﻿A hardly known factorization

# A hardly known factorization

If you get the command to factorize a4 + b4 most people with a little sense of mathematics tell you
that it cannot be done because the sum of two squares cannot be factorized and
a4 and b4 are a square each for sure.

We will be brave and make a guess. Imagine that it would work as follows.
(a2 - pab + b2)(a2 + pab + b2). Here p is a number to be determined.
Now we hope a number of terms will cancel each other out.

For the moment we call the second factor F. Then (a2 + pab + b2) = F.
(a2 - pab + b2)(a2 + pab + b2) =
(a2 - pab + b2)F =
a2F - pabF + b2F =
a2(a2 + pab + b2) - pab(a2 + pab + b2) + b2(a2 + pab + b2) =
(a4 + pa3b + a2b2) - (pa3b + p2a2b2 + pab3) + (a2b2 + pab3 + b4)

The F was not necessary but it could help.
Many brackets were unnecessary. However, they were placed for clarity.
The last nine-term we can simplify:

a4 + a2b2 - p2a2b2 + a2b2 + b4 =
a4 + (2-p2)a2b2 + b4

If we want it equal to a4+ b4 always then should be p2 = 2.
So the factorizing succeeds if p = √2 or if p = -√2.

Summary of the termination:
a4 + b4 = (a2 - ab√2 + b2)(a2 + ab√2 + b2)
To many people this will be a surprise.

## A faster method.

a4 + b4 =
a4 + 2(ab)2 + b4 - 2(ab)2 =
(a2 + b2)2 - (ab√2)2 =
(a2 - ab√2 + b2)(a2 + ab√2 + b2)

On this site is not discussed, whether the found two factors can be factorized.

## Higher exponents of a en b

As well is right: a4n + b4n = (a2n - (ab)n √2 + b2n)(a2n + (ab)n √2 + b2n)
The n is a natural number.