Concerning Dickson's Conjecture Dickson's Conecture says the following:
If $f_1(n)=a_1+b_1n, f_2(n)=a_2+b_2n, \ldots, f_k(n)=a_k+b_kn$, with $b_j \geq 1$, $1 \leq j \leq k$ satisfy a congruence condition, then there exist infinitely many positive integers $n$ for which every $f_j(n)=a_j+b_jn$ is a prime number, $1 \leq j \leq k$.

(1) What is the congruence condition?

In Schinzel's hypothesis H, a condition is mentioned, which I suspect is the congruence condition that I am looking for; it says:
"For every prime $p$, there is an $n$ such that all the polynomial values at $n$ are not divisible by $p$".
This condition obviously excludes the case $f_1(n)=3+n$ and $f_2(n)=4+n$, 
which is a trivial counterexample to Dickson's conjecture/Schinzel's hypothesis H, since always one of $\{3+n,4+n\}$ is an even number, so cannot be a prime number (this counterexample does not satisfy the above condition, since for $p=2$ there is no $n$ with $2 \nmid 3+n$ and $2 \nmid 4+n$). 

(2) If the congruence condition is indeed: "For every prime $p$, there is an $m$ such that all the polynomial values at $m$, $\{f_j(m)\}$, are not divisible by $p$", then it seems to me that in practice it is very difficult to check (except trivial cases as $f_1(n)=3+n, f_2(n)=4+n$); Am I missing something?

I am interested in the special case where $\gcd(a_j,b_j)=1$, $1 \leq j \leq k$,
$k=2$.
In that case, if I wish to check the congruence condition, then, for a given prime $p$, all I know (by Dirichlet's theorem on arithmetic progressions) that there exists $m$ such that $p \nmid f_1(m)=a_1+b_1m$ (just take large enough $m$ such that $a_1+b_1m$ is a prime number greater than $p$, so clearly $p$ does not divide a greater prime number), but $p$ may divide $f_2(m)$. 
Therefore, if we assume that Dickson's Conjecture is true, then even in the special case $k=2$, $\gcd(a_1,b_1)=1$, $\gcd(a_2,b_2)=1$, it seems not applicable.
Edit: This is a relevant paper that I have just found. In that paper an interesting result due to Maynard-Tao is mentioned ("Even more impressively... it was shown that one can take $k_2=50$").
However, they deal with $b_1=\cdots=b_k=1$.
Thank you very much!
 A: I saw it, in addition to $a_j > 0$ and $(a_j, b_j) = 1$ ($1 \leq j \leq k$), in the following form: for any prime $\color{red}{p \leq k}$, the mapping $n \mapsto P(n) := \displaystyle\prod_{j = 1}^{k}(a_j + b_j n) \bmod p$ is not identically zero.
For $p > k$ this holds automatically, as any polynomial over $\mathbb{Z}/p\mathbb{Z}$ in $n$, which is identically zero as a function of $n$, must be a multiple of $n(n-1)\ldots(n-(p-1))\color{lightgray}{\equiv n^p-n}$ with degree $p>k$ (and in our case it cannot be identically zero as a polynomial).
Thus it's not that hard to check ($n^p - n$ should not divide $P$ over $\mathbb{Z}/p\mathbb{Z}$ for each prime $p \leq k$).
A: In practice it is not that hard to check (2).  Unless $b$ is itself divisible by $p$, there is exactly one congruence class of $m$ for which $a+bm$ is divisible by $p$.  That means that in the generically-worst case, there are at most $k$ congruence classes of $m$ mod $p$ that could cause any of the $\{f_j(m)\}_j$ to be divisible by $p$.
If $p > k$ then there will always be some congruence class available that makes all the $\{f_j(m)\}$ non-zero, so we only need to be concerned with $p \le k$, at least when $p$ does not divide any $b_j$.
But there are only finitely many possible primes that divide any $b_j$, so those can be checked separately (as metamorphy's answer suggests, they are in fact completely harmless unless $p$ also divides $a_j$).
