Thursday, 28 April 2016

real analysis - There is no polynomial f(x) with integer coefficients such that f(n) is prime for all non-negative integers n.



I am currently working on a homework problem from my Real Analysis class:



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So far, I was able to prove part(a), that is, f(kp)p. I'm quite confused as to how I can use this result to prove part (b). Also, I don't see the intuition in how f(x)p would be of any help to reach a contradiction. Any suggestions would be helpful.


Answer



Like paw88789, I'm assuming that the problem asks you to prove that there is no nonconstant f that satisfies the described property. So suppose otherwise that there is a nonconstant f(x)=ri=0aixi such that f(k) is prime for all nonnegative integers k.



From (a), f(kp) is divisible by p. But for k0, f(kp) is prime so that f(kp)=p for all k0. But this means the polynomial g(x)=f(xp)p has infinitely many roots x=0,1,2, By the theorem that you are allowed to assume, we infer that g(x) is identically 0. But
g(x)=ri=1(aipi)xi


so we have ai=0 for all i1. This means f(x) is the constant polynomial f(x)=p. Contradiction.



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