Application of derivatives

Critical points occur when $f'(x) = 0$ . Therefore

$\begin{aligned} f(x) & = 2x^3 - 9ax^2 + 12a^2 x + 1 & \small \blue{\text{Differentiate both sides w.r.t. }x} \\ f'(x) & = 6x^2 - 18ax + 12a^2 & \small \blue{\text{Putting }f'(x)=0} \\ x^2 - 3ax + 2a^2 & = 0 \\ (x-a)(x-2a) & = 0 \end{aligned}$

Therefore, the critical points are at $x=a$ and $x=2a$ . Since $p$ and $q$ are natural numbers and $p^2 = q$ , we assume $p < q$ and hence $p=a$ and $q=2a$ , then

$\begin{aligned} a^2 & = 2a \\ a^2 - 2a & = 0 \\ a(a-2) & = 0 & \small \blue{\text{Since }a \text{ is a natural number.}} \\ \implies a & = \boxed 2 \end{aligned}$

$f'(x) =0\implies x^2-3ax+2a^2=0$

Roots of this equation are $2a,a$

So $2a=a^2\implies a=\boxed {0,2}$ (since $a$ is a natural number).

If only positive integer solution is needed, then $a=\boxed 2$ .