A (real) quadratic form is a homogeneous plynomial of degree 2 (with real cofficients), in any number of variables. A quadratic form is positive definite if it is takes only nonnegative values. Now assume that $Q$ is a positive definite quadratic form of $n$ variables, then what I want to show is that $Q(\mathbb{Z}^n)$ is dicrete.
My idea is to assume that $Q(\mathbb{Z}^n)$ is not discrete then for any $\epsilon>0$ we can always find two distinct points $x,y\in\mathbb{Z}^n$ such that $0<|Q(x)-Q(y)|<\epsilon$ and then use this to make a contradiction with the fact that $Q$ only takes nonnegative value. But I do not know how to proceed. Thanks for any help.
Hint. Let $\lambda=\lambda_\min(Q)$. Then $\lambda\|v\|^2 \le Q(v)$ for every vector $v$. If some sequence $\{Q(y_n)\}$ converges to some $Q(x)$, eventually we must have $Q(y_n)\le 2\max(1,Q(x))$ and in turn $$ \|y_n\|^2 \le \frac{2\max(1,Q(x))}{\lambda}. $$ How many integer vectors are there inside any sphere?