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Math Help - Positive integer solutions

  1. #1
    Super Member dhiab's Avatar
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    Positive integer solutions

    Find the total number of positive integer solutions of:
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    MHF Contributor red_dog's Avatar
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    (x-y)(x+y)=33

    Then \left\{\begin{array}{ll}x-y=1\\x+y=33\end{array}\right.

    or \left\{\begin{array}{ll}x-y=3\\x+y=11\end{array}\right.

    Now find x and y.
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by red_dog View Post
    (x-y)(x+y)=33

    Then \left\{\begin{array}{ll}x-y=1\\x+y=33\end{array}\right.

    or \left\{\begin{array}{ll}x-y=3\\x+y=11\end{array}\right.

    Now find x and y.
    Hello,Thank you,
    Can not have :

    BECAUSE ????????
    Last edited by dhiab; July 2nd 2009 at 09:13 AM.
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  4. #4
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    Quote Originally Posted by dhiab View Post
    Find the total number of positive integer solutions of:
    Have you tried graphing the function?

    You would probably have to enter it as

    y = \pm\sqrt{x^2 - 33}.


    I can see straight away that x^2 - 33 \geq 0 so x \geq \sqrt{33}.

    Since x needs to be a positive integer, we know therefore that x \geq 6.

    Now you just need x^2 - 33 to be a perfect square. x = 7 works, since 7^2 - 33 = 16, so y = 4.


    See if you can come up with any other solutions...
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  5. #5
    Super Member dhiab's Avatar
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    Quote Originally Posted by Prove It View Post
    Have you tried graphing the function?

    You would probably have to enter it as

    y = \pm\sqrt{x^2 - 33}.


    I can see straight away that x^2 - 33 \geq 0 so x \geq \sqrt{33}.

    Since x needs to be a positive integer, we know therefore that x \geq 6.

    Now you just need x^2 - 33 to be a perfect square. x = 7 works, since 7^2 - 33 = 16, so y = 4.


    See if you can come up with any other solutions...
    Hello: the solutions is

    In graphic of the function , we find the Point cordinates integer.
    LOOK HERE
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  6. #6
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    Quote Originally Posted by dhiab View Post
    Hello: the solutions is

    In graphic of the function , we find the Point cordinates integer.
    LOOK HERE
    Are there any more?
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  7. #7
    Super Member dhiab's Avatar
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    Quote Originally Posted by Prove It View Post
    Are there any more?
    Hello : yes in this question

    Hello,Thank you,
    Can not have :

    BECAUSE

    AND IN THIS ANSWER :
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  8. #8
    Senior Member pacman's Avatar
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    as the graph suggests, the values are symetric, mirror image of each other,
    x = -/+ 17, y = -/+ 16
    x = /+ 7 ; y -/+ 4
    Attached Thumbnails Attached Thumbnails Positive integer solutions-aa.gif  
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  9. #9
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    Hello, dhiab!

    Here is a primitive solution . . .


    Find the number of positive integer solutions of: . x^2 - y^2 \:=\:33
    Consecutive squares differ by an odd number: . (n+1)^2 - n^2 \:=\:2n+1


    We have: . x^2 - y^2 \:=\:33

    . . Then: . 2n+1 \:=\:33 \quad\Rightarrow\quad n \:=\:16

    . . Hence: . \boxed{17^2 - 16^2 \:=\:33}



    \text{We also have: }\;x^2-y^2 \:=\:\underbrace{9 +11+13}_{4^2\quad5^2\quad6^2\quad7^2}

    . . Hence: . \boxed{7^2 - 4^2 \:=\:33}


    Therefore, there are exactly two solutions.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I came up with this procedure while in college.


    Example: . a^2-b^2 \:=\:105



    105 is the difference between \left(\tfrac{105-1}{2}\right)^2 and \left(\tfrac{105+1}{2}\right)^2

    . . Therefore: . 53^2 - 52^2 \:=\:105



    We have: . 105 \:=\:35 + 35 + 35 \:=\:33 + 35 + 37

    . . \text{Hence: }\;a^2-b^2 \:=\:\underbrace{33 + 35 + 37}_{16^2\quad17^2\quad18^2\quad19^2}

    . . Therefore: . 19^2 - 16^2 \:=\:105



    We have: . 105 \:=\:15 + 15 + 15 + 15 + 15 + 15 + 15 \:=\:9 + 11 + 13 + 15 + 17 + 19 + 21

    . . \text{Hence: }\;a^2-b^2 \:=\:\underbrace{9 + 11 + 13 + 15 + 17 + 19 + 21}_{4^2\quad\;\;5^2\quad\;\;6^2\quad\;\;\;7^2\qua  d\;\;\;8^2\quad\;\;9^2\quad\;\;10^2\quad\;\;11^2}

    . . Therefore: . 11^2 - 4^2 \:=\:105



    \text{The next combination is: }\;105 \:=\:\underbrace{7+7+7+ \hdots + 7}_{\text{15 terms}}

    But this cannot be expressed as a sum of conscutive positive odd numbers.



    Therefore, there are three solutions: . (53,52),\;(19,16),\;(11,4)

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  10. #10
    Super Member dhiab's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, dhiab!

    Here is a primitive solution . . .

    Consecutive squares differ by an odd number: . (n+1)^2 - n^2 \:=\:2n+1


    We have: . x^2 - y^2 \:=\:33

    . . Then: . 2n+1 \:=\:33 \quad\Rightarrow\quad n \:=\:16

    . . Hence: . \boxed{17^2 - 16^2 \:=\:33}



    \text{We also have: }\;x^2-y^2 \:=\:\underbrace{9 +11+13}_{4^2\quad5^2\quad6^2\quad7^2}

    . . Hence: . \boxed{7^2 - 4^2 \:=\:33}


    Therefore, there are exactly two solutions.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I came up with this procedure while in college.


    Example: . a^2-b^2 \:=\:105



    105 is the difference between \left(\tfrac{105-1}{2}\right)^2 and \left(\tfrac{105+1}{2}\right)^2

    . . Therefore: . 53^2 - 52^2 \:=\:105



    We have: . 105 \:=\:35 + 35 + 35 \:=\:33 + 35 + 37

    . . \text{Hence: }\;a^2-b^2 \:=\:\underbrace{33 + 35 + 37}_{16^2\quad17^2\quad18^2\quad19^2}

    . . Therefore: . 19^2 - 16^2 \:=\:105



    We have: . 105 \:=\:15 + 15 + 15 + 15 + 15 + 15 + 15 \:=\:9 + 11 + 13 + 15 + 17 + 19 + 21

    . . \text{Hence: }\;a^2-b^2 \:=\:\underbrace{9 + 11 + 13 + 15 + 17 + 19 + 21}_{4^2\quad\;\;5^2\quad\;\;6^2\quad\;\;\;7^2\qua  d\;\;\;8^2\quad\;\;9^2\quad\;\;10^2\quad\;\;11^2}

    . . Therefore: . 11^2 - 4^2 \:=\:105



    \text{The next combination is: }\;105 \:=\:\underbrace{7+7+7+ \hdots + 7}_{\text{15 terms}}

    But this cannot be expressed as a sum of conscutive positive odd numbers.



    Therefore, there are three solutions: . (53,52),\;(19,16),\;(11,4)
    Hello :
    Thank you for your efforts, but génerale case is based on the divisores of a number
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