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Consider a nontrivial polynomial over the integers \(f(x) = xg(x) + a\) with nonzero \(a\) and \(g\). Let \(n\) be larger than any root of \(g\) and for any given prime \(p\), evaluate

\[
f(anp!) = a(np!g(anp!) + 1).
\]

The value \(np!g(anp!) + 1\) is not divisible by any prime less than \(p\). Therefore, all of its divisors must be larger, and thus the integer values of a nontrivial polynomial over the integers are divisible by infinitely many primes.

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(I've seen other proofs of this that seem more complicated, so I feel like I may have made a dumb mistake, please point it out if you see it.)

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