Find all primes $p$ such that the equation
$$x^2(y+z)+y^2(z+x)+z^2(x+y)+3xyz=p^2$$
has positive integers solutions.
Proposed by Alessandro Ventullo, Milan, Italy
Solution:
Observe that $$x^2(y+z)+y^2(z+x)+z^2(x+y)+3xyz=(x+y+z)(xy+yz+zx),$$
so the equation can be rewritten as $$(x+y+z)(xy+yz+zx)=p^2.$$
Since $x,y,z$ must be positive integers, then $x+y+z>1$ and $xy+yz+zx>1$. This gives $x+y+z=xy+yz+zx=p$. Therefore, $$x(y-1)+y(z-1)+z(x-1)=0.$$ Note that each term on the left hand side is non-negative, so $$x(y-1)=y(z-1)=z(x-1)=0,$$ which means that $x=y=z=1$. Thus, $p=3$ is the only prime which satisfies the required property.
Observe that $$x^2(y+z)+y^2(z+x)+z^2(x+y)+3xyz=(x+y+z)(xy+yz+zx),$$
so the equation can be rewritten as $$(x+y+z)(xy+yz+zx)=p^2.$$
Since $x,y,z$ must be positive integers, then $x+y+z>1$ and $xy+yz+zx>1$. This gives $x+y+z=xy+yz+zx=p$. Therefore, $$x(y-1)+y(z-1)+z(x-1)=0.$$ Note that each term on the left hand side is non-negative, so $$x(y-1)=y(z-1)=z(x-1)=0,$$ which means that $x=y=z=1$. Thus, $p=3$ is the only prime which satisfies the required property.
