pohoatza wrote: From $S = pr = \frac{p^{2}}{n}$ and $S=\sqrt{p(p-a)(p-b)(p-c)}$, we have that $p^{3}= n^{2}(p-a)(p-b)(p-c)$. And now by denoting $p-a=x$, $p-b=y$, $p-c=z$, the above equality turns out into $(x+y+z)^{3}= n^{2}xyz$. Now the problems turns out to show that the above equation has a solution in natural numbers for infinitely many $n$. For that assume $z=k(x+y)$, $\forall k \in \mathbb{Z_{+}}$. Thus the equation becomes $(k+1)^{3}(x+y)^{2}=kn^{2}xy$, and now by setting $n=3(k+1)$, we have that $(k+1)(x+y)^{2}=9kxy$. Denote $t=\frac{x}{y}$. Thus the above equation has solutions in positive integers if and only if $(k+1)(t+1)^{2}=9kt$ has a rational solution, which is equivalent in proving that its discriminant $D=k(5k-4)$ is a perfect square. Thus the problem turns out into proving that the equation $5k^{2}-4k=t^{2}$ has infinitely many integer solutions, which looks like Pell.

The equation $5k^2 - 4k = t^2$ indeed looks like Pell, smells like Pell, and reduces to Pell, since it may be written as $(5k-2)^2 - 5t^2 = 4$, or $A^2 - 5B^2 = 4$ (of course, out of its infinitely many solutions, one needs to ascertain that infinitely many will have $A \equiv -2 \pmod{5}$, for example the primitive solution $(A_0,B_0) = (3,1)$).