A square polynomial?

First of we establish a relation between $p,q,1$ .

Let the roots be $a,-a, b$ . Thus, by the Vieta's theorem,

$a+(-a)+b=-p$ $\implies -b=p$

Also, $a(-a)+(-a)b+b(a)=q$ $\implies -a^{2}=q$

And, $a(-a)b=-1$ $\implies a^{2}b=1$

Here, observe that $pq=(-b)(-a^{2})=a^{2}b=1$

Now, since $p, q \in Z$ , thus $(p,q)=(1,1)$ or $(p,q)=(-1,-1)$

But $(p,q)=(1,1)$ $\implies A(x)=x^{3}+x^{2}+x+1=(x+1)(x^{2}+1)$ but here the condition that two of the roots are equal and of the opposite sign is not satisfied.

Thus, $(p,q)=(-1,-1)$ (We can check and find that the condition is satisfied here.)

$\implies A(x)=x^{3}-x^{2}-x+1=(x-1)^{2}(x+1)$

We are almost there. Now, $A(x)$ is the square of an integer $\implies A(x) \in {0,1,4,9,...}$ . Now, since $(x-1)^{2}$ is always a perfect square, we consider $(x+1)$ so that $A(x)$ is a perfect square.

[There is one case when $(x-1)^{2}$ is not to be ignored, i.e. when $x=1$ , which yield $A(x)=0^{2}$ . But we are given that $x>1$ , so this case is to be ignored]

Thus, this entire problem reduces to:

Find the smallest positive integer $r$ such that $(r+1)$ is the square of an integer.

It is not hard to see that the solution to this problem is $\boxed {r=3}$