How many rational?

Let the rational number be $x$ then, for any integer $y$ $2x^2=y^2$ $x\sqrt{2}=y$ $\frac{y}{x}=\sqrt{2}$ Now as $\sqrt{2}$ is irrational therefore $\frac{y}{x}$ must also be irrational.

As $y$ is integer therefore $x$ must be irrational but $x$ was considered to be rational, therefore this assumption that such a rational number exists is wrong

Therefore no such rational number exists.

The number $0$ when squared yields $0$ . On doubling this we get $0$ , which is a perfect square. Hence the answer is $\boxed 1$ .