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Let A be an nxn matrix with integer entries, and determinant -1. Let b be an n-vector with integer entries. Then the solution of Ax = b is necessarily a vector with integer entries. ex. ...,-3, -2, -1, 0, 1, 2, 3,...

How can I prove this either to be true or false? Would a proof by contradiction work best?

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$-1$ is invertible in $\mathbb Z$. There is a formular of the inverse matrix in terms of the "adjugate matrix": $A^{-1}=\frac 1 {\text{det}(A)} A^{ad}$.

You obtain $A^{ad}$ by calculating the determinants of submatrices of A. Hence $A^{ad}$ too consists of integer entries. But then $A^{-1}$ consists of integer entries, so the solution $x$ must too.

Sincerely, slinshady

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The best proof would be a constructive one. Try to explicitly find the solution (in terms of $A $ and $b $) and show it has only integer values.

It may be of interest to note that $A $ has determinant $-1$. What does that mean?

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