1. **Problem Statement:** Given that $[1 + i]^n = x + iy$ where $x$ and $y$ are real numbers and $n$ is an integer, prove that $x^2 + y^2 = 2^n$. 2. **Recall the formula:** For any complex number $z = a + bi$, its magnitude (or modulus) is given by $|z| = \sqrt{a^2 + b^2}$. 3. **Important rule:** The magnitude of a product of complex numbers equals the product of their magnitudes, i.e., $|z_1 z_2| = |z_1| |z_2|$. 4. **Calculate the magnitude of $1 + i$:** $$|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$$ 5. **Calculate the magnitude of $[1 + i]^n$:** Using the property of magnitudes, $$|[1 + i]^n| = |1 + i|^n = (\sqrt{2})^n = 2^{\frac{n}{2}}$$ 6. **Express $[1 + i]^n$ as $x + iy$:** Given $[1 + i]^n = x + iy$, the magnitude is $$|x + iy| = \sqrt{x^2 + y^2}$$ 7. **Equate magnitudes:** Since magnitudes must be equal, $$\sqrt{x^2 + y^2} = 2^{\frac{n}{2}}$$ 8. **Square both sides:** $$x^2 + y^2 = \left(2^{\frac{n}{2}}\right)^2 = 2^n$$ **Final answer:** $$\boxed{x^2 + y^2 = 2^n}$$ This proves the required identity.