared numbers on one side and 0 on the other.


    Quote Originally Posted by kiwijoey View Post

    Both sides of the equation must also be equal to 2e^2, which means they must be equal to a number that is 2 times a perfect square.
    So 2e^2 = 2e^2 ?
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  3. #3
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    Re: Is it possible to show this equation has no solutions

    in other words,

    3a^2 + a + 3b^2 + b = 3c^2 + c + 3d^2 + d = 2e^2.

    My first instinct is to rearrange to

    3a^2 + 3b^2 - 3c^2 - 3d^2 = c + d - a - b

    Now the right hand side must both be even (from the 2e^2 part) and divisible by 3 (from the left hand side), which implies that

    c + d - a - b = 0 (mod 6).

    But that's going to give a lot of cases to consider if I just start considering a, b, c and d mod 6. Is there a better way to proceed?
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  4. #4
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    Re: Is it possible to show this equation has no solutions

    Finding integer solutions to the equation

    3a^2+a+3b^2+b=3c^2+c+3d^2+d

    isn't difficult. ( a=41,b=8,c=40,d=12 is a solution). A reasonable technique is to complete the squares on each of the four expressions, (you find that the numbers introduced cancel out), and then switching a pair from one side of the equation to the other. That leads to a 'difference of two squares' situation and hence an equation of the form w.x=y.z. Then find sets of values for w,x,y,z that produce integer values for a,b,c,d.

    The real problem is to find (at least two ) different solutions of the equation

    3a^2+a+3b^2+b=2n^2

    for some value(s) of n.
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