Find all triplets $(x,y,p)$ such that $\frac{xy^3}{x+y}=p$

Question

Find all triplets $(x,y,p)$ such that $\frac{xy^3}{x+y}=p$, where $x$ and $y$ are positive integers and $p$ is prime. I don't have much experience with Diophantine equations such as this, and I don't know how to fully solve it. After a bit of fiddling around I found that $(n^4-n,n,n^3-1)$ is a solution if $n^3-1$ is prime, so $(14,2,7)$ is one solution. I suspect this is the only solution since $n^3-1$ is not prime for $n>2$, and I don't see any other obvious ways to generate solutions.

Answer 1

We have: $y = \dfrac{xy^3}{p} - x$. Thus $p \mid xy^3$. So either $p \mid x$ or $p \mid y^3$ , hene $p \mid y$ since it is a prime number. If $p \mid x \implies x = kp\implies y = ky^3 - kp\implies kp = ky^3-y = y(ky^2-1)$. thus $p \mid y$ or $p \mid ky^2-1$. We have again two cases: if $p \mid y$, then $y = np \implies kp = np(ky^2-1)\implies k = n(ky^2-1)\implies n = \dfrac{k}{ky^2-1}\le \dfrac{k}{k-1}< 2\implies n = 1\implies k = ky^2-1\implies k = \dfrac{1}{y^2-1}\notin \mathbb{N}$. Thus $p \nmid y$, and we must have $p \mid ky^2-1\implies ky^2-1 = pm\implies kp = y(pm)\implies k = my\implies my^3 - 1 = pm\implies m(y^3-p) = 1\implies m = 1 = y^3-p\implies p = y^3-1 = (y-1)(y^2+y+1)\implies y - 1 = p, y^2+y+1 = 1\implies y = 0$ which is absurd, so $y-1 = 1, y^2+y+1 = p\implies p = 2^2+2+1 = 7, k = my = 1(2) = 2, x = kp = 2(7) = 14$. Thus $(x,y,p) = (14,2,7)$. The second case is $p \mid y$, thus $y = rp\implies rp = \dfrac{xr^3p^3}{p} - x\implies rp = xr^3p^2-x\implies x = (xr^3p-r)p\implies p \mid x \implies x = tp\implies r = tr^3p^2 -t=t(r^3p^2-1)\implies t \mid r$. Also, $t = r(tr^2p^2-1)\implies r \mid t\implies r = t\implies 1 = t^3p^2-1\implies 2 = t^3p^2$, and this equation has no solution in $\mathbb{N}$. Thus the only solution to the given equation is: $(x,y,p) = (14,2,7)$.