32. It is given that $p$ is a prime number such that
$$
x^{3}+y^{3}-3 x y=p-1
$$
for some positive integers $x$ and $y$. Determine the largest possible value of $p$.
32. Answer: 5
Suppose $x$ and $y$ are positive integers satisfying $x^{3}+y^{3}-3 x y=p-1$. That is $(x+y+1)\left(x^{2}+y^{2}-x y-x-y+1\right)=p$. Since $x, y \geq 1$, we must have $x+y+1=p$ and $x^{2}+y^{2}-x y-x-y+1=1$. Suppose $x=y$. Then the second equation gives $x=y=2$. Thus $p=x+y+1=5$. Next we may suppose without loss of generality that $x>y \geq 1$. Thus $x-y \geq 1$. Then the equation $x^{2}+y^{2}-x y-x-y+1=1$ can be written as $x+y-x y=(x-y)^{2} \geq 1$. That is $(x-1)(y-1) \leq 0$. It follows that $x=1$ or $y=1$. Since we assume $x>y$, we must have $y=1$. Then from $x^{2}+y^{2}-x y-x-y+1=1$, we get $x=2$. But then $x+y+1=4$ is not a prime. Consequently there is no solution in $x$ and $y$ if $x \neq y$. Therefore the only solution is $x=y=2$ and $p=5$.