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  1. #1
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    Algebra help

    find all integer x and y that fulfils

    231x^2 +528y^2=16896
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  2. #2
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    Quote Originally Posted by johnnyboiy View Post
    find all integer x and y that fulfils

    231x^2 +528y^2=16896
    This is an eqn of an ellipse. Please see attached graph. The integers that satisfy this eqn are shown RED points in graph.
    Attached Thumbnails Attached Thumbnails Algebra help-g5.jpg  
    Last edited by Shyam; February 2nd 2009 at 05:55 PM.
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  3. #3
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    Quote Originally Posted by johnnyboiy View Post
    find all integer x and y that fulfils

    231x^2 +528y^2=16896
    Notice that if (x,y) is a solution then (\pm x,\pm y) is a solution. This means we can only focus on positive values x and y. If y\geq 6 then 528y^2 \geq 19008 > 16896, and so no solution exists. Thus, you need to see for which 0\leq y\leq 5 we have \sqrt{ \frac{16896 - 528y^2}{231} } is an integer.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Notice that if (x,y) is a solution then (\pm x,\pm y) is a solution. This means we can only focus on positive values x and y. If y\geq 6 then 528y^2 \geq 19008 > 16896, and so no solution exists. Thus, you need to see for which 0\leq y\leq 5 we have \sqrt{ \frac{16896 - 528y^2}{231} } is an integer.
    Thanks for this, but here is another theory, that some made, but i dont get some of the steps he makes...

    Let x^2=a, y^2=b

    231a+528b=16896

    16896=32*528 that will give (a,b)=(0,32) {why does he take 32*528?}

    a=0+528n {how does he get this?}
    b=32-231n {how does he get this?}

    231a+528b = 231(0+528n)+528(32-231n) = 231*528n+528*32-528*231n

    x^2=528n {how does he get this from the equation above?}
    y^2=32-231n {how does he get this from the equation above?}

    n<0 gives 528n<0, n>0 gives 32-231n<0

    x^2 =0
    y^2=32

    Is then the answer x=0,y=sqrt(32)=5,6 {but that aint a Z (integer?)}??
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