Prime polynomals

Find all polynomals $P$ where if $x$ is a prime then $P(x)$ is also a prime.

Trivial solutions are $P(x) = p$ for $p$ any prime and $P(x)=x$.  Now we will prove that these are the only solutions.

As out polynomial has an infinite number of rational points, all coefficients are rational. 

Let $q$ be the common multiple of all denominators and $Q(x) = qP(x)$. Now $Q$ has only integer coefficients. 

There are two possible cases:

Case 1: $P(p)=p$ for all but finitly many primes $p$.

As to non-identical polynomials can only intersect a finite number of times, and this polynomial intersects $P(x)=x$ an infinite number of times, the polynomial is identical to $P(x)$, a trivial solution.

Case 2: $\exists^{\infty} p: P(p)=x$ with $x\neq p$

This is equivalent to $Q(p)=qx$.

As there exists infinitely many $p$, take a $p$ bigger than $2q$, 

This means that $p$ is coprime with $2q$, as it is with $x$, so $p$ is coprime with $2qx$.

According to Dirichlet's theorem, $p+2qxk$ hits infinitely many primes.

$Q(p+2qxk)$ can be written as $Q(p) + 2qx\times c$ with c a constant. 

So if $Q(p)=qx$, then $Q(p+2qx)$ is a multiple of $qx$, so $P(p+2qx)$ is a multiple of $x$.

Now there are two cases:

Case 1: The polynomial is constant. This gives rise to the trivial solutions.

Case 2: The polynomial is unbounded, $\forall a,\exists r \forall v > r, |P(v)| > a$.

Take $a=x$, take $v$ a prime of the form $p+2qxk$ which is larger than $r$.

It has the following properties:

1. $|P(v)| > x$
2. $|P(v)|$ is a multiple of $x$
3. $|P(v)|$ is prime.

These 3 properties are unsatisfiable. (the only multiple of $x$ that is prime is $x$ itself, which does not larger than $x$). Therefore this leads to a contradiction.

This means that the trivial solutions are the only solutions.