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Let $P(x,y)$ be the statement $x + 2y = xy$, where $x,y$ are integers. Is the statement $\forall x \exists y \text{ s.t.} P(x,y)$ true?

From what I understand I think it is true because if I choose any value for $x$, such as $x = 3$, $P(x,y)$ becomes $3 + 2y = 3y$, which gives me $y = 3$. So I get at least $1$ value for $y$ that makes this true. Am I looking at this the wrong way? Please explain. The book says it is False.

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To prove the statement true, you have to prove it true for all possible cases. To prove it untrue, you just need a single counterexample.

Still, in this case, the statement is saying that for any $x$, there exists a $y$ such that the equation is true, with $x$ and $y$ both being integers, of course. So for a given $x$, you can solve the equation for $y$:

$\begin{eqnarray}x + 2y & = & xy \\ xy - 2y & = & x \\ y(x - 2) & = & x \\ y & = & \frac{x}{x-2}\end{eqnarray}$

So it's solvable, but is that solution an integer? Well, only if $x-2$ is a factor of $x$. What happens if $x = 5$? Then $x - 2 = 3$, and $y = \frac{5}{3}$ isn't an integer. Counter-example right there.

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Let $x=2$, then there is no $y$ that satisfies $2+2y=2y$.

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