# $x \cdot y$ is an integer and $x - y$ is an integer. Do $x$ and $y$ both have to be integers?

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.

• 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. – Mark Bennet Oct 16 at 12:51

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.

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

• 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. – virchau13 Oct 16 at 12:48

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)

• This is not correct. Let $x = \phi$, the golden ratio, and $y = \phi ^ {-1}$. Then $xy = x - y = 1$. – Mikhail Hogrefe Oct 18 at 15:36

Any complex number $$x +iy$$ and its conjugate $$x-iy$$ where $$x$$ and $$y$$ are non zero integers provides a counter example.

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.