1. **Stating the problem:** We want to factorize the expression $$x^4 + 4y^4$$. 2. Notice that $$x^4 + 4y^4$$ is a sum of squares: $$x^4 + (2y^2)^2$$. 3. Use the Sophie Germain identity for sum of fourth powers: $$a^4 + 4b^4 = (a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab)$$. 4. Substitute $$a = x$$ and $$b = y$$ in the identity: $$x^4 + 4y^4 = (x^2 + 2y^2 - 2xy)(x^2 + 2y^2 + 2xy)$$. 5. Both factors are quadratic expressions that cannot be factored further over the real numbers easily. **Final answer:** $$x^4 + 4y^4 = (x^2 + 2y^2 - 2xy)(x^2 + 2y^2 + 2xy)$$.