2011 JBMO Problems/Problem 2

Problem

Find all primes $p$ such that there exist positive integers $x,y$ that satisfy $x(y^2-p)+y(x^2-p)=5p$.

Solution

Rearrange the original equation to get \begin{align*} 5p &= xy^2 + x^2 y - xp - yp \\ &= xy(x+y) -p(x+y) \\ &= (xy-p)(x+y) \end{align*} Since $5p$, $xy-p$, and $x+y$ are all integers, $xy-p$ and $x+y$ must be a factor of $5p$. Now there are four cases to consider.

Case 1: $xy-p = 5p$ and $x+y = 1$

Since $x$ and $y$ are positive integers, $x+y \ne 1$, so the first case can not happen.

Case 2: $xy-p = 1$ and $x+y = 5p$

Rearranging the first equation results in $5xy-5 = 5p$. By substitution, we get \begin{align*} x+y &= 5xy - 5 \\ 5 &= 5xy - x - y \\ 25 &= 25xy - 5x - 5y. \end{align*} Applying Simon's Favorite Factoring Trick results in $$26 = (5x-1)(5y-1)$$. From this equation, $5x-1$ and $5y-1$ can equal 1, 2, 13, or 26. Since no value of $x$ or $y$ work, there are no solutions, so the second case can not happen.

Case 3: $xy-p = 5$ and $x+y = p$

Rearranging the first equation results in $xy-5 = p$. By substitution and Simon's Favorite Factoring Trick, we get \begin{align*} xy-5 &= x+y \\ xy-x-y &= 5 \\ (x-1)(y-1) &= 6 \end{align*} From this equation, $x-1$ and $y-1$ can equal 1, 2, 3, or 6. The ordered pairs of $(x,y)$ can be $(2,7)$, $(7,2)$, $(4,3)$, or $(3,4)$. Since $p$ is prime, $p$ can equal $7$ in this case.

Case 4: $xy-p = p$ and $x+y = 5$

Rearranging the first equation results in $xy = 2p$. Since $x,y$ are positive integers, we can simply take the few ordered pairs $(x,y)$ that satisfy the second equation and plug the values into the first equation to check for values of $p$.

If $(x,y)$ is $(1,4)$ or $(4,1)$, then $2p = 4$, so $p = 2$. If $(x,y)$ is $(2,3)$ or $(3,2)$, then $2p = 6$, so $p = 3$. Both $2$ and $3$ are prime numbers, so both of the numbers work for $p$.

In summary, the values of $p$ that satisfy the original conditions are $\boxed{2, 3, 7}$.