#ProofInAToot!

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.

Follow

@JordiGH sorry, can you point me to the question or the statement this is trying to prove. Sorry if this is too noobish :)

· · Web · 1 · 0 · 0

@j605 The final statement, that the values of a polynomial over the integers are divisible by infinitely many primes.

Sign in to participate in the conversation
ieji.de

The social network of the future: No ads, no corporate surveillance, ethical design, and decentralization! Own your data with Mastodon!