Minimize a quadratic form over $\mathbb{Z}$, where the matrix is positive-definite and symmetric I'm currently dealing with some quadratic forms over $\mathbb{Z}$ like this one: $$D(x_1,x_2)=(a^2-2b)(x_1^2+x_2^2)+4b(x_1x_2),$$ where $a,b\in\mathbb{R}$ and $a^2-2b>0$. Let us assume that this is always $>0$, unless $x_1=x_2=0$, in which case, $D(x_1,x_2)=0$.
I know I can translate $D$ to $x^TAx$, where $$A=\begin{pmatrix}a^2-2b & 2b\\ 2b & a^2-2b\end{pmatrix}.$$
Notably, the matrix is symmetric.
Now, it really seems like the minimum of this form is always achieved in some $(x_1,x_2)$ with $\Vert x_i\Vert \leq 1, \ \forall i=1,2$. In other words, every time I compute $D(x_1,x_2)$ with $\Vert x_i\Vert > 1$ for some $i$, the result seems to be always larger than $D(1,0)$ or $D(-1,1)$ or $D(1,1)$. It is definitely doable to prove that by brute force, but is there any result that makes this immediate? For sometimes I encounter similar quadratic forms but with more variables, and the result seems to hold still.
The results regarding minimizing quadratic forms I know always involve the form being over $\mathbb{R}$, but the form in this case is over $\mathbb{Z}$, which should result, I think, in a less difficult optimization.
But is there any result that says something regarding the minimum ($\neq 0$) of a quadratic form over $\mathbb{Z}$ when the matrix is symmetric and positive-definite? I can't find something of this sort anywhere. Please let me know if my question is inapropriate in some way.
 A: When $A$ is symmetric, one can decompose it as $A = G^TG$ (say via the Cholesky decomposition). One can then write the form as:
$$Dx = x^T Ax = x^T G^TGx = \lVert Gx\rVert_2^2$$

The results regarding minimizing quadratic forms I know always involve the form being over R, but the form in this case is over Z, which should result, I think, in a less difficult optimization.

This is incorrect.
In the above equivalent form of the problem, you want to minimize $Dx = \lVert Gx\rVert_2^2$ where $x\in\mathbb{Z}^2$ (or $\mathbb{Z}^n$ in general).
This is equivalent to solving the Shortest Vector Problem in the lattice
$$\Lambda(G) = \{Gx\mid x\in\mathbb{Z}^n\}$$
The lattice $G$ is often known as a basis (or generating set - I am being imprecise) for the lattice.
The shortest vector problem on a lattice $L$ is to find the shortest non-zero vector $\ell\in L\setminus\{0\}$, e.g. find the $x\in\mathbb{Z}^n\setminus\{0\}$ such that the non-zero value $D(x) = \lVert Gx\rVert_2^2$ is minimized.
This is precisely the question you are asking about (except in the language of lattices, rather than quadratic forms).
There are many things you can say about this setting.
For a few:

The results regarding minimizing quadratic forms I know always involve the form being over $\mathbb{R}$, but the form in this case is over $\mathbb{Z}$, which should result, I think, in a less difficult optimization.

It is still plenty hard --- the problem is NP-hard in general (as the dimension $n$ of the underlying lattice grows asymptotically).

Now, it really seems like the minimum of this form is always achieved in some $(x_1,x_2)$ with $\lVert x_i\rVert \leq 1$, $\forall i=1,2$. In other words, every time I compute $D(x_1,x_2)$ with $\lVert x_i\rVert >1$ for some $i$, the result seems to be always larger than $D(1,0)$ or $D(−1,1)$ or $D(1,1)$. It is definitely doable to prove that by brute force, but is there any result that makes this immediate? For sometimes I encounter similar quadratic forms but with more variables, and the result seems to hold still.

