Let $A$ and $B$ be 2 × 2 matrices with integer entries such that $A$, $A + B$, $A + 2B$, $A + 3B$, and $A + 4B$ are all invertible with integral entries. I want to prove that $A + tB$ is invertible and that its inverse has integral entries, assuming that $t ∈ Z$.
I started by proving that det$(A) = \pm 1$. Let matrix $A = [a_{ij}]$. Since $A$ and $A^{-1}$ both have integral values, det$(A)$ and det$(A^{-1})$ must both be integers. Also, $\det(AA^{-1})=1$. Therefore, $\det(A)\det(A^{-1})=1$. We already established that det$(A^{-1})$ had to be an integer. Thus, the only possible solutions for $\det(A)$ are $\pm 1$.
Now det$(A + tB) = a + tc + t^2b$ where $a =$ det$(A)$, $c = (a_{11}b_{22} + a_{22}b{11}) - (a_{12}b_{21} - a_{21}b_{12})$, and $b =$ det$(B)$. From the problem statement, since $A$, $A + B$, $A + 2B$, $A + 3B$, and $A + 4B$ all have integral entries, it follows that: $$a = \pm 1$$ $$ a + c + b = \pm1$$ $$ a + 2c + 4b = \pm1$$ $$ a + 3c + 9b = \pm1$$ $$ a + 4c + 16b = \pm1$$
I'm not quite sure how to proceed to complete the proof. Any guidance is much appreciated!
Edit: Solving the system of equations yields: $$a = \pm 1, b = 0, c = 0$$
Thus, det$(A + tB) = a + tc + t^2b = a =$ det$(A)$ for any integer $t$ and $A + tB$ is invertible and has integer entries for all integers $t$.
(With the assumption that $ (A + tB)^{-1}$ must all have integer entries for $ t = 0, 1, 2, 3, 4$.)
(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)
Note: You forgot $ a = \pm 1$, so we have 5 equations. This is crucial.
Hint: Using the Pigeonhole principle, conclude that $ b = c = 0$.
Corollary: Hence, $det(A + tB) = \pm 1$, and so we are done.