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I have the feeling that I'm missing something very obvious: I'm looking for a counterexmple for the following statement for some $n>1$ (it is trivially true for $n=1$):

Let $A,B\in\mathbb Z^{n\times n}$ symmetric with determinant $\pm 1$ such that there exists an invertible matrix $C\in\mathbb Q^{n\times n}$ with $C^{-1}AC=B$. Then there exists $S\in\mathbb Z^{n\times n}$ with determinant $\pm 1$ such that $S^{-1}AS=B$.

The motivation behind it is that I want to give an example for two quadratic integral forms of discriminant $\pm 1$ that are not equivalent as integral forms but as rational forms. This then gives that the notion of integral form equivalence is "finer" than that of rational form equivalence (a statement from Serre's book "A course in arithmetic" Chapter 5, 1.1).

It may also be that my approach is clumsy, then I would be very thankful for inspiration for a different one.

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You mean $C^T$ and $S^T$, not $C^{-1}$ and $S^{-1}$. – David Speyer Oct 27 '14 at 23:53

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