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.
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.
