Gazeta Matematica 4/2014, Problem E:14641

Problem:
Find all primes $p$ such that the equation $$x^3+y^3+3xy=p^2+1$$ has integer solutions.

Proposed by Alessandro Ventullo

Solution:
Observe that $x^3+y^3-1+3xy=(x+y-1)(x^2+y^2+1-xy+x+y)$. So, the given equation can be rewritten as
$$(x+y-1)(x^2+y^2+1-xy+x+y)=p^2.$$ Since $x^2+y^2+1-xy+x+y=\dfrac{1}{2}[(x-y)^2+(x+1)^2+(y+1)^2] \geq 0$, then $x+y-1\geq0$, i.e. $x+y \geq 1$. This gives $x^2+y^2+1-xy+x+y\geq 2|xy|+1-xy+1=2|xy|-xy+2 \geq 2$. Moreover, $x^2+y^2+1-xy+x+y>x+y-1$. So, it must be
$$\begin{array}{rll} x+y-1&=&1 \\ x^2+y^2+1-xy+x+y&=&p^2. \end{array}$$
The first equation gives

$x+y=2$                                (1)

the second equation gives $(x+y)^2-3xy+(x+y)+1=p^2$, i.e.

$xy=\dfrac{7-p^2}{3}.$                                (2)

So, $$t^2=\left(\dfrac{x-y}{2}\right)^2=\dfrac{(x+y)^2-4xy}{4}=1-\dfrac{7-p^2}{3},$$ i.e. $3t^2=p^2-4$. If $p>2$, we have that $3t^2 \equiv 0,3 \pmod{4}$ and $p^2-4 \equiv 1 \pmod{4}$, a contradiction. So, $p=2$. This value is acceptable because, from equations (1) and (2), we get $(x,y)=(1,1)$.