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Is there an easy way to see the following:

Suppose Q is an integral quadratic form in $n$ variables that is positive definite, that is $Q(x) \geq 1$ for all $0 \neq x \in \mathbb{Z}^n$. Then the number of solutions to the equation $Q(x)=m$, for some fixed $n \in \mathbb{N}$, is finite.

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1 Answer 1

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Suppose that $Q$ is represented by an $n \times n$ matrix that I will by abuse of notation refer to as $Q$, and that $Q$ as a positive definite matrix satisfies a "coercivity" condition:

$$ x^T Q x \ge c x^T x $$

for some constant $c \gt 0$.

Then the existence of at most finitely many solutions to $Q(x) = x^T Q x = m$ follows from the existence of at most finitely many $x$ such that $c x^T x \le m$, counting on the integer components of $x$.

Added: The basic definition of an integer n-ary quadratic form is a homogeneous of degree 2 polynomial $Q(x_1,\ldots,x_n) = \sum_{i \le j} a_{ij}x_i x_j$. As alluded to this can be expressed in matrix form $x^T Q x$ where $x$ is a column vector $(x_1,\ldots,x_n)$, whereby the off-diagonal entries become $\frac{1}{2} a_{ij}$ for $i \neq j$ to obtain real symmetric matrix $Q$ (indeed $Q$ is rational).

Now real symmetric matrices have a complete set of eigenvalues and eigenvectors, and the task is to show that quadratic form $Q(x) \geq 1$ for integer vectors implies matrix $Q$ is positive definite in the sense that all eigenvalues are bounded below by positive constant $c \gt 0$, from which the coercivity property is immediate.

We really just need, since the number of eigenvalues of $Q$ is finite, to eliminate the possibility of negative and/or zero eigenvalues. If there were negative eigenvalues, then taking $x$ to be an appropriate rational approximation to a corresponding eigenvector would produce a negative value for $Q(x)$, and by multiplying through by any denominator, a corresponding integer $x$ that also gives a negative value for the quadratic form.

The case of zero eigenvalues is even simpler to dispose of. If matrix $Q$ were to have a nontrivial nullspace, then $Qx = 0$ would have rational nontrivial solutions (since $Q$ is rational), and again some integer multiple of such a solution $x$ would give $Q(x) = 0$ over the integer vectors, contradicting the asserted property of $Q(x) \ge 1$.

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Could you explain this coercivity a little bit, I don't see how $Q(x) \geq 1$ implies that your equation holds. –  Boris Datsik Jun 14 '13 at 13:40
    
Sure, I'll add more details. –  hardmath Jun 14 '13 at 15:24

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