Factorize $a^{4}+4 b^{4}$ (this is the Sophie Germain identity)

Hint: we have $a^{4}+4 b^{4}=a^{4}+4 b^{4}+4 a^{2} b^{2}-4 a^{2} b^{2}$.
We recognize one after another two remarkable identities (it seems it was done on purpose): $a^{4}+4 b^{4}=a^{4}+4 b^{4}+4 a^{2} b^{2}-4 a^{2} b^{2}=\left(a^{2}+2 b^{2}\right)^{2}-(2 a b)^{2}=\left(a^{2}+2 a b+2 b^{2}\right)\left(a^{2}-2 a b+2 b^{2}\right)$.

Here we can see that sometimes we need to make the missing terms "appear" so that we can factorize as needed, and that is where the difficulty of such problems lies. Another example for the road:

## Example 5.

$a^{3}-b^{3}=a^{3}+a^{2} b+b^{2} a-a^{2} b-b^{2} a-b^{3}=a\left(a^{2}+a b+b^{2}\right)-b\left(a^{2}+a b+b^{2}\right)=(a-b)\left(a^{2}+a b+b^{2}\right)$

By the way... We have $a^{1}-b^{1}=(a-b)(1), a^{2}-b^{2}=(a-b)(a+b), a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$, isn't there a pattern here?

The answer is yes (wow!) :

## Proposition 6.

$a^{n}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2} b+a^{n-3} b^{2}-\ldots+a^{2} b^{n-3}+b^{n-2} a+b^{n-1}\right)$

Proof. It suffices to expand the right-hand side, everything simplifies magically!

To show that it is indeed correct, it suffices to start from the right-hand side and expand, everything simplifies magically!

If we now assume in addition that $n$ is odd, we can replace $b$ by $-b$ in the previous formula to get

$a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2} b+a^{n-3} b^{2}-\ldots+a^{2} b^{n-3}-b^{n-2} a+b^{n-1}\right)$.