Show that a polynomial $f(x)$ over a field $k$ is irreducible if the polynomial $f(x+1)$ is irreducible

Question

I was thinking of using contradiction by assuming that that $f(x)$ is reducible but I really don't know how to continue from that idea.

Answer 1

Assume that $f(x)$ is reducible. Then:

$\exists \ h(x), \ g(x) \in \mathbb K[x]$ such that $f(x) = g(x)h(x)$.

Then: $f(x + 1) = g(x+1)h(x+1)$, and so reducible. A contradiction.

Answer 2

Let's prove the contrapositive: That if $f(x)$ is reducible then $g(x) \equiv f(x+1)$ is also reducible. That is logically equivalent to the statement to be proven.

Since $f(x)$ is reducible there exists some $x_0\in K:f(x_0)=0$. Then since $1 \in K$ and $K$ is a field, $y\equiv x_0-1 \in K$. And $$ g(y) = f((x_0-1)+1) = f(x_0) = 0 $$ so $g(x)$ is reducible.

EDITED in response to Bill's comment:

This "proof" is invalid. Yes, if $f(x)$ has a zero then $g(x)$ is irreducible. But we don't know that $f(x)$ has a zero; all we know is that it is irreducible. So "step 1.5" in the above proof is hogwash. Sorry.