Problem involving polynomial function and prime numbersLet f be a polynomial function, with integer coefficients, strictly increasing on N such that f ( 0 ) = 1. Show that it doesn't exist any arithmetic progression of natural numbers with ratio r &gt; 0 such that the value of function f in every term of the progression is a prime number. I noticed that the last term of f is 1, due to the fact that f ( 0 ) = 1. I don't know how to continue.

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Problem involving polynomial function and prime numbersLet f be a polynomial function, with integer coefficients, strictly increasing on N such that   f  (  0  )  =  1. Show that it doesn&#039;t exist any arithmetic progression of natural numbers with ratio   r  &amp;gt;  0 such that the value of function f in every term of the progression is a prime number. I noticed that the last term of f is 1, due to the fact that   f  (  0  )  =  1. I don&#039;t know how to continue.

reinmelk3iu · Accepted Answer

Step 1Let f(x) be a polynomial with integer coefficents. Then for any   a  ,  b  ∈      Z  ,   g  (  x  )  =  f  (  a  x  +  b  ) is also a polynomial with integer coefficients. Assume g(n) is prime for all   n  ∈      N   and let   p  =  g  (  1  ). Note that g(n)modp is periodic with period p. We conclude that   p  ∣  g  (  n  p  +  1  ) for all   n  ∈      N  . It follows that   g  (  n  p  +  1  )  −  p has infinitely many roots and hence (provided   a  ≠  0) also   f  (  x  )  −  p has inifinitely many roots, hence must be the zero polynomial, i.e., f(x) is constant - but it was given that f(x) is strictly increasing!Step 2Remark: The fact   f  (  0  )  =  1 was not used. And instead of strictly increasing, non-constant suffices.

Problem involving polynomial function and prime numbers. Let f be a polynomial function, with integer coefficients, strictly increasing on N such that f(0)=1.

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