$A^{-1}$ has integer entries if and only if the ${\rm det}\ (A) =\pm 1$

So, $A$ is a $n \times n$ matrix with integer entries. The question is to prove that $A^{-1}$ has all integer entries if and only if ${\rm det}\ (A) =\pm 1$ .

I know that $A^{-1}= {\rm adj}(A)/{\rm det}(A)$ but I have no idea where to go from there for the forward direction. Any help here would be greatly appreciated.

For the backwards direction, I think I am okay. I plugged the 1 and the -1 options into $ {\rm adj}(A)/{\rm det}(A)$ So, I know that $A^{-1}= {\rm adj}(A)$ or $-{\rm adj}(A)$. Then I said that because the ${\rm adj}(A)$ is simply the matrix of co-factors of $A$, and because $A$ has all integer entries then ${\rm adj}(A)$ will have all integer entries, which will mean that $A^{-1}$ will have all integer entries. Is that okay? Or am I missing something else?

Thank you!!``

The direction you've done is correct to a t and is exactly how I'd do it. As for the other direction, you might be tempted to try to reverse your argument but that definitely isn't going to go your way. Here's a hint: suppose that $A^{-1}$ has integer entries, then we know that

$$\det(A^{-1}A) = \det(I) = 1$$

but we also have that $\det(A^{-1}A) = \det(A^{-1})\det(A)$. Can you take it from here?

You did one direction well. Here the other direction:

If $A$ only contains integers then $\det(A)$ is an integer (computing a determinant only uses $+$ and $\cdot$). Assume $A^{-1}$ is also an integer matrix, then $\det(A^{-1})$ is also integer. Then note that

$$\Bbb Z\ni\det(A^{-1})=\frac 1{\det(A)}$$

implies $\det(A)\in\{-1,1\}$ because no other integers then $\pm 1$ have integer reciprocals.