Algebraic equation from determinant of 2x2 positive definite matrix of even diagonal I have a positive definite matrix $A^TA=\begin{pmatrix} a^2+c^2 & ab+cd \\ ab+cd & b^2+d^2 \end{pmatrix}, \ a,b,c,d\in\mathbb{R}$, where $a^2+c^2,b^2+d^2\in2\mathbb{Z}$ and $ab+cd\in\mathbb{Z}$. 
I need to find at least one (it could be unique!) such matrix whose determinant equals 3, and it is really getting the best of me. I am unfamiliar with these kinds of questions and so am unaware of any tricks/techniques. 
I tried setting one of $a,b,c,d$ to zero but that doesn't work (in particular I am unable to satisfy the diagonal being even). I guess one could also try experimenting with what seems to me to be the simplest equations, namely 
\begin{align*}
a^2b^2+a^2d^2+c^2b^2+c^2d^2=4,\ a^2b^2+2abcd+c^2d^2=1.
\end{align*}
Is there another way or do I have to resort to computer search?
 A: Any positive definite matrix $B$ is of the form $A^TA$ for some $A$. For a $2\times 2$ symmetric matrix $B$ to be positive definite it is enough to have positive diagonal entries and positive determinant. So your matrix $B=A^TA$ will be of the form 
$$
B=\begin{bmatrix}  2n & s \\ s & 2m\end{bmatrix}
$$
with $4nm-s^2=3$. So, as long as $mn\ne0$, you will always have $4nm>3$ and the equation will have a solution: thus
$$
B=\begin{bmatrix}  2n & \pm\sqrt{4nm-3} \\ \pm\sqrt{4nm-3} & 2m\end{bmatrix},\ \ \ m,n\in\mathbb N
$$
(where both square roots have the same sign) gives all possible solutions. If you want $s$ to be an integer, you are looking for the cases where $4mn-3$ is a square. We can get infinitely many examples by looking at odd squares, since
$$
(2k+1)^2+3=4(k^2+k+1). 
$$
So, taking $m=1$ and $n=k^2+k+1$  (or viceversa) gives an example. 
As KB Dave, mentioned, taking $n=7$ (when $k=2$), $m=1$, gives 
$$
\begin{bmatrix} 14 & 5\\ 5&2\end{bmatrix}. 
$$
Or taking $n=3$ (that is, $k=1$), $m=1$ we get the example
$$
\begin{bmatrix} 6 &3\\ 3&2\end{bmatrix}. 
$$
So 
$$
\begin{bmatrix} 
2(k^2 + k+1) & 2k+1 \\ 2k+1 & 2
\end{bmatrix},\ \ k\in\mathbb N
$$
gives an infinite family of examples. 
A: Let the positive-definite matrix in question be
$$\begin{bmatrix}a & b \\ b & d
\end{bmatrix}\text{.}$$
Our starting point is ($a,d$ even)
$$\begin{align}
ad-b^2 &= 3\\
a + d &> 0\text{.}
\end{align}$$
Since $a$ and $d$ are even, let $a=2u$, $b=2v$ and eliminate in favor of $u,v$. Then equivalently we have
$$\begin{align}
4uv-b^2 &= 3\\
u + v &> 0\text{.}
\end{align}$$
Hence $b$ is odd. Let $b=2w+1$ and eliminate in favor of $w$. Then equivalently we have
$$\begin{align}
uv &= w^2 + w + 1 \\
u+v &> 0\text{.}
\end{align}$$
Hence $uv$ is odd by Fermat's little theorem, so let $u=2x+1$, $v=2y+1$ and eliminate in favor of $x,y$. Then equivalently we have
$$\begin{align} 4xy + 2x + 2y & = w^2 + w \\
x+y &\geq 0 \text{.}
\end{align}$$
The equality has no constant terms, so $(w,x,y)=(0,0,0)$ is a solution. (And therefore $(-1,0,0)$ is also a solution.) Substituting backwards,
$$\begin{bmatrix}2&1 \\ 1 & 2\end{bmatrix}$$
is a solution. To find more solutions, you might try either


*

*stereographically projecting from this one solution to find the whole family of rational solutions and searching for integer solutions there or

*continuing the Hensel lift at $2$ to find more restrictions on integer solutions


or some combination of these two strategies.
