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:


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


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*$y \equiv 0\mod 2\Rightarrow y+1=3 \Leftrightarrow z=15$, 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).

$144 = 2^4*3^2$
If $x,y,z$ are prime then $a=x+1,b=y+1,c=z+1 \ge 3$ so we will only consider factors at least $3$.  If, wolog, $x+1, y+1 \ge 3$ then $z+1 \le \frac {144}9 = 16$.
So we need to only consider triplets of factors between $3$ and $16$.
The factors of $144$ are of the form $2^4*3^2$ and are $1,2,4,8,16, 3,6,12,24,48,9,18, 72, 144$ and in order of size we are considering only the factors $3,4,6,8,9,12,16$. As we want these to be one more than primes, we can't have $16$ or $9$.
So the factors we may have are $3,4,6,8,12$.  Now lets find the triplets by listing them in order.  wolog $a \le b \le c=\frac {144}{ab}$.
We have $a,b,c =$
$3,3,16$ no good! $16$ not in our list
$3,4,12$
$3,6,8$ 
$3,8, *erk*$ we have $c = \frac {144}{ab} < b$ so that's it for $a = 3$.  $a=4$ next.
$4*4*9$ no good.  $9$ not on our list.
$4,6,6$ And that's it. We've "hit the middle" for $a=4$.  Should we try $a=6$ next?
$6,6,*erk*$ we have $c = \frac {144}{ab} < b$ so we've hit the wall.
So ignoring permutations $\{x,y, z\} = \{2,3,11\},$ or $\{2,5,7\}, \{3,5,5\}$
