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!!!!!!
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!!!!!!