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I have some difficulty to do this problem:

Find real number $x$ such that both $x+\sqrt{2022}$ and $\frac{3}{x}-\sqrt{2022}$ is an integer

My attempts: I tried to addition and multiple $x+\sqrt{2022}$ and $\frac{3}{x}-\sqrt{2022}$ together but it seems doesn't help

Anyone have idea on how to solve this? Thank you so much

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    $\begingroup$ $x = n - \sqrt{2022}$ for some $n \in \mathbb{Z}$. Maybe putting this in $3/x - \sqrt{2022}$ will help? $\endgroup$
    – L. F.
    Jun 17, 2022 at 3:01
  • $\begingroup$ @L.F. ohhh, thank you so much for your idea!!! I solved it with your idea $\endgroup$
    – Lini
    Jun 17, 2022 at 3:06

1 Answer 1

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We know that $2022$ doesn't contain a square (as $2022=2*3*337$). So the solution must be of the form $x=n-\sqrt{2022}$ to satisfy the first equation, where $n\in \mathbb Z$.

Injection in the 2nd equation: $$\frac3{n-\sqrt{2022}}-\sqrt{2022}\in\mathbb Z$$ We multiply numerator/denominator by $n+\sqrt{2022}$: $$\frac{3(n+\sqrt{2022})}{n^2-2022}-\sqrt{2022}\in\mathbb Z$$ The denominator must be 3 so that the two square roots cancel, so $n^2-2022=3$ and then $n=45$ or $n=-45$.

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    $\begingroup$ I think that $n^2-2022=3$ then n=45 or -45 $\endgroup$
    – Lini
    Jun 17, 2022 at 3:14
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    $\begingroup$ @Lini this is correct, I modified. $\endgroup$
    – PC1
    Jun 17, 2022 at 3:16

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