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  1. #1
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    integral points

    the number of integral points on the hyperbola{x}^{2}-{y}^{2}={2000}^{2} are

    can we factorize 2000 in 2 and 5
    help
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: integral points

    Quote Originally Posted by prasum View Post
    the number of integral points on the hyperbola{x}^{2}-{y}^{2}={2000}^{2} are

    can we factorize 2000 in 2 and 5
    help

    Hint:

    x^2-y^2=(x-y)(x+y)

    (x-y)(x+y)=2000^2=2^8 5^6=2^22^65^6

    Now, what can you tell me about (x-y) and (x+y)...?
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  3. #3
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    Re: integral points

    Hello, prasum!

    The number of integral points on the hyperbola x^2-y^2\:=\:2000^2 is __.

    We have: . x^2 - y^2 \:=\:2000^2
    n . . (x+y)(x-y) \:=\:4,\!000,\!000


    As ASZ pointed out, 2000^2 \:=\:2^8\cdot5^6
    . . which has 32 pairs of factors.

    . . \begin{array}{ccc} 1&\times&4,\!000,\!000 \\ 2& \times& 2,\!000,\!000 \\ 4 &\times& 1,\!000,\!000 \\ 5 & \times& 800,\!000 \\ 10&\times& 400,\!000 \\ 16 &\times& 250,\!000 \\ 20 &\times& 200,\!000 \\ 25 &\times& 160,\!000 \\ 32 &\times& 125,\!000 \\ \vdots \\ 1280 &\times& 3125 \\ 1600&\times& 2500 \\ 2000 &\times& 2000 \end{array}


    We have: . x^2-y^2 \:=\:P\cdot Q

    Hence: . \begin{Bmatrix}x + y &=& P \\ x -y &=& Q\end{Bmatrix}\;\text{ where }P \ge Q.

    Then: . \begin{Bmatrix}x &=& \dfrac{P+Q}{2} \\ \\[-3mm] y &=& \dfrac{P-Q}{2} \end{Bmatrix}


    Each of the 32 pairs of factors will provide us with a solution.

    \text{For example, the pair }(1250,\,3200)\text{ gives us: }\:\begin{Bmatrix}x &=& \dfrac{3200 + 1250}{2} &=& 2225 \\ \\[-3mm] y &=& \dfrac{3200-1250}{2} &=& 975 \end{Bmatrix}

    We see that: . x\,=\,\pm2225,\;y \,=\,\pm975
    So each of these factorings gives us four solutions.

    Except the last pair (2000,\,2000)\!:\; \begin{Bmatrix}x &=& \frac{2000+2000}{2} &=& 2000 \\ y &=& \frac{2000-2000}{2} &=& 0 \end{Bmatrix}

    . . which produces only two solutions: . (\pm2000,\,0)


    Therefore, there are: . 4\cdot31 + 2 \:=\:124 lattice points on the hyperbola.

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  4. #4
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    Re: integral points

    In Soroban's solution, you need to exclude the seven factorisations of the form (2^8\cdot5^k)\times5^{6-k}, where one factor is even and the other factor is odd. Those will not lead to integer solutions for x and y. Thus the correct answer is that the number of integral points on the hyperbola is 7\cdot24+2=98.
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  5. #5
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    Re: integral points


    Absolutely right, Opalg!

    I had that on my to-do list . . . and forgot!

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