# A hardly known factorization

If you get the command to factorize *a*^{4} + *b*^{4} 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

*a*^{4} and *b*^{4} are a square each for sure.

We will be brave and make a guess. Imagine that it would work as follows.

(*a*^{2} - *pab* + *b*^{2})(*a*^{2} + *pab* + *b*^{2}). 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 (*a*^{2} + *pab* + *b*^{2}) = *F*.

(*a*^{2} - *pab* + *b*^{2})(*a*^{2} + *pab* + *b*^{2}) =

(*a*^{2} - *pab* + *b*^{2})*F* =

*a*^{2}*F* - *pab**F* + *b*^{2}*F* =

*a*^{2}(*a*^{2} + *pab* + *b*^{2}) - *pab*(*a*^{2} + *pab* + *b*^{2}) + *b*^{2}(*a*^{2} + *pab* + *b*^{2}) =

(*a*^{4} + *pa*^{3}*b* + *a*^{2}*b*^{2}) - (*pa*^{3}*b* + *p*^{2}*a*^{2}*b*^{2} + *pab*^{3}) + (*a*^{2}*b*^{2} + *pab*^{3} + *b*^{4})

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:

*a*^{4} + *a*^{2}*b*^{2} - *p*^{2}*a*^{2}*b*^{2} + *a*^{2}*b*^{2} + *b*^{4} =

*a*^{4} + (2-*p*^{2})*a*^{2}*b*^{2} + *b*^{4}

If we want it equal to *a*^{4}+ *b*^{4} always then should be *p*^{2} = 2.

So the factorizing succeeds if *p* = √2 or if *p* = -√2.

Summary of the termination:

*a*^{4} + *b*^{4} = (*a*^{2} - *ab*√2 + *b*^{2})(*a*^{2} + *ab*√2 + *b*^{2})

To many people this will be a surprise.

## A faster method.

*a*^{4} + *b*^{4} =

*a*^{4} + 2(*ab*)^{2} + *b*^{4} - 2(*ab*)^{2} =

(*a*^{2} + *b*^{2})^{2} - (*ab*√2)^{2} =

(*a*^{2} - *ab*√2 + *b*^{2})(*a*^{2} + *ab*√2 + *b*^{2})

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

## Higher exponents of *a* en *b*

As well is right: *a*^{4n} + *b*^{4n} = (*a*^{2n} - (*ab*)^{n} √2 + *b*^{2n})(*a*^{2n} + (*ab*)^{n} √2 + *b*^{2n})

The *n* is a natural number.