We have $p | xy^3$ so either $p | x$ or $p | y $. If $p | x $. Let $x=bp$ then the equation becomes $y(by^2-1)=bp$ and since $b$ and $by^2-1$ are coprimes then $b | y$. Let $y = \alpha b$. Since $y | pb $ then we conclude that $ \alpha | p$ so $y=b$ or $y=pb$ which it is equivalent to $y=x$ or $x=py$. If $p | y $. Let $y=ap$ then the equation becomes $a^3xp^2=x+ap$ so $p | x$ and we prove as above that $x=y$ or $x=py$ If $x=y$ then $x^3=2p$ is even so $x$ is even ($=2x'$) and the equation becomes $4x'^3=p$ which not have a solution. If $x=py$ then $y^3=(1+p)$ thus $p=y^3-1=(y-1)(y^2+y+1)$ is a prime then $y=2$ , $p=7$ and $x=py=14$

First note that we have $ x + y | xy^3 $ and $ x + y | xy^2 + y^3 $ and so $ x + y | xy^2 $. But we again also have $ x + y | x^2 y + xy^2 $ and that gives $ x + y | x^2 y $. Finally this gives $ x + y | x^2 y - x y^2 = xy(x - y) (1)$. But from the question we see that if a prime $ q $ divides both $ x $ and $ y $, then surely assuming the power of $ q $ dividing x and y to be $ n $ and $ m $, we get that the exponent of power of $ q $ dividing the L.H.S is $ m + 3n - min {m,n} \geq 1 $ . So $ q = p $, which on a similar argument gives that the power of p dividing L.H.S is more than 1. Hence we conclude that $ g.c.d{x,y} = 1 $. Now, in (1), $ g.c.d (x + y, x - y) = g.c.d { x, 3y} $ which is atmost 3 and that $ g.c.d {x + y, xy} = 1 $. So our only hope for any solution is $ x + y = 3 $ which however gives no solutions.

Hello!The equation does have a solution.That is,$(x,y,p)=(14,2,7)$. I will let you find out your mistakes on your own.

Yes there was a little mistake at the end of my proof $y^3=p+1$ instead of $y^3=p(p+1)$. So $p=y^3-1=(y-1)(y^2+y+1)$ is a prime then $y=2$ , $p=7$ and of course $x=py=14$

We get $xy^3=p(x+y)$, which taken modulo $x$ gives $x|py$, and modulo $y$ gives $y|px$. Hence for any prime $q\neq p$, $v_q(x)\le v_q(y)\le v_q(x)$, hence $v_q(x)=v_q(y)$, and also $|v_p(x)-v_p(y)|=1$. This gives us three cases: $x=py$, $x=y$, $y=px$. Case 1: $x=py$, then $py^4=p(py+y)$, $y^3=p+1$, $p=y^3-1=(y-1)(y^2+y+1)$. Here $y-1$ is the smaller factor, hence $y-1=1$ and $y^2+y+1=p$. Hence $y=2$, $p=7$, $x=14$. Case 2: $x=y$, then $x^4=p\cdot 2x$, $x^3=2p$, absurd because $v_2(x^3)$ can now be either $1$ or $2$, not divisible by $3$. Case 3: $y=px$, then $p^3x^4=p(x+px)$, $p^2x^3=1+p$, which is impossible even mod $p$. Hence the only solution: $(x,y,p)=(14,2,7)$.

One of the numbers x and y must not be odd $\longrightarrow 2| x$ or $ 2|y$ . $(***)$ We have from $p | x*y^3$ that: $1.$ $p|x$ . Let $x=mp$ so $m*y^3=mp+y $$\Longleftrightarrow$ $ y(m*y^2-1)=mp \Longrightarrow gcd(m, m*y^2-1)=1 \Longrightarrow m|p$, so let $y=mn$, because $y|mp \Longrightarrow n|p$, so $n=p$ or $n=1$ $\Longrightarrow$ so: $a) y=mp=x$ $\Longleftrightarrow$ $\frac{x^3}{2}=p $ . This equation doesn't has a solution because $2*p$ can't be cube of any natural number. $b) y= \frac{x}{p}$ so $ x=py$, starting equation transform in $y^3=p+1 $$\Longleftrightarrow$ $ p=(y-1)(y^2+y+1)$ so $g=y-1=p$ or $f=y^2+y+1=p$.Because $f>g \Longrightarrow p=y^2+y+1$ and $y-1=1$. so $\boxed{y=2}$,$\boxed{p=7}$,$\boxed{x=14}$ . $2.$ $p|y$. Let $y=p*q$ $\Longrightarrow$ $pq(xp*q^2-1)=x$, because $gcd(xp*q^2-1, x)=1$ $\Longrightarrow$ $x|pq$ and $pq|x$ from $pq(xp*q^2-1)=x$ equation. So $x=y$. We already done this case above. Hence the only solution is $\boxed{(x,y,p)=(14, 2, 7)}$

Let me make the contribution of a shortest proof. Take $d=\gcd(x,y)$, $x=da$, $y=db$, with $\gcd(a,b)=1$, which also forces $\gcd(a,a+b)=\gcd(b,a+b)=1$. Then $\dfrac {d^3ab^3}{a+b}=p$ forces $\bullet$ either $a=b=1$, $d^3=p(a+b)=2p$; impossible; $\bullet$ or $a=p$, $b=1$, $a+b=d^3$, thus $p=(d-1)(d^2+d+1)$, leading to $d=2$, thus $\boxed{(x,y,p) = (14,2,7)}$;

Just take $(xy^2-p)(y^3-p)=p^2$ The rest is easy

Clearly, $p\mid xy^3$, so either $p\mid x$ or $p\mid y$. Case 1: $p\mid x$ Let $x=ap$. Now $y(ay^2-1)=ap$, so since $\gcd(y,ay^2-1)=\gcd(a,ay^2-1)=1$, we have $a\mid y$. Let $y=ab$, then $b(a^3b^2-1)=p$. Case 1.1: $b=1$ and $a^3b^2-1=p$ Then $p=(a-1)(a^2+a+1)$, so $a-1=1$ since $a^2+a+1>1$. Then $a=2$, so $p=7$ and we get the solution $(x,y)=(14,2)$. Case 1.1: $b=p$ and $a^3b^2-1=1$ Then $x=ap=y$, which leads to (in the original equation) $2p=x^3$. This is impossible, since the number of factors of two in the LHS is always $1$ or $2$ while the number of factors of $2$ in the RHS is always divisible by $3$. Case 2: $p\mid y$, $p\nmid x$ Let $y=ap$. Now $x(a^3p^2-1)=ap$, but this is impossible as $p\nmid x$ and $p\nmid a^3p^2-1$. In the end, the only solution is $(x,y)=\boxed{(14,2)}$, which indeed fits giving $p=7$.