Suppose $f(x)$ is an polynomial of integer coefficients. If for infinitely many integers $x$, $f(x)$ is prime. Show that $f(x)$ is irreducible in $\Bbb Q[x]$.

Suppose $f(x)$ is is reducible in $\Bbb Q[x]$, then what...

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    $\begingroup$ 1. Gauss's Lemma. 2. Zircht's Answer. 3. If a polynomial $f$ with coefficients in $\mathbb{Z}$ vanishes at infinitely many points, then $f=0$. 4. Apply step 3 to either $g(x)-1$ or $h(x)-1$ in Zircht's answer. $\endgroup$ – Prism Dec 16 '14 at 3:09

Then $f(x)=g(x)h(x)$ where $\deg g,\deg h< \deg f$. If $f(x)$ is a prime for infinitely many integers $x$, then either $g(x)$ or $h(x)$ is $1$ for infinitely many $x$, so...


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