Solve $(x+1)(y+1)(z+1)=144$ in primes "Solve $(x+1)(y+1)(z+1)=144$ in primes".
So far, I have concluded that the solutions are $(x,y,z)=(2,3,11)$ or $(2,5,7)$ and their permutations. I worked like this:


*

*$x \equiv 0\mod 2\Rightarrow x+1=3, 144=2^4*3^2 \Leftrightarrow (y+1)(z+1)=48=2^4*3$


*

*$y \equiv 0\mod 2\Rightarrow y+1=3 \Leftrightarrow z=15$, contradiction since $15$ is not prime

*$y \equiv 1\mod 2\Rightarrow y+1\geq 2^2 \Leftrightarrow y+1=2*3, z+1=2^3$ or $y+1=2^2*3, z+1=2^2 \Leftrightarrow y=5, z=7$ or $y=11, z=3$
And since the equation is symmetric the solutions are the permutations of the latter ones. Similarly, through casework, we find the same solutions if $x \equiv 1\mod 2$ (if I haven't made a mistake). My question is if there exists a more simple way to solve this problem besides lots of casework (and if there exists another triplet that satisfies the given equation).

Assume x <= y <= z. The factors can be 3, 4, 6, 8, 12, 18, 24 and 72 (primes where p+1 divides 144). The smallest factor cannot be 6 or larger since 6^3 >= 216, so we get 3x48 or 4x36. 48 = 4x12 or 6x8, 36 = 6x6. So the solutions (x, y, z) are (2, 3, 11), (2, 5, 7) and (3, 5, 5). 
And of course all the permutations. 
You can also see that the product of the two smallest factors is at least 9, so the last one cannot be more than 16. That could be useful if you solve the same problem for much larger numbers. 
