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).
 A: $144 = 2^4*3^2$
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 $\{2,5,7\}, \{3,5,5\}$
A: Denote $x' := x + 1$ and analogously. As you point out, solutions are closed under permutation, so we may suppose w.l.o.g. that $x' \leq y' \leq z'$.
Since the factors $x', y', z'$ are all at least $3$, we must have $z + 1 \leq \frac{144}{3^2} = 16$. But the factors of $144 = 2^4 \cdot 3^2$ no larger than $16$ are
$1, 2, 3, 4, 6, 8, 9, 12, 16$, and the numbers among these $1$ larger than a prime are $\{3, 4, 6, 8, 12\}$. We can reduce the number of cases to check by observing that $5^3 < 144 < 6^3$, which implies that $x' \in \{3, 4\}$ and $z' \in \{6, 8, 12\}$.


*

*If $z' = 6$, then $x' y' = \frac{144}{6} = 24$ and $\sqrt{24} \leq y' \leq 6$, so $y' = 6$: $(x', y', z') = (4, 6, 6)$.

*If $z' = 8$, then $x' y' = \frac{144}{8} = 18$ and $\sqrt{18} \leq y' \leq 8$, but the only factor of $18$ in that range is $6$: $(x', y', z') = (3, 6, 8)$.

*If $z' = 12$, then $x' y' = \frac{144}{12} = 12$, so $x' \leq \sqrt{12}$ and thus $x' = 3$: $(x', y', z') = (3, 4, 12)$.


The solutions are: $$(x, y, z) = (2, 3, 11), (2, 5, 7), (3, 5, 5) .$$ 
It seems the error in the original post was in the unwritten details of the case $x \equiv 1 \pmod 2$.
A: 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. 
