I am currently working on a homework problem from my Real Analysis class:
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 k≥0, f(kp) is prime so that f(kp)=p for all k≥0. 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)=r∑i=1(aipi)xi
so we have ai=0 for all i≥1. This means f(x) is the constant polynomial f(x)=p. Contradiction.
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