2021/098) Find integers x,y z, such that $x + y +z = xyz$

 Because of symmetry if (x,y,z) is a solution then any permutation of (x,y,z) is also a solution

without loss of generality let us assume that $x<=y<=z$ 

So $x+y+z <= 3z$ 

puttying in the given equation we get

$3z >= xyz$

or $3>=xy$

This gives the following set in (x,y)

(1,1) giving 2 + z = z which does not have a solution

(1,2) giving 3 + z = 2z or z = 3

(1,3) giving 4 + z = 3z giving z = 2 which is a contradiction as it should not be less than 3

so solution set (1,2,3) or a permutation of the same.