The condition of a matrix $Q\in \mathbb{R}^{n\times n}$ to be positive definite is that $x^TQx>0$ for all $x\neq 0$.
I would like to know if the following is an equivalent condition: $x^TQx>0$ for all $0\neq x\in \mathbb{Z}^n$ and $Q$ is invertible.
It is necessary to include the condition that $Q$ is invertible since $\begin{bmatrix} 1 & -\pi \\ -\pi & \pi^2 \end{bmatrix}$ is a $2$-dimensional example of a matrix that has $x^TQx>0$ over $\mathbb{Z}$ but is clearly not positive definite in the real sense.
Does somebody know of a conterexample that shows that my "equivalent" does not hold?
Yes, the condition $x^TQx > 0$ for all non-zero $x \in \Bbb Z^n$ is sufficient.
On proof is as follows. Consider an arbitrary $x \in \Bbb R^n$. Let $x_k$ be a sequence in $\Bbb Q^n$ with $x_k \to x$. For any $x_k$, there exists an intger $N>0$ such that $Nx_k \in \Bbb Z^n$ (for instance, take $N$ to be the least common denominator among all entries of $x$). With that, we have $$ (Nx_k)^TQ(Nx_k) > 0 \implies x_k^TQx_k = \frac{(Nx_k)^TQ(Nx_k)}{N^2} > 0. $$ So, $x_k^TQx_k \geq 0$ for all $k$, and by the continuity of $x \mapsto x^TQx$ we have $x_k^TQx_k \to x^TQx$. It follows that $x^TQx \geq 0$.
Because $Q$ is invertible and $x^TQx \geq 0$ for all $x$, we can conclude that $Q$ is positive definite, as was desired.