Positive integer $2\times 2$ matrix with two positive integer square root matrices So suppose I have this matrix:
$$M=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$$
I am sure that every entry is a positive integer. I am trying to figure out if it has 2 square roots in which every entry is also a positive integer. I took a look at http://en.wikipedia.org/wiki/Square_root_of_a_2_by_2_matrix . It basically says that, setting $s = \pm \sqrt{\det M}$ and $t = \pm \sqrt{a + d + 2s}$ you get that the square roots of $M$ are all of the form $$R = \frac{1}{t} \begin{pmatrix}a+s & b \\ c & d+s\end{pmatrix}$$
So, in order for every entry of $R$ to be a positive integer, am I correct in thinking $\det{M}$ must be a perfect square, $a + d \gt 2s$, $a>s$, $d>s$, $t>0$, t is an integer, and all enteries must be divisible by t? Or are there wholes in these conditions?

The Wikipedia article is not very clearly written. Let us derive the result by ourselves. Suppose
$$
M=\begin{pmatrix}A&B\\C&D\end{pmatrix}
=R^2={\underbrace{\begin{pmatrix}a&b\\c&d\end{pmatrix}}_{R}}^2
=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&bc+d^2\end{pmatrix}.
$$
Then $A+D+2(\det R) = (a+d)^2$ and
$$
\begin{pmatrix}A+\det R&B\\C&D+\det R\end{pmatrix}
=(a+d)\begin{pmatrix}a&b\\c&d\end{pmatrix}.
$$
Therefore, if $A+D+2\delta\neq0$ for some $\delta\in\{-\sqrt{\det M},\sqrt{\det M}\}$, then
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix}
=\frac{1}{\pm\sqrt{A+D+2\delta}}
\begin{pmatrix}A+\delta&B\\C&D+\delta\end{pmatrix}\tag{1}
$$
is a (perhaps complex) square root of $M$. Also, if $A+D+2\delta\neq0$ for both $\delta=-\sqrt{\det M}$ and $\delta=\sqrt{\det M}$, then all square roots of $M$ are given by $(1)$.
When $A+D+2\delta=0$ for some $\delta\in\{-\sqrt{\det M},\sqrt{\det M}\}$, $M$ can have other square roots that do not take the form of $(1)$. For instance, as pointed out by Samuel, the identity matrix has infinitely many square roots: apart from $\pm I$, there are also square roots of the form
$$
\begin{pmatrix}a&b\\c&-a\end{pmatrix}
$$
where $a\in\mathbb{C}$ is arbitrary and $bc=1-a^2$. (Putting $a=0$ and $b=c=1$ gives the well known one; my favourite one is obtained by taking $a=b=1$ and $c=0$.)

Now, go back to your question. If $M=R^2$ for some positive integer matrix $R$, then $\det M=(\det R)^2$ must be a perfect square.
Also, we have
$$A+D+2\sqrt{\det M}>A+D-2\sqrt{\det M}> A+D-2\sqrt{AD}=(\sqrt{A}-\sqrt{D})^2 \ge0.$$
Therefore, every positive square root matrix of $M$, if any, takes the form of $(1)$.
It follows that for each $\delta\in\{-\sqrt{\det M},\sqrt{\det M}\}$ that produces an integer square root matrix from $(1)$, $\sqrt{A+D+2\delta}$ must be an integer that divides $A+\delta,\,B,\,C,\,D+\delta$. Note that it may not divide $A$ or $D$ (so your claim that it divides all entries of $M$ is false). For a counterexample, consider
$$
M=\begin{pmatrix}2&5\\5&17\end{pmatrix}
=\begin{pmatrix}1&1\\1&4\end{pmatrix}^2,\ \sqrt{\det M}=3.
$$
We have $\sqrt{A+D+2\sqrt{\det M}}=5$ divides $A+3=B=C=5$ and $D+3=20$, but not $A=2$ or $D=17$. This counterexample also shows that the condition $A>\sqrt{\det M}$ does not necessarily hold. By permuting the rows and columns of $M$, it follows that the condition $D>\sqrt{\det M}$ also is not necessarily true.
