$$x^3-bx^2+4hx-ph=0$$
Taking the equation above, is there away to prove that for every positive prime p:
There are positive integer values for b and h so that the cubic has 3 positive integer roots
Or 2 positive integer roots with one being repeated.
Intuitively, it is obvious that this is the case and to help visualize it graphically I used desmos. Looking at specific cases of primes is fine, however I am not sure how to go about the general case.
Even proving that there are always 2 or 3 positive rational roots would be helpful.
Thanks in advance for any help.
Assume that your cubic has 3 real roots, $kp, m, n \text{ with } k,m,n \in \mathbb{N}$, so that
$$\begin{align*}f(x)&= (x-kp)(x-m)(x-n) \\ \\ &= x^3-(kp+m+n)x^2+\left(kpm + kpn +mn\right)x-kpmn \\ \\ &= x^3-bx^2+4hx-ph \end{align*}$$
Which yields
$$\begin{align*}b &= kp+m+n\\ h &= kmn\\ 4h &= kp(m+n) +mn\\ \end{align*}$$
So your conjecture is equivalent to the following statement about how every prime, $p$, must be able to be expressed with $ k,m,n \in \mathbb{N}$:
$$\begin{align*} p&=\dfrac{(4k-1)mn}{k(m+n)}\\ \\ p&= \left(4-\dfrac{1}{k}\right)\dfrac{1}{\dfrac{1}{m}+\dfrac{1}{n}} \end{align*}$$
If you can prove every positive prime can be expressed like that, then your conjecture is true.