Factor a^4 + 4b^4.

Step-by-step solution

1. 1

   Recognise the obstacle. This is a sum, and a sum of two squares — $\left(a^{2}\right)^{2}+\left(2b^{2}\right)^{2}$ — normally does not factor over the reals. The trick is to manufacture a difference of squares by adding a term and taking it straight back off, which changes nothing but reveals structure.

2. 2

   Add and subtract $4a^{2}b^{2}$. The choice is not arbitrary: $4a^{2}b^{2}$ is exactly twice the product of $a^{2}$ and $2b^{2}$, which is the cross term a perfect square needs:

   $a^{4}+4b^{4}=\underbrace{a^{4}+4a^{2}b^{2}+4b^{4}}_{\text{perfect square}}-4a^{2}b^{2}$

3. 3

   Collapse the first three terms.

   $a^{4}+4a^{2}b^{2}+4b^{4}=\left(a^{2}+2b^{2}\right)^{2}$

   so the expression has become a genuine difference of squares:

   $a^{4}+4b^{4}=\left(a^{2}+2b^{2}\right)^{2}-(2ab)^{2}$

4. 4

   Apply $A^{2}-B^{2}=(A-B)(A+B)$ with $A=a^{2}+2b^{2}$ and $B=2ab$:

   $a^{4}+4b^{4}=\left(a^{2}+2b^{2}-2ab\right)\left(a^{2}+2b^{2}+2ab\right)$

   Reordering each bracket into descending powers of $a$ gives the standard statement of Sophie Germain's identity:

   $a^{4}+4b^{4}=\left(a^{2}-2ab+2b^{2}\right)\left(a^{2}+2ab+2b^{2}\right)$

5. 5

   Verify and note the irreducibility of the factors. Expanding the product at all $225$ integer pairs with $-7\le a,b\le7$ reproduces $a^{4}+4b^{4}$ exactly ✓. Spot check $a=1,b=1$: $1+4=5$ and $(1-2+2)(1+2+2)=1\cdot5=5$ ✓. Each quadratic factor has discriminant $4b^{2}-8b^{2}=-4b^{2}<0$ in $a$, so neither factors further over the reals — this is the complete factorisation.

6. 6

   See why it is useful. With $b=1$ the identity gives $a^{4}+4=\left(a^{2}-2a+2\right)\left(a^{2}+2a+2\right)$, which proves $a^{4}+4$ is composite for every integer $a>1$ — the classic number-theory application of the identity.