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Prove that there is no polynomial $$P(x)=a_nx^n+a_{n-1}x^{n-1}+ \dots+ a_0$$ with integer coefficients and of degree at least $1$ with the property that $P(0), P(1), P(2), \dots$ are all prime numbers.

How should one approach this? Contradiction seems plausible if we would assume it we would get that $P(0), P(1), P(2) \dots$ would all equal some primes. Also from $P(0) = q$, where $q$ is some prime we would get that $a_0=q.$ From here on I don’t quite know how to continue... Any hints would be appreciated.

1 Answers1

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The key point here is that $a-b|P(a)-P(b)$ (if you haven’t seen this already you should try proving it, it’s a nice exercise).

From here, if $P(n) = p$, then $p$ divides $P(n+kp)$ for all positive $k$. But since all of these are prime, we get that $P(n+kp)=p$, and so $P$ takes the same value infinitely many times, and hence is constant, contradicting the degree condition.

hdighfan
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  • Hmm I don’t think I follow with the first result. Isn’t $P(a)-P(b) = p - q$ for some primes $p, q$? If so, how do we get the result that $a-b | p-q$? –  Jun 04 '20 at 18:46
  • It suffices to show that $a-b|a^n-b^n$, since $P$ is a sum of powers. But $a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots+b^{n-1})$, as required. – hdighfan Jun 04 '20 at 18:48
  • Just to clarify, the statement $a-b \vert P(a) - P(b)$ holds for any polynomial $P$ with integer coefficients. Not just the one in the question. – hdighfan Jun 04 '20 at 18:49