Eukleidis wrote: Problem 2. Find all primes $p$ such that there exist positive integers $x,y$ that satisfy $x(y^2-p)+y(x^2-p)=5p$ Solution: $x(y^2-p)+y(x^2-p)=5p$ ==> $(x+y)(xy-p)=5p$ because $5$ and$p$ are prime we have $x+y=5,xy=2p$ then $p=2,x=4,y=1$or $p=3,x=2,y=3$ $xy-p=5,x+y=p$==>$(x-1)(y-1)=6$ so $x+y=7$or $x+y=9$but $9$ isn't prime so$p=7$ and the third case is $xy-p=1, x+y=5p$ $x=1$ than $y=p+1$ and $y=5p-1$ in this case we don't have solution $x,y\geq 2$ than $xy\geq x+y$ so also in this case we don't have solution.

Eukleidis wrote: Problem 2. Find all primes $p$ such that there exist positive integers $x,y$ that satisfy $x(y^2-p)+y(x^2-p)=5p$ Expanding a bit we have $xy(x+y) -p(x+y)=5p$ or $(x+y)(xy-p)=5p$. There are two cases to consider: 1. $x+y=5$ and $xy=2p$, where we find $(x,y)=(4,1)$ as a solution which gives $p=2$; or $(x,y)=(3,2)$ which gives $p=3$ as a solution. 2. $x+y=p$ and $xy-p=5$, or $xy=p+5=x+y+5$ or $x(y-1)=y-1 +6$ or $(y-1)(x-1)=6$. Solving this gives $(x,y)=(7,2)$ or $(x,y)=(4,3)$. In the first case $x+y=9$ is not prime; in the second case $x+y=7$, so $p=7$. Therefore the only such numbers are $p=2$, $p=3$ and $p=7$.

enndb0x wrote: Eukleidis wrote: Problem 2. Find all primes $p$ such that there exist positive integers $x,y$ that satisfy $x(y^2-p)+y(x^2-p)=5p$ Expanding a bit we have $xy(x+y) -p(x+y)=5p$ or $(x+y)(xy-p)=5p$. There are two cases to consider: 1. $x+y=5$ and $xy=2p$, where we find $(x,y)=(4,1)$ as a solution which gives $p=2$; or $(x,y)=(3,2)$ which gives $p=3$ as a solution. 2. $x+y=p$ and $xy-p=5$, or $xy=p+5=x+y+5$ or $x(y-1)=y-1 +6$ or $(y-1)(x-1)=6$. Solving this gives $(x,y)=(7,2)$ or $(x,y)=(4,3)$. In the first case $x+y=9$ is not prime; in the second case $x+y=7$, so $p=7$. Therefore the only such numbers are $p=2$, $p=3$ and $p=7$. I think you have to consider the third case like arber did: $xy-p=1,x+y=5p$ which is easy, but necessary.

In principle, even the fourth case $x+y = 1$, $xy-p = 5p$ should be mentioned, and immediately dismissed since $x+y$ must be at least $2$. The third case $x+y = 5p$, $xy-p = 1$ leads to $4xy + (x-1)(y-1) = 6$, whence $4xy \leq 6$, so $x=y=1$, not good.

p-primes x,y-positive integer x(y2-p)+y(x2-p)=5p xy2-xp+yx2-yp=5p => xy(x+y)-p(x+y)=5p =>(x+y) (xy-p)=5p system 1)x+y=5 xy-p=p =>(x=2,y=3,p=3);( x=3,y=2,p=3);( x=1,y=4,p=2);( x=4,y=1,p=2) 2)x+y=p xy-p=5 => no solution

Rewrite the given expression as $(x+y)(x y-p)=5 p$. As both $5$ and $p$ are prime, there are only 4 possibilities to consider. The first one, $x+y=5$, $x y-p=p \rightarrow$ $x y=2 p$. When $(x, y)=(2,3),(3,2), p=3$, when $(x, y)=(1,4),(4,1)$, $p=2$. In the second case, $x+y=5 p$, $x y-p=1$. Putting $x, y=1$ doesn't yield solution, and for $x, y \geq 2$ we know $x y \geq x+y \rightarrow 1+p \geq 5 p$ which yields no solution. Also, $x+y>1$, hence the only possibility left is $x+y=p, x y-p=5$. Now, $p=x y-5$, $x+y=x y-5 \Rightarrow x y-x-y+1=6 \Rightarrow(x-1)(y-1)=6$ $\Rightarrow (x,y)= (3,4),(4,3) \rightarrow p=7$. So $\boxed {p=2,3,7}$.

Note that everything I am working with here is WLOG, since the equation is symmetric (that is, no difference in value of p in switching x with y and vice versa.) $x(y^2-p)+y(x^2-p)=5p$ => $(x+y)(xy-p)=5p$. Then $x+y=5,xy=2p$ => $p=2,x=4,y=1$or $p=3,x=2,y=3$ $xy-p=5,x+y=p$ => subtracting first eq. - second gives $xy-x-y+1-p=5-p+1$, or $(x-1)(y-1)=6$ so either x=3 y=4 or x=2 y=7, representing either $x+y=7$, or $x+y=9$, but $9$ isn't prime so $p=7$. $xy-p=1, x+y=5p$ If $x=1$ than $y=p+1$ and $y=5p-1$, or $5p-1=p+1$, absurd. $x,y\geq 2$ than $xy\geq x+y$, so subtracting eq. $x+y-xy+p=5p-1$, or $x+y-xy=4p-1$, but LHS must be nonpositive and RHS is positive, contradiction. Finally, if $xy-p=5p, x+y=1$; x and y cannot be positive integers. Our final solutions are $\boxed{p=2,3,7}$

(x+y)(xy -p) = 5p After this just simple casework So ,p = 2,3,7 Seems easy for a JBMO P2