Factorization of $112^4+1$

Question

So I read answers like this one For which $n$ the polynomial $x^n + y^n$ will be irreducible over $\mathbb Z$?

and this one:

Factoring $x^n + y^n$ over the integers

They seem to imply (correctly) that $x^{2^k}+1$ could not be factorised. I'm just wondering how one could explain the factorization of $$112^4+1 = (2^4*7)^4+1= (112*418+1)(112*30+1)$$.

Is there a way to factorize $112^4+1$ using like a cube facor formula (which obviously cannot be used here) or something else?

Answer 1

As $x^4+1$ is irreducible over $\mathbb{Z}[x]$, this factorization doesn't generalize to all values of $x^4+1$. But it does to some:

This factorization is based on the observation that

$$3\cdot 112^2+4=194^2$$

(in other words, $3x^2+4=y^2$ has a solution at $x=112$; this is a Pell equation and as such there are infinitely many such pairs $(x,y)$ of integers.) So, we have, letting $x=112,y=194$,

$$x^4+1=(4x^4+4x^2+1)-(3x^4+4x^2)=(2x^2+1)^2-x^2(3x^2+4);$$

as $3x^2+4=y^2$ this reduces to

$$x^4+1=(2x^2+1)-(xy)^2= (x(2x-y)+1)(x(2x+y)+1),$$

which is the factorization you gave.

Answer 2

When they say expression $x^4+1$ does not factor over rational field they mean we can not find polynomials with rational coefficients to divide $x^4+1$

The story of integer factorization is a different story because an integer could be written in many different forms. For example $ 82= 3^4+1$ and $ 82=2(41)$ so it does not factor in one form but it does in another form. similarly $5^4 +1 = 626 = 2(313)$