Find all $p \in \mathbb{Z}$ such that $ p^2+ 4p + 16 $ is a perfect square I've tried some things involving modular residues but they don't seem to work. Anyone know how to do this? 

So, we need $(p+2)^2+12=m^2$ for some integer $m$
$-12=(p+2-m)(p+2+m)$
As $p+2-m+p+2+m$ is even, the multiplicands have the same parity $\implies $ both must be ven
Consequently, $$\dfrac{p+2-m}2\cdot\dfrac{p+2+m}2=-3=1(-3)=(-1)3$$

Alternatively let $p^2+4p+16=(p+a)^2$ for some integer $a$
$\implies p=\dfrac{a^2-16}{4-2a}$
$\iff-2p=\dfrac{a^2-16}{a-2}=a+2-\dfrac{12}{2-a}$
$\iff12=(2-a)(a+2+2p)$
As $a+2+2p+2-a$ is even, $a+2+2p,2-a$ must have the same parity which must even as $12$ is even
$\implies\dfrac{2-a}2\cdot\dfrac{2+a+2p}2=3=(-3)(-1)=3\cdot1$
