TITLE: Are there infinitely many natural numbers not covered by one of these 7 polynomials?
QUESTION [11 upvotes]: Consider the following polynomials:
$$
f_1(n_1, m_1) = 30n_1m_1 + 23n_1 + 7m_1 + 5\\
f_2(n_2, m_2) = 30n_2m_2 + 17n_2 + 13m_2 + 7\\
f_3(n_3, m_3) = 30n_3m_3 + 23n_3 + 11m_3 + 8\\
f_4(n_4, m_4) = 30n_4m_4 + 11n_4 + 29m_4 + 11\\
f_5(n_5, m_5) = 30n_5m_5 + 29n_5 + 17m_5 + 16\\
f_6(n_6, m_6) = 30n_6m_6 + 19n_6 + 7m_6 + 4\\
f_7(n_7, m_7) = 30n_7m_7 + 31n_7 + 13m_7 + 13\\
$$
where each $n_1, m_1,..., n_7, m_7 \in \mathbb{N}$
How can I prove that  $\left\vert{\:\mathbb{N} \setminus (f_1 \cup f_2\: \cup\: ... \cup \:f_7)\:}\right\vert = \infty$?

REPLY [18 votes]: Notice that
$$30\cdot f_1(n_1,m_1) = (30\cdot m_1+23)\cdot (30\cdot n_1+7) - 11\\
30\cdot f_2(n_2,m_2) = (30\cdot m_2+17)\cdot (30\cdot n_2+13) - 11\\
30\cdot f_3(n_3,m_3) = (30\cdot m_3+23)\cdot (30\cdot n_3+11) - 13\\
30\cdot f_4(n_4,m_4) = (30\cdot m_4+11)\cdot (30\cdot n_4+29) + 11\\
30\cdot f_5(n_5,m_5) = (30\cdot m_5+29)\cdot (30\cdot n_5+17) - 13\\
30\cdot f_6(n_6,m_6) = (30\cdot m_6+19)\cdot (30\cdot n_6+7) - 13\\
30\cdot f_7(n_7,m_7) = (30\cdot m_7+31)\cdot (30\cdot n_7+13) - 13
$$
These representations imply that the polynomials cannot represent integers $k>0$ such that $30k \pm 11$ and $30k + 13$ are primes. There are likely infinitely many such prime constellations. See http://mathworld.wolfram.com/PrimeConstellation.html 
P.S. The polynomial $f_4(n_4,m_4)$ is the only one, whose representation has $+11$ as a free term. If it were $-11$ or $-13$ as in the other representations, then we would need existence of infinitely many twin primes of the form $(30k+11,30k+13)$.