This result definitely does not hold in general.
It is relatively simple to find a counterexample --- let $Dx = \lVert Gx\rVert_2^2$ for some $G$.
Imagine that $D$ is such that there exists unique non-zero $x\in\{0,1\}^n$ such that $D(\pm x) = \lVert Gx\rVert_2^2$ is a minimum.
Then, by choosing a suitable change-of-basis $U\in\mathsf{SL}_n(\mathbb{Z})$, one can write $D'x = \lVert GUx\rVert_2^2$ to write the quadratic form "in a different basis".
The minima of this quadratic form will occur at $\pm U^{-1}x$ for $x\in\{0,1\}^n$ though.
By choosing $U$ correctly, it should be relatively straightforward to ensure that $U^{-1}x\not\in\{-1,0,1\}^n$.
This leads to the natural question --- for every quadratic form/lattice, is there a "good basis" such that the shortest vector can be written as $Bx$ for $x\in\{0,1\}^n$ (or $\{-1,0,1\}^n$)?.
There exist certain classes of lattices for which this is known, specifically a "Voronoi's First Kind" lattice has such a basis (known as an "obtuse superbasis", see definitions in this paper).
VFK lattices are incredibly common in low dimensions (all lattices of rank at most 4 are VFK, this follows from cor. 2.16 of this), but there exist non-VFK lattices as well in high dimensions (one can directly show that a certain family of lattices, namely the "$D_n$ lattices", are not VFK. Alternatively, CVP on VFK lattices is easy, but CVP with preprocessing is hard, so obtuse superbasis cannot exist for all lattices unless $P = NP$).
Of course, being VFK implies your conjecture, but it is not clear that it is equivalent to it.
I would personally suspect that your conjecture fails in large dimensions, but don't have an explicit example in mind.
The above general theory does state that (at a minimum) your conjecture is true (provided the form is reduced appropriately) in up to 4 variables though.
A: Note that
\begin{align}
D(x,y)&=\big(a^2-2b\big)\big(x^2+y^2)+4bxy\\
&=\frac{a^2(x+y)^2}{2}+\frac{\big(a^2-4b\big)(x-y)^2}{2}\ .
\end{align}
If $\ 4b>a^2\ $, then $\ D(x,-x)=-2\big(4b-a^2\big)x^2\ $, the matrix $\ A\ $ is not positive definite, and $\ D\ \ $ has no minimum.
If $\ a^2\ge4b\ $, though, then
\begin{align}
 D(x,y)&\ge 
a^2-2b&\text{for} \ x,y\in\mathbb{Z}\setminus\{0\}\ ,
\end{align}
and
$$
D(1,0)=D(0,1)=D(-1,0)=D(0,-1)=a^2-2b\ .
$$
Thus, when $\ a^2\ge4b\ $ the minimum of $\ D\ $ over $\ \mathbb{Z}\times\mathbb{Z}\setminus\{(0,0)\}\ $ is always achieved at the points $\ (1,0),\ (0,1),\ (-1,0),\ $ and $\ (0,-1)\ $.
A: Let $$f:\mathbb{Z}^n\to \mathbb Z,\quad x\mapsto \frac12 x^T A x -b^T x +c$$ be the function to minimize.
Let $A$ be symmetric positive definite.
If you can find vectors $p_1,...,p_n\in \mathbb Z^n$ with $i\neq j \Rightarrow p_i^T A p_j =0$ so that $\{\sum_{i=1}^n \alpha_i p_i\mid \alpha_i\in\mathbb Z\}=\mathbb Z^n$, then you can use the theory of CG-Method to show that
$$\min_{x\in\mathbb Z} f(x) = \sum_{i=1}^n \min_{\alpha_i\in\mathbb Z} f(\alpha_i p_i)$$
. 
If $x^*$ is the minimizer of $\min_{x\in\mathbb Z} f(x)$ and $\alpha_i^*$ the minimizer of $\min_{\alpha_i\in\mathbb Z} f(\alpha_i p_i)$, then we further have $x^*= \sum_{i=1}^n \alpha_i^* p_i$.
Given that each of the minimization problems $ \min_{\alpha_i\in\mathbb Z} f(\alpha_i p_i)$ is strictly convex, they are simple to solve: Find the global minimum over $\mathbb R$. Now either the next point right or left to it in $\mathbb Z$ is the minimum over $\mathbb Z$..
However, I do not know whether $p_i$ as required always exist, and if they do, how hard it is to find them.
