i thought that this problem was in mathlinks but when i searched i didn't find it.so here it is:
Find all positive integers m for which for all $\alpha,\beta \in \mathbb{Z}-\{0\}$
\[ \frac{2^m \alpha^m-(\alpha+\beta)^m-(\alpha-\beta)^m}{3 \alpha^2+\beta^2} \in \mathbb{Z} \]
1. **Define the polynomial \( P(x) \):**
   \[
   P(x) := 2^m \alpha^m - (\alpha + x)^m - (\alpha - x)^m
   \]
   We need to show that \( P(x) \) is divisible by \( 3\alpha^2 + x^2 \).

2. **Evaluate \( P(x) \) at \( x = i\sqrt{3}\alpha \):**
   \[
   P(i\sqrt{3}\alpha) = 2^m \alpha^m - (\alpha + i\sqrt{3}\alpha)^m - (\alpha - i\sqrt{3}\alpha)^m
   \]
   Simplify the terms inside the polynomial:
   \[
   \alpha + i\sqrt{3}\alpha = \alpha(1 + i\sqrt{3})
   \]
   \[
   \alpha - i\sqrt{3}\alpha = \alpha(1 - i\sqrt{3})
   \]
   Therefore,
   \[
   P(i\sqrt{3}\alpha) = 2^m \alpha^m - (\alpha(1 + i\sqrt{3}))^m - (\alpha(1 - i\sqrt{3}))^m
   \]

3. **Express the complex numbers in exponential form:**
   \[
   1 + i\sqrt{3} = 2 \left( \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right) \right) = 2 e^{i\pi/3}
   \]
   \[
   1 - i\sqrt{3} = 2 \left( \cos\left(\frac{\pi}{3}\right) - i \sin\left(\frac{\pi}{3}\right) \right) = 2 e^{-i\pi/3}
   \]
   Thus,
   \[
   (\alpha(1 + i\sqrt{3}))^m = (\alpha \cdot 2 e^{i\pi/3})^m = \alpha^m \cdot 2^m \cdot e^{im\pi/3}
   \]
   \[
   (\alpha(1 - i\sqrt{3}))^m = (\alpha \cdot 2 e^{-i\pi/3})^m = \alpha^m \cdot 2^m \cdot e^{-im\pi/3}
   \]

4. **Substitute back into \( P(i\sqrt{3}\alpha) \):**
   \[
   P(i\sqrt{3}\alpha) = 2^m \alpha^m - \alpha^m \cdot 2^m \cdot e^{im\pi/3} - \alpha^m \cdot 2^m \cdot e^{-im\pi/3}
   \]
   Factor out \( 2^m \alpha^m \):
   \[
   P(i\sqrt{3}\alpha) = 2^m \alpha^m \left( 1 - e^{im\pi/3} - e^{-im\pi/3} \right)
   \]

5. **Simplify using Euler's formula:**
   \[
   e^{im\pi/3} + e^{-im\pi/3} = 2 \cos\left(\frac{m\pi}{3}\right)
   \]
   Therefore,
   \[
   P(i\sqrt{3}\alpha) = 2^m \alpha^m \left( 1 - 2 \cos\left(\frac{m\pi}{3}\right) \right)
   \]

6. **Set \( P(i\sqrt{3}\alpha) = 0 \):**
   \[
   2^m \alpha^m \left( 1 - 2 \cos\left(\frac{m\pi}{3}\right) \right) = 0
   \]
   Since \( 2^m \alpha^m \neq 0 \), we must have:
   \[
   1 - 2 \cos\left(\frac{m\pi}{3}\right) = 0
   \]
   \[
   \cos\left(\frac{m\pi}{3}\right) = \frac{1}{2}
   \]

7. **Solve for \( m \):**
   \[
   \cos\left(\frac{m\pi}{3}\right) = \frac{1}{2} \implies \frac{m\pi}{3} = \pm \frac{\pi}{3} + 2k\pi \quad \text{for some integer } k
   \]
   \[
   m = \pm 1 + 6k
   \]
   Therefore,
   \[
   m \equiv \pm 1 \pmod{6}
   \]

8. **Verify the converse:**
   If \( m \equiv \pm 1 \pmod{6} \), then \( P(x) \) is divisible by \( 3\alpha^2 + x^2 \). This can be checked by substituting back into the original polynomial and verifying the divisibility.

The final answer is \( \boxed{ m \equiv \pm 1 \pmod{6} } \).