1. **Problem statement:** Factorise $a^4 + 4b^4$ by adding and subtracting $4a^2 b^2$. 2. **Step 1: Add and subtract $4a^2 b^2$ inside the expression:** $$a^4 + 4b^4 = a^4 + 4a^2 b^2 - 4a^2 b^2 + 4b^4$$ 3. **Step 2: Group terms to form a perfect square and a difference:** $$= (a^4 + 4a^2 b^2 + 4b^4) - 4a^2 b^2$$ 4. **Step 3: Recognize the perfect square:** $$= (a^2 + 2b^2)^2 - (2ab)^2$$ 5. **Step 4: Apply the difference of squares formula $x^2 - y^2 = (x - y)(x + y)$:** $$= \bigl(a^2 + 2b^2 - 2ab\bigr) \bigl(a^2 + 2b^2 + 2ab\bigr)$$ 6. **Step 5: Rewrite the factors in a more standard form:** $$= (a^2 - 2ab + 2b^2)(a^2 + 2ab + 2b^2)$$ --- 7. **Part (b) Prove that $545^4 + 4^{545}$ is not prime:** 8. **Step 1: Substitute $a = 545$ and $b = 2^{\frac{545}{2}}$ (since $4^{545} = (2^2)^{545} = 2^{1090}$):** Note: We use $b^4 = 4^{545}$ so $b = 2^{\frac{1090}{4}} = 2^{272.5}$ but since the factorization is algebraic, we keep $b$ symbolic. 9. **Step 2: Using the factorization from part (a):** $$545^4 + 4^{545} = (545^2 - 2 \cdot 545 \cdot b + 2b^2)(545^2 + 2 \cdot 545 \cdot b + 2b^2)$$ 10. **Step 3: Since both factors are integers greater than 1, the number is composite, hence not prime.** **Final answer:** $$a^4 + 4b^4 = (a^2 - 2ab + 2b^2)(a^2 + 2ab + 2b^2)$$ and $$545^4 + 4^{545}$$ is not prime because it factors as above.