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Let $f$ be a polynomial function, with integer coefficients, strictly increasing on $\Bbb N$ such that $f(0)=1$. Show that it doesn't exist any arithmetic progression of natural numbers with ratio $r>0$ such that the value of function $f$ in every term of the progression is a prime number. I believe that the solution includes a reductio ad absurdum, but I don't know how to solve it.

I. Stefan
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  • See http://math.stackexchange.com/questions/304330/proof-of-lack-of-pure-prime-producing-polynomials It is not difficult to adapt the answer to your case. – almagest May 16 '16 at 08:21
  • Well, except that you don't specify the length of your progression. Is it meant to be infinite? – almagest May 16 '16 at 08:23

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If progression is $a+nb$, and $f(a)=p$, then $f(a+pnb)\equiv 0\pmod p$ and absolute value is more then $p$ for large $n$.

Mikhail Ivanov
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