G3.2 Let $p$ be a prime and $m$ be an integer. If $p(p+m)+2 p=(m+2)^{3}$, find the greatest possible value of $m$.
$$
p(p+m+2)=(m+2)^{3}
$$

If $m$ is even and $p$ is odd, then odd $\times($ odd + even +2$)=(\text { even }+2)^{3} \Rightarrow$ LHS $\neq$ RHS !!!
If $m$ is odd and $p$ is odd, then odd $\times($ odd + odd +2$)=(\text { odd }+2)^{3} \Rightarrow$ LHS $\neq$ RHS !!!
In all cases, $p$ must be even.
$\because$ the only even prime is $2 \therefore p=2$
$$
2(m+4)=(m+2)^{3}
$$

LHS is even $\Rightarrow(m+2)^{3}$ is even $\Rightarrow m+2$ is even $\Rightarrow$ RHS is divisible by $8 \Rightarrow$ LHS is divisible by 8
$\Rightarrow m+4=4 n$, where $n$ is an integer $\Rightarrow m+2=4 n-2$
Put $m+2=4 n-2$ into the equation: $2(4 n)=(4 n-2)^{3}$
$$
n=(2 n-1)^{3}
$$
$\Rightarrow n=1, m=0$ (This is the only solution, $n1$ and $n>(2 n-1)^{3}$ for $n<1$ )