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Math Help - Find all integer solutions to an equation

  1. #1
    Senior Member Mukilab's Avatar
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    Find all integer solutions to an equation

    Whenever I'm confronted with a question bearing the label of 'find all solutions' I almost never know how to solve it algebraically (apologies if I spelt that incorrectly), I always have to resort to manual methods, which usually are a bore and I know that is not the true mathematical way to solve them.

    Now I've been shown this equation:
    x^2+y^2+x+y=1001

    and I have to find all integer solutions.

    I have no idea of how to solve this because I think the left half of the equation cannot be simplified (or I simply do not know of the method). Until necessary, I'll refrain from using the calculator so please could anyone help me?

    ~Cyfer
    Last edited by mr fantastic; September 8th 2010 at 12:53 PM. Reason: Copied information form title into main body of post.
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  2. #2
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    e^(i*pi)'s Avatar
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    You can only solve for x in terms of y or vice versa (and you can use the quadratic formula in this case)

    To solve for x and y you need another equation
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  3. #3
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    Quote Originally Posted by Mukilab View Post
    Whenever I'm confronted with a question bearing the label of 'find all solutions' I almost never know how to solve it algebraically (apologies if I spelt that incorrectly), I always have to resort to manual methods, which usually are a bore and I know that is not the true mathematical way to solve them.

    Now I've been shown this equation:
    x^2+y^2+x+y=1001
    and I have no idea of how to solve this because I think the left half of the equation cannot be simplified (or I simply do not know of the method). Until necessary, I'll refrain from using the calculator so please could anyone help me?
    If x^2+y^2+x+y=1001 then 4x^2+4y^2+4x+4y=4004, and so (2x+1)^2 + (2y+1)^2 = 4006 = 2\times2003. Now 2003 is a prime that is congruent to 3 mod 4 (or in everyday language, it leaves a remainder 3 when divided by 4). It is known (see for example here) that an integer can only be expressed as a sum of two squares if each of its prime factors that is congruent to 3 mod 4 occurs to an even power. That is not the case with 4006, which therefore cannot be expressed as a sum of two squares. So there are no integer solutions (x,y) to your equation.
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  4. #4
    Senior Member Mukilab's Avatar
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    EDIT: Sorry didn't see the previous post before I wrote this. Will create a reply for the previous post when ready. Apologies.

    Surely it would go to 4002 and not 4004 because the two left over 1s from the factorising are positive, so they become negative when brought over.
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    Notice that Opalg is assuming you are asked to find all integer solutions, which was not said in the original post.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Notice that Opalg is assuming you are asked to find all integer solutions, which was not said in the original post.
    But it was in the title of the post!
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  7. #7
    Senior Member Mukilab's Avatar
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    Opalg please refer to my last post, I think you missed it
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  8. #8
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    Quote Originally Posted by Mukilab View Post
    Surely it would go to 4002 and not 4004 because the two left over 1s from the factorising are positive, so they become negative when brought over.
    No, you have to add 2 to both sides: (2x+1)^2 + (2y+1)^2 = 4x^2+4x+1+4y^2+4y+1 = 4(x^2+x+y^2+y) + 2 = 4(1001) + 2 = 4006.
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  9. #9
    Senior Member Mukilab's Avatar
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    why do you add 2 to both sides? I understand from creating (2x+1)^2 + (2y+1)^2 you have 2 left over from the previous expression but how do you get 2 from the right side?
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  10. #10
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    Quote Originally Posted by Mukilab View Post
    why do you add 2 to both sides? I understand from creating (2x+1)^2 + (2y+1)^2 you have 2 left over from the previous expression but how do you get 2 from the right side?
    If an equation is to balance then you always have to do the same thing to both sides of it. Starting with the given equation x^2+y^2+x+y=1001, you can multiply both sides by 4 to get 4x^2+4y^2+4x+4y=4004. You can then add 2 to both sides to get 4x^2+4y^2+4x+4y +2 =4004 + 2 = 4006. Finally, factorise the left side to get (2x+1)^2 + (2y+1)^2 = 4006.
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    Quote Originally Posted by Mukilab View Post
    Whenever I'm confronted with a question bearing the label of 'find all solutions' I almost never know how to solve it algebraically (apologies if I spelt that incorrectly), I always have to resort to manual methods, which usually are a bore and I know that is not the true mathematical way to solve them.

    Now I've been shown this equation:
    x^2+y^2+x+y=1001

    and I have to find all integer solutions.

    I have no idea of how to solve this because I think the left half of the equation cannot be simplified (or I simply do not know of the method). Until necessary, I'll refrain from using the calculator so please could anyone help me?

    ~Cyfer
    Here is another approach.

    x^2 + x = x (x+1) is the product of two consecutive integers, hence is even.

    The same can be said of y^2 + y.

    So their sum must be...
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