Let m and n be a integer. Show that for all values of n there is a polynomial such that P(n) equals toma prime number. For instance for the polynomial $$x^{2}+1$$ for x=1 the result is equal to 2. Question is finding a polynomal that is not equals to a prime number for all values of x.
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$\begingroup$ What is the purpose of $m$? Do ask if for each $n$ there exists a polynomial $P$ such that $P(n)$ is not prime or if there exists a polynomial $P$ such that $P(n)$ is not prime for all $n$? $\endgroup$– Paul FrostJul 18, 2018 at 9:18
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$\begingroup$ it has no purpose it is extra $\endgroup$– Demir EkenJul 18, 2018 at 9:22
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$\begingroup$ $p(x) = x^2$ never produces prime numbers. $\endgroup$– Paul FrostJul 18, 2018 at 9:25
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Assuming that the question is the following - Is there a polynomial $P(x)$ such that for any integer $n$, $P(n)$ is not prime? - yes, there are very simple examples: $P(x)=4x$ or $P(x)=x^2$ would both do the job.
