Please explain this particular proof of no polynomial can only produce primes I understand that similar questions have been asked and answered in the past. I need someone to break the proof down to a level where I can see it explicitly. I am not a math graduate, just a hobbyist, so certain things which are apparent to somebody from a math background may not be that apparent to me. Below is the proof and inline my comments on what I specifically do not get. 
Let f(0)=p be a prime. OK!
Obviously, f(kp)=f(0)=0(mod p), so p|f(kp). I suppose it can be proved that f(kp) contains a factor of p. So p|f(kp)
Since f is a Prime-Generating Polynomial, f(kp) can only be p, 0, or -p. Did not understand this.
But there's at most 3n values of k which f(kp)=0,p,-p(n is the degree of f)
Contradiction.  Did not follow this
 A: The second line does not say that $f(kp) = f(0)$, it says that $$f(kp)=f(0)\mod p$$
which is obvious if you write out $f(x) = a_0 + a_1 x + \dots + a_nx^n$ and then see that $f(x) = a_0 + x\cdot g(x)$ for some function $g(x)$. Then,
$$f(kp) = a_0 + kp\cdot(g(kp)) = a_0 + p\cdot C$$
for some constant $C$, meaning that $f(kp)=a_0\mod p$.

The second thing is that since $f$ is a prime generating polynomial, you know that $f(kp)$ is either a prime, a negative prime or $0$. Also, $p$ divides $f(kp)$, so if $f(kp)$ is a prime, it must be $p$ or $-p$ because that is the only prime divisible by $p$.

The last line follows from the fact that a polynomial of degree $n$ can only have $n$ zeroes. So, the polynomial $f+(x) = f(x) - p$ can also only have $n$ zeroes (therefore, $f(x) = p$ for only $n$ values of $x$) and $f_-(x) = f(x) + p$ can also only have $n$ zeroes (therefore, $f(x) = -p$ for only $n$ values of $x$).

Also, sorry to say this, but if this proof is not obvious to you, you are obviously reading a book that is (currently) above your skill level.
