TITLE: Is there an 11-term arithmetic progression of primes beginning with 11?
QUESTION [29 upvotes]: i.e. does there exist an integer $C > 0$ such that $11, 11 + C, ..., 11 + 10C$ are all prime?

REPLY [2 votes]: There seems to be an interest in $p$-app ($p$-long prime arithmetic progressions) which start with prime $p$. Thus, I've decided to put some $\LaTeX$ sweat into more information, as elementary as it is.



A $p$-app is an arithmetic progression
     $\ (p+t\!\cdot\! d\ :\ t=0\ldots n\!-\!1)\ $ such that all its terms are primes, and integer $\ d>1.\ $ Then a simple theorem assures us that
$$ \prod \mathbf P(p-1)\ |\ d $$
Where $\ \mathbf P(x)\ $ is the set of all primes $\ \le\ x.$
Given any prime $p$ we would like to find all we can about the $p$-app's (do they exist, etc.).



Let $\ D:=\prod\mathbf P(n_1)\ $ as above. In particular, we would like to know everything about $\ r\ $ such $\ d=r\cdot D,\ $ where $\ d\ $ is the difference of arbitrary $p$-app.
Thus, let $\ q\ $ be any prime not in the said $p$-app. Then
$$ d\not\equiv 0\mod q\quad \Longrightarrow\quad \forall_{0\le k< p}\quad p+k\cdot d\ \not\equiv 0 \mod q $$
Notations:   Let $\ \ /_n\ \ $ be the $\mod n\ $ division by non-$0$ integers which are not factors of $\ n,\ $ with the division value in $\ \{0\ldots\ n\!-\!1.$
Thus,
$$  d\not\equiv 0\mod q\quad \Longrightarrow\quad \forall_{0\le k< p}\quad k\ \ne (-p)\,\ /_q\,\ d   $$
In other words, looking at the division remaining options (and under the established notation),
THEOREM 1 
$$  d\not\equiv 0\mod q\quad