Determine positive integer solutions

[anemone]

Determine all triples of positive integers $x,\,y,\,z$ such that $z\ge y\ge x$ and $x+y+z+xy+yz+xz=xyz+1$.

 

[kaliprasad]

Determine all triples of positive integers $x,\,y,\,z$ such that $z\ge y\ge x$ and $x+y+z+xy+yz+xz=xyz+1$.

Spoiler

add $1 + xyz$ on both sides to get $(1+x)(1+y)(1+z) = 2(1+xyz)$

for upper bound on x letting $x =y = z$ we get $(1+x)^3 =2(1+x^3)=> x <4$

so we need to check for 3 cases $x = 1, 2 , 3$

$x=1 => 2(1+y)(1+z) = 2(1+yz) => 1+y+z + yz = 1 + yz=> y+ z = 0$ or no solution$x=2=> 3(1+y)(1+z) = 2(1+2yz)$
or $3 + 3y + 3z+ 3yz = 2 + 4yz$
or $ yz-3y-3z =1$
or $(y-3)(z-3) = 10$
giving one solution $z-3=10,y-3=1=> z = 13,y=4$
or another solution $z-3 = 5,y-3 =2 => z= 8,y = 5$

$x=3=> 4(1+y)(1+z) = 2(1+3yz)$
or $4 + 4y + 4z + 4yz = 2 + 6yz$
or $ yz - 2y - 2z = 1$
or $(y-2)(z-2) = 5$ giving $z-2=5, y-2 = 1$ or $z = 7, y = 3$

so 3 sets of solution $(x,y,z) = (2,5,8),(2,4,13),(3,3,7)$

 

[anemone]

Well done kaliprasad!