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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