Find all integers $a$ and $b$ satisfying
\begin{align*}
a^6 + 1 & \mid b^{11} - 2023b^3 + 40b, \qquad \text{and}\\
a^4 - 1 & \mid b^{10} - 2023b^2 - 41.
\end{align*}
1. We start with the given conditions:
   \[
   a^6 + 1 \mid b^{11} - 2023b^3 + 40b
   \]
   \[
   a^4 - 1 \mid b^{10} - 2023b^2 - 41
   \]

2. Note that \(a^2 + 1 \mid a^6 + 1\) and \(a^2 + 1 \mid a^4 - 1\). This is because:
   \[
   a^6 + 1 = (a^2)^3 + 1 \quad \text{and} \quad a^4 - 1 = (a^2)^2 - 1
   \]
   By the factor theorem, \(a^2 + 1\) divides both \(a^6 + 1\) and \(a^4 - 1\).

3. Therefore, \(a^2 + 1\) must also divide both:
   \[
   b^{11} - 2023b^3 + 40b
   \]
   \[
   b^{10} - 2023b^2 - 41
   \]

4. Multiply the second expression by \(b\):
   \[
   b \cdot (b^{10} - 2023b^2 - 41) = b^{11} - 2023b^3 - 41b
   \]

5. Subtract this from the first expression:
   \[
   b^{11} - 2023b^3 + 40b - (b^{11} - 2023b^3 - 41b) = 81b
   \]
   Thus, \(a^2 + 1 \mid 81b\).

6. Notice that \(3 \nmid a^2 + 1\) because \(a^2 + 1\) is always greater than 1 and not divisible by 3. Therefore, \(a^2 + 1 \mid b\).

7. Since \(a^2 + 1 \mid b\), we can substitute \(b = k(a^2 + 1)\) for some integer \(k\).

8. Substitute \(b = k(a^2 + 1)\) into the second condition:
   \[
   a^2 + 1 \mid (k(a^2 + 1))^{10} - 2023(k(a^2 + 1))^2 - 41
   \]
   Simplifying, we get:
   \[
   a^2 + 1 \mid k^{10}(a^2 + 1)^{10} - 2023k^2(a^2 + 1)^2 - 41
   \]
   Since \(a^2 + 1\) divides both terms involving \(k\), it must also divide 41.

9. Since 41 is a prime number, the only divisors of 41 are 1 and 41. Therefore, \(a^2 + 1 = 1\) or \(a^2 + 1 = 41\).

10. If \(a^2 + 1 = 1\), then \(a^2 = 0\) which implies \(a = 0\).

11. If \(a^2 + 1 = 41\), then \(a^2 = 40\). However, 40 is not a perfect square, so there are no integer solutions for \(a\) in this case.

12. Therefore, the only solution is \(a = 0\). For \(a = 0\), the conditions reduce to:
    \[
    1 \mid b^{11} - 2023b^3 + 40b
    \]
    \[
    0 \mid b^{10} - 2023b^2 - 41
    \]
    which are trivially satisfied for any integer \(b\).

The final answer is \( \boxed{ (a, b) = (0, b) } \) where \(b\) is any integer.