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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 $\left\vert{\:\mathbb{N} \setminus (f_1 \cup f_2\: \cup\: ... \cup \:f_7)\:}\right\vert = \infty$?

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  • $\begingroup$ The $30$ in those polynomials is suspicious and the fact that the degree one terms have coefficients which are all units modulo $30$ is also peculiar. Have you tried reducing the equations $\pmod{30}$? $\endgroup$ – Stijn Apr 30 '14 at 3:30
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    $\begingroup$ This reads a bit like a homework problem. Please read meta.mathoverflow.net/questions/70, and think about moving your question to our sister site, Math.StackExchange. $\endgroup$ – Theo Johnson-Freyd Apr 30 '14 at 3:46
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    $\begingroup$ @stijn: if you reduce modulo 30, the first polynomial becomes $7(m_1-n_1)+5$ which is surjective on $\mathbb Z/30 \mathbb Z$ since $(7,30)=1$. So you can't get anything this way. $\endgroup$ – Joël Apr 30 '14 at 11:13
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    $\begingroup$ In my opinion the problem is interesting and suitable for MO. I don't know where these polynomials come from, and maybe I am missing something that those voting to close have seen. The general problem of understanding what numbers are not of the form $axy+bx+cy$ where $a$, $b$ and $c$ are given and $x$ and $y$ run over ${\Bbb N}$ seems interesting to me, and I don't know the answer to it. I vote to reopen. $\endgroup$ – Lucia Apr 30 '14 at 13:59
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    $\begingroup$ Crossposting: math.stackexchange.com/questions/775663/…. $\endgroup$ – Dietrich Burde Apr 30 '14 at 14:25
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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 $(30k+11,30k+13)$.

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    $\begingroup$ Excellent! And for the general problem of $axy+bx+cy$, one multiplies through by $a$ and gets $(ax+c)(ay+b)-bc$, so that the numbers missed by such polynomials are again closely related to shifted primes. $\endgroup$ – Lucia Apr 30 '14 at 21:18
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    $\begingroup$ This factoring trick is called "completing the rectangle". $\endgroup$ – Jeremy Rouse Apr 30 '14 at 21:25
  • $\begingroup$ Great response @Max! $\endgroup$ – joebloggs Apr 30 '14 at 21:27
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    $\begingroup$ To go a step away from the prime triple conjecture. $f_1$ and $f_2$ allow that $30n+11$ contains prime factors $p\equiv 1,11,19,29 \mod 30$. $f_3,f_5,f_6,f_7$ together are more restrictive, All possible prime factors are forbidden, For $f_4$: $30n-11$ may contain prime factors $p\equiv 1, 7, 19, 13\mod 30$. (Note: 1,11,19,29 and 1,7,19,13 are both multiplicative subgroups mod 30). Hence there is one prime condition, and 2 half-prime conditions, (sieve dimension 2), which is still undoable. (For sieve dimension 3/2 there is sometimes hope, Iwaniec half dimensional sieve). $\endgroup$ – Christian Elsholtz Apr 30 '14 at 21:50

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