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Sorry if this question is kind of stupid, but it randomly came into my mind and I've been thinking about it all day.

$xy \in \mathbb{Z}$ and $x - y \in \mathbb{Z}$. Are there any solutions for $x$ and $y$ where either $x$ or $y$ are not integers?

I tried writing a (badly written) small script to test this, and it detected no solutions for $-10 \leq x, y \lt 10$, although I'm pretty sure that's wrong.

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    $\begingroup$ Note that this means that $(x-1)(y+1)=xy+(x-y)-1$ is also an integer as is $(x-1)-(y+1)=x-y-2$ so you can reduce $x$ if you like. Or you can make additional solutions out of any given solution. $\endgroup$ – Mark Bennet Oct 16 at 12:51
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If $p=xy$ and $d=x-y$ are these integers, then $x$ and $-y$ are the solutions of $$ X^2-dX-p=0.$$ These are rational (and, by the rational root theorem, automatically integer!) if and only if the discriminant $d^2+4p$ is a perfect square.

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Take $x=y=\sqrt2$.${}{}{}{}{}{}$

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  • $\begingroup$ Well, consider me stupid. I'll accept your answer if a better one doesn't come along suggesting a general case other than just sqrt(k) where k is an integer. $\endgroup$ – virchau13 Oct 16 at 12:48
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Assume that x and y (are different) are both not integers. We use the canonical representation of real numbers as decimal fractions. Then both have a nontrivial fractional part and with are canonical multiplication on R we find that this number can never be an integer. If one of the two is no integer we can assume that x*y is an integer, but their subtraction can never be (as they have a nontrivial decimal fractional part and the other has not). Vice versa is never possible. The only cases are shown below but x and y then have to be the same number.

(This is a "non-formal" proof but you can sketch it more precisely)

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  • $\begingroup$ This is not correct. Let $x = \phi$, the golden ratio, and $y = \phi ^ {-1}$. Then $xy = x - y = 1$. $\endgroup$ – Mikhail Hogrefe Oct 18 at 15:36
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Any complex number $x +iy$ and its conjugate $x-iy $ where $x$ and $y$ are non zero integers provides a counter example.

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Well, $\sqrt 2\cdot \sqrt 2 = 2\in\Bbb Z$ and $\sqrt 2 - \sqrt 2 = 0\in\Bbb Z$, but $\sqrt 2$ is not a rational number.

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