Application of the Schönemann-Eisenstein criterion 
Let $P\in \mathbb{Z}[x]$ be a polynomial of degree 2011. Can you show that there is an infinite amount of such polynomials, such that $P$ and $P+2$ remain irreducible? 

$x^{2011}+15ax+3$ is a polynomial of degree 2011. By Schönemann-Eisenstein, it follows that it must be irreducible since: 3 doesn't divide $1$, 9 doesn't divide 3 but $3|15$. So this is irreducible over $\mathbb{Q}[x]$. It also follows that 25 respectively 5 doesn't divide $5$ respectively $1$, but does divide 15 for all $a\in \mathbb{N}$, so by using 5 as a prime it follows that P+2 is also irreducible. 
The book asks for polynomials in $\mathbb{Z}[x]$, I have shown it for $\mathbb{Q}[x]$. Gauss lemma states that if it is for integers only over $\mathbb{Q}[x]$, it also counts for $\mathbb{Z}[x]$
 Is my reasoning correct? Please do tell 
 A: To make things harder, we prove a result about polynomials that have shape very much like the one described by the OP. 
We will show that there are infinitely many irreducible polynomials $x^{2011} +p(p+2)x+p$ with $p$  prime and $x^{2011} +p(p+2)x+p+2$  irreducible. For each of these two polynomials, the irreducibility proof uses  the Eisenstein Criterion. 
A possible approach is to show that $p+2$ is prime for infinitely many primes $p$. Then, as in the OP's example, $x^{2011} +p(p+2)x+p$ 
and
$x^{2011} +p(p+2)x+p+2$ are both irreducible by Eisenstein's Criterion. However, this requires proving the Twin Prime Conjecture, so it is not the easy way to go.
Instead, recall that a number $n$ is called powerful(l) if for every prime $q$ that divides $n$, its square $q^2$ divides $n$. Powerful numbers are relatively scarce. There is a constant $C$ such that the number of powerful numbers less than $x$ is less than $Cx^{1/2}$ (see for example the Wikipedia article).
Since the primes have much greater asymptotic density than the powerful numbers, there are infinitely many primes $p$ such that $p+2$ is not powerful. Indeed, for "most" primes $p$, the number $p+2$ is not powerful.  Let $p$ be any prime such that $p+2$ is not a powerful number. 
Then 
$x^{2011}+p(p+2)+p$ and $x^{2011}+p(p+2)+p+2$ are both irreducible. For the second polynomial, we use the Eisenstein Criterion with $q$ any prime which divides $p+2$ but whose square does not divide $p+2$. There is such a prime $q$ since $p+2$ is not powerful.
Comment: The proof in principle is of irreducibility over $\mathbb{Q}[x]$. But irreducibility over $\mathbb{Z}[X]$ is an immediate consequence, since our polynomial is monic, and consequently the $\gcd$ of the coefficients is $1$.
By using Chen primes, we can find infinitely many primes $p$ such that $p+2$ is prime or has at most two prime factors. That is the nearest we can come to imitating your $3$, $5$ example.
A: Dear VVV your example is very nice, and your polynomials are irreducible over $\mathbb Z$.
There is a subtlety involved in the Schönemann-Eisenstein criterion:
1) The Schönemann-Eisenstein criterion only gives irreducibility in $\mathbb Q[X]$.
 For example you can apply it to $f(X)=2X^2+6$ with $p=3$: the criterion will tell you that $f(X)$ is irreducible in $\mathbb Q[X]$.
However  $f(X)$ is not irreducible in $\mathbb Z[X]$ because there you have a genuine factorization $2X^2+6=(2).(X^2+3)$.
The apparent  paradox is explained by the fact that $2$ is not invertible in $\mathbb Z[X]$, but it is invertible in $\mathbb Q[X]$
2) However in your case, your  polynomials are also irreducible over $\mathbb Z$ because of the following result:   

Given a polynomial $f(X)\in A[X]$ where $A$ is a UFD with fraction field $K$, we have the equivalence:
  $ f(X)$ is irreducible in  $A[X]$
   $ \iff  $
     $f(X)$ is irreducible in  $K[X]$  and no prime element in $A $ divides all the coefficients of $f(X) $

