Let $P$ be an integer valued polynomial with degree $> 1$, is it a theorem that the values $P(n)$ have arbitrarily large prime factors?
polynomial values take on arbitrarily large prime factors?
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1 Answers
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Yes suppose that all values of $P(n)$ are divisible by the prime numbers $p_1,p_2,...,p_r$ only. Now consider $P(P(0)p_1p_2...p_r)$, if $P(0)$ is different from zero we find that none of the primes $p_1,p_2,...,p_r$ divides $P(P(0)p_1p_2...p_r)/P(0)$.(a contradiction). If $P(0)=0$, then $P(0)$ is divisible by arbitrary large primes.
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2[Here](http://qchu.wordpress.com/2009/09/02/some-remarks-on-the-infinitude-of-primes/) is an interesting alternate approach to this problem. – 2012-12-08
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0Thank you for the nice link – 2012-12-08
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1Dear @Amr: I am not sure if your statement "none of the primes $p_1, \dots, p_r$ divides $P(P(0)p_1\dots p_r)$" is true unless you divide by $P(0)$ when $P(0) \neq 0$. – 2012-12-08
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0Yes you are right. I will edit my answer – 2012-12-08
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0why /P(0) and how do you know p_1 doesn't divide that? – 2012-12-11
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0I did not understand. Could you rephrase your question ? – 2012-12-11