@Sachin Mittal 1
of course if $p=p^2$ and $q=q^2$ then $(p^2,q^2)$ will also satisfy the equation because $(p,q)$ satisfies the equation but in this way you will be imposing the constraint as $p=1$ and $q=1.$
Suppose in the real domain you have $f(x,y)=0$
and you have only one point $(x_0,y_0)$ satisfies $f(x,y)=0$ i.e. $f(x_0,y_0)=0$
Now, if you say $x=x^2$ and $y=y^2$ for all the points $(x,y)$ it means $x_0^2=x_0$ and $y_0^2=y_0$ are true and that implies $x_0=0,1$ and $y_0=0,1$.
So, in this way you are imposing some constraints on the points that satisfy the given equation.