Find all integer values can take $n$ such that $$\cos(2x)=\cos^nx - \sin^nx$$ for every real number $x$.
1. We start with the given equation:
   \[
   \cos(2x) = \cos^n x - \sin^n x
   \]
   for every real number \( x \).

2. To find the integer values of \( n \), we first set \( x = -\frac{\pi}{4} \). At this value, we have:
   \[
   \cos\left(2 \left(-\frac{\pi}{4}\right)\right) = \cos\left(-\frac{\pi}{2}\right) = 0
   \]
   and
   \[
   \cos\left(-\frac{\pi}{4}\right) = \sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}
   \]
   Therefore,
   \[
   \cos^n\left(-\frac{\pi}{4}\right) - \sin^n\left(-\frac{\pi}{4}\right) = \left(-\frac{\sqrt{2}}{2}\right)^n - \left(-\frac{\sqrt{2}}{2}\right)^n = 0
   \]
   This implies that \( n \) must be even, so let \( n = 2m \).

3. Next, we set \( x = 0 \):
   \[
   \cos(2 \cdot 0) = \cos(0) = 1
   \]
   and
   \[
   \cos^n(0) - \sin^n(0) = 1^n - 0^n = 1
   \]
   This is valid for any \( n \geq 1 \).

4. Now, we need to solve the equation for \( n = 2m \):
   \[
   \cos^{2m} x - \sin^{2m} x = \cos^2 x - \sin^2 x
   \]
   for all \( x \). Let \( y = \cos^2 x \), then \( \sin^2 x = 1 - y \). The equation becomes:
   \[
   y^m - (1 - y)^m = 2y - 1
   \]
   for all \( y \in [0, 1] \). Since this is a polynomial equation, it must hold for all \( y \in \mathbb{R} \).

5. Setting \( y = 2 \):
   \[
   2^m - (1 - 2)^m = 2^m - (-1)^m = 2 \cdot 2 - 1 = 3
   \]
   This simplifies to:
   \[
   2^m = 3 + (-1)^m
   \]
   If \( m \geq 3 \), then \( 2^m \geq 8 \), which is greater than 4, so \( m \) must be 1 or 2.

6. Checking \( m = 1 \):
   \[
   2^1 = 3 + (-1)^1 \implies 2 = 2
   \]
   This is true.

7. Checking \( m = 2 \):
   \[
   2^2 = 3 + (-1)^2 \implies 4 = 4
   \]
   This is also true.

8. Therefore, the possible values of \( n \) are \( 2 \) and \( 4 \).

The final answer is \( \boxed{n \in \{2, 4\}} \).