$f(x)=x^3 -px^2 + qx$ and $g(x)=3x^2-2px+q$
I want to prove that there are infinitely many pairs of positive integer $p,q$ such that $f(x)$ has three distinct integer roots and $g(x)$ has two distinct integer roots. I can show that if $f,g$ have the required roots, then $p$ is multiple of $3$, $q$ is multiple of $9$, $p^2-3q$ is positive perfect square and $p^2-4q$ is positive perfect square. But i do not know how to show there are infinitely many required pairs. We also require that $gcd(p,q)=3$
Best Answer
There are an infinite number of solution pairs. That's a general result for quadratic Diophantine equations: they either have no solution (except perhaps for trivial solutions involving zero), or they have an infinite number of solutions, although those solutions will fall into a finite number of families. IIRC, Lagrange gave the first proof of that result (along with various other important proofs on Pell's equation and quadratic forms).
As mentioned by Daniyar, for both equations to have integer solutions, both $p^2-4q$ and $p^2-3q$ must be perfect squares.
Let $$p^2-4q=r^2$$ $$p^2-3q=s^2$$ Subtracting, $$q=s^2-r^2$$ and substituting $$p^2=4s^2-3r^2$$
Using Brahmagupta's identity, products of integers of the form $4s^2-3r^2$ are also of that form if the second terms are even:
$$(4s^2-3r^2)(4x^2-3y^2)$$ $$=(16s^2x^2+9r^2y^2) - 3(4s^2y^2+4r^2x^2)$$ $$=(16s^2x^2+9r^2y^2 \pm 24sxry) - 3(4s^2y^2+4r^2x^2 \pm 8syrx)$$ $$=(4sx \pm 3ry)^2 - 3(2sy \pm 2rx)^2$$
If either of $r$ or $y$ are even, $4sx \pm 3ry$ will also be even.
Now we could search for numbers of the form $4s^2-3r^2$, looking for perfect squares. But we can use a minor modification of the above identity to construct such squares.
Let $$s=(u^2+3v^2)/4$$ where $u$ & $v$ are coprime and both odd, hence $$u^2\equiv v^2\equiv 1\pmod 4$$ and $$u^2+3v^2\equiv 0\pmod 4$$ (so $s$ is an integer), and let $$r=uv$$
So $$4s^2-3r^2$$ $$=(u^2+3v^2)^2/4-3u^2v^2$$ $$=(u^4+9v^4+6u^2v^2 - 12u^2v^2)/4$$ $$=(u^4+9v^4-6u^2v^2)/4$$ $$=(u^2-3v^2)^2/4$$
Note that $(u^2-3v^2)\equiv 2 \pmod 4$, i.e., it's twice an odd number, so $(u^2-3v^2)^2$ is an even square, and $$p=(u^2-3v^2)/2$$ is an odd integer. We do need to choose appropriate $(u, v)$ pairs to ensure that $p$ and $q$ are positive.
Now sometimes $\gcd(p,q)=3$, but if that's not the case they are coprime, so we can easily multiply them both by 3 to achieve the desired GCD constraint. I'll leave the proof of that as an exercise for the reader. ;)
Here are some solutions constructed using the above procedure.
$$\begin{array}{|r|r|} \hline p & q \\ \hline 33 & 72 \\ 39 & 360 \\ 69 & 360 \\ 111 & 3024 \\ 141 & 840 \\ 177 & 2520 \\ 183 & 5616 \\ 213 & 2640 \\ 219 & 11880 \\ 249 & 5040 \\ 291 & 12240 \\ 321 & 3168 \\ 327 & 22176 \\ 363 & 32760 \\ 393 & 10920 \\ 429 & 8568 \\ 429 & 15120 \\ 501 & 18480 \\ 537 & 23760 \\ 573 & 7920 \\ 681 & 28728 \\ 717 & 19872 \\ 753 & 43680 \\ 789 & 51480 \\ 897 & 15960 \\ 897 & 62832 \\ 933 & 72072 \\ 1041 & 59400 \\ 1149 & 94248 \\ 1221 & 118560 \\ 1257 & 131040 \\ \hline \end{array}$$
You can see more solution pairs using this small interactive Python script. The script itself is encoded into the URL, and it runs on a SageMath, Inc, server. I use a Farey sequence generator to produce coprime $(u, v)$ pairs. The script parameter $m$ selects the Farey sequence.