A proof for a (non-constant) polynomial can't take only primes as value I know a proof of this statement, see How to demonstrate that there is no all-prime generating polynomial with rational cofficents?
My question is that, in the book 
Introduction to Modern Number Theory - Fundamental Problems, Ideas and Theories

by Manin, Yu. I., Panchishkin, Alexei A,
it says (p16)

I would like to know if this is correct. If I understand it correctly, then $x^2+1$ is a counterexample of this, by considering the Legendre symbol $(\frac{-1}{p}) = (-1)^{\frac{p-1}{2}}$.

suppose the polynomial P(x) takes prime value for each integer k.assume r is a root of P(x).so P(r)=0


*

*now k -r|P(k) -P(r)$ \Rightarrow $k -r|P(k)=m=a prime. so clearly k -r=m or 1

*so clearly k-n=a prime or 1 for all roots n of P(x)

*consider P(x)=a(x-r)(x-c)(x-d)....................(x-v)[r,c,d,.........,v are roots]

*now P(k)=a(k-r)(k-c)(k-d)................(k-v)=a(a prime or 1)(a prime or1)....................(a prime or 1)=a constant

*but P(x) is non constant.so a contradiction!!!!!!

