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Thread: the number of solutions in N

  1. #1
    Super Member dhiab's Avatar
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    the number of solutions in N

    Determine the number of solutions in N of the equation:
    x + y + 2xy-2019x-2019y-2020 = 0
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  2. #2
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    Re: the number of solutions in N

    Factor it. You should find either 2019 or 2020 solutions in $\mathbb{N}^2$ depending on whether zero is a natural number or not. Some books say it is, some say it is not.
    Last edited by SlipEternal; Sep 14th 2018 at 02:53 AM.
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    Super Member dhiab's Avatar
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    Re: the number of solutions in N

    Quote Originally Posted by SlipEternal View Post
    Factor it. You should find either 2019 or 2020 solutions in $\mathbb{N}^2$ depending on whether zero is a natural number or not. Some books say it is, some say it is not.
    Thanks ...if zero is a natural number then you have" n" solutions whether zero is not a natural number "n-1" solutions..
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    Re: the number of solutions in N

    \begin{align}(x+y)^2 - 2019(x+y) - 2020 &= 0 &\text{now set $u=x+y$ for clarity} \\ u^2 - 2019u -2020 &= 0 \\ (u-2020)(u+1) &= 0 &\text{and replace $u=x+y$} \\ (x+y-2020)(x+y+1) &= 0\end{align}
    That's two solutions to be interpreted.
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  5. #5
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    Re: the number of solutions in N

    Quote Originally Posted by Archie View Post
    \begin{align}(x+y)^2 - 2019(x+y) - 2020 &= 0 &\text{now set $u=x+y$ for clarity} \\ u^2 - 2019u -2020 &= 0 \\ (u-2020)(u+1) &= 0 &\text{and replace $u=x+y$} \\ (x+y-2020)(x+y+1) &= 0\end{align}
    That's two solutions to be interpreted.
    It is one solution for $u$ (over natural numbers) but 2019 or 2020 over $x,y$. Example: (1,2019), (2,2018), (3,2017), etc. are all distinct solutions over the natural numbers.
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