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Thread: Help finding all ordered pairs satisfying?

  1. #1
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    Help finding all ordered pairs satisfying?

    Help finding all ordered pairs satisfying?-2m.png

    I don't really know where to start, I know 9000 is 2^3*3^2*5^3, and m and n have to be of different parities.
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  2. #2
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    Re: Help finding all ordered pairs satisfying?

    m,n
    4,898
    7,559
    8,496
    14,293
    23,176
    24,168
    39,93
    44,78
    71,27
    74,23
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  3. #3
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    Re: Help finding all ordered pairs satisfying?

    Quote Originally Posted by Ilikebugs View Post
    Click image for larger version. 

Name:	2m.png 
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ID:	38548
    I don't really know where to start, I know 9000 is 2^3*3^2*5^3, and m and n have to be of different parities.
    Quote Originally Posted by DenisB View Post
    m,n
    4,898
    7,559
    8,496
    14,293
    23,176
    24,168
    39,93
    44,78
    71,27
    74,23
    Denis, thank you for that solution. However, don't you think if a person can program then such a solution is simple?
    Question: is there a solution without using a program?
    If not then what is the point of the question?
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  4. #4
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    Re: Help finding all ordered pairs satisfying?

    Why do you think Denis got that from a program?

    Since we know the prime factorization of 9000, we can write pairs a, b such that ab= 9000 then get m= a+ 1, n= (b- m)/2.

    2(4500) so m= 1, n= (4500- 1)/2 is not an integer.
    3(3000) so m= 2, n= (3000- 2)/2= 2998/2= 1497. (m, n)= (2, 1497).
    4(2250) so m= 3, n= (2250- 3)/2 is not an integer.
    5(1800) so m= 4, n= (1800- 5)/2 is not an integer.
    6(1300) so m= 5, n= (1300- 5)/2 is not an integer.
    8(1125) so m= 7, n= (1125- 7)/2= 559. (m, n)= (7, 559).
    etc.
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  5. #5
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    Re: Help finding all ordered pairs satisfying?

    Quote Originally Posted by HallsofIvy View Post
    Why do you think Denis got that from a program?
    Since we know the prime factorization of 9000, we can write pairs a, b such that ab= 9000 then get m= a+ 1, n= (b- m)/2.

    2(4500) so m= 1, n= (4500- 1)/2 is not an integer.
    3(3000) so m= 2, n= (3000- 2)/2= 2998/2= 1497. (m, n)= (2, 1497).
    4(2250) so m= 3, n= (2250- 3)/2 is not an integer.
    5(1800) so m= 4, n= (1800- 5)/2 is not an integer.
    6(1300) so m= 5, n= (1300- 5)/2 is not an integer.
    8(1125) so m= 7, n= (1125- 7)/2= 559. (m, n)= (7, 559).
    etc.
    Because it was just a listing.
    You gave the reason.
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  6. #6
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    Re: Help finding all ordered pairs satisfying?

    with $a=m+1$ , $b=2 n+m$ , and $a b = 9000$

    Case $a$ even, $b$ must be odd since $a+b$ is odd

    now $9000=2^3 3^2 5^3 $ so $a=8 c$ where $c$ is a divisor of $3^2 5^3 $

    $b>a$ implies $\frac{9000}{8c}>8c$ which implies $1\leq c\leq 11$ and therefore $c=1,3,5,9$

    so in this case we have four solutions

    Case $a$ odd
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    Re: Help finding all ordered pairs satisfying?

    Or turn into a quadratic:

    m = {-2n - 1 +- sqrt[(2n + 1)^2 - 8n + 36000]} / 2

    ...so [(2n + 1)^2 - 8n + 360000] = a square
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  8. #8
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    Re: Help finding all ordered pairs satisfying?

    Quote Originally Posted by Idea View Post
    with $a=m+1$ , $b=2 n+m$ , and $a b = 9000$

    Case $a$ even, $b$ must be odd since $a+b$ is odd

    now $9000=2^3 3^2 5^3 $ so $a=8 c$ where $c$ is a divisor of $3^2 5^3 $

    $b>a$ implies $\frac{9000}{8c}>8c$ which implies $1\leq c\leq 11$ and therefore $c=1,3,5,9$

    so in this case we have four solutions

    Case $a$ odd
    So, DenisB missed:
    2, 1499
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  9. #9
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    Re: Help finding all ordered pairs satisfying?

    Quote Originally Posted by SlipEternal View Post
    So, DenisB missed:
    2, 1499
    Yes. I did have it at top of list, then went to add (m,n) headings and
    somehow erased the 2,1499...I'll go stand in the corner for 14.99 minutes...
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