How do we come up with the idea in the proof of $x^4+1$ is irreducible over $\Bbb Z$ by tranfer into $(x+1)^4+1$?

Question

$x^4+1$ is irreducible over $\Bbb Z$ (hence $\Bbb Q$). The proof I saw is to try transforming it into $(x+1)^4+1=x^4+4x^3+6x^2+4x+2$ and use the Eisenstein criterion with prime $2$. I can understand the proof. However, how do we come up with this idea? How do we know we should try to transfer $x$ into $x+1$, rather than, say $3x-5$? And by the way, I also wonder is there other way to proof the irreducibility?

Answer 1

This is a trick that one sees over and over again, and perhaps in a year or two, you’ll see it as perfectly natural.

However, let me explain how I see it, from a $p$-adic (in this case, $2$-adic) viewpoint. The roots of $f(X)=X^4+1$ are all close to $1$ in the $2$-adic sense, namely for a root $\rho$ of $f$, $\vert1-\rho\vert_2<1$, while they are not close to zero: $\vert\rho-0\vert_2=\vert\rho\vert_2=1$. The Eisenstein criterion deals exclusively with small roots of a polynomial, so if we have any hope of using it, we need to look not at the numbers $\rho$, but at the numbers $\rho-1$. It’s $f(X+1)$ that has these for its roots.

Now, I’m guessing that at your stage of mathematical experience, you haven’t seen the $p$-adic numbers or the $p$-adic absolute value on the rationals. Maybe as you get more experienced, you’ll see another justification for the trick, and maybe you’ll see things through a lens similar to mine. Throughout mathematics, there are many ways to look at a phenomenon.

Answer 2

Another way: the polynomial $x^4+1$ doesn't have an integral root, so $x^4+1 \ne (x-a)(x^3+bx^2+cx+d)$.

we assume that $x^4+1=(x^2+ax+b)(x^2+cx+d)$, then $a+c=0, ac+b+d=0, ad+bc=0, bd=1$, and so $x^4+1=(x^2-\sqrt2 x+1)(x^2+\sqrt2 x+1)$. i.e. $x^4+1$ is irreducible over $\mathbb Z$.