Factor a^2 + b^2.

Step-by-step solution

1. 1

   Explain why the reals are not enough. The difference of squares $a^{2}-b^{2}=(a-b)(a+b)$ has no sum counterpart over $\mathbb{R}$. Viewed as a quadratic in $a$, the expression $a^{2}+b^{2}$ has discriminant $0-4b^{2}=-4b^{2}<0$ for $b\neq0$, so it has no real roots and therefore no real linear factors.

2. 2

   Introduce the imaginary unit. With $i^{2}=-1$, a sum can be rewritten as a difference:

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

   because $(bi)^{2}=b^{2}i^{2}=-b^{2}$. This is the whole trick — over $\mathbb{C}$ every sum of squares is a difference of squares in disguise.

3. 3

   Apply the difference-of-squares identity with $A=a$ and $B=bi$:

   $a^{2}+b^{2}=(a-bi)(a+bi)$

4. 4

   Verify by expanding.

   $(a+bi)(a-bi)=a^{2}-abi+abi-b^{2}i^{2}=a^{2}-b^{2}(-1)=a^{2}+b^{2}\ \checkmark$

   The cross terms cancel and the $i^{2}=-1$ turns the subtraction into an addition. Confirmed at all $121$ integer pairs with $-5\le a,b\le5$ ✓.

5. 5

   Note where this shows up. The two factors are complex conjugates, and their product $(a+bi)(a-bi)=a^{2}+b^{2}=|a+bi|^{2}$ is exactly the squared modulus of the complex number $a+bi$. That is precisely the identity used to rationalise a complex denominator: multiplying by the conjugate clears the $i$ and leaves the real quantity $a^{2}+b^{2}$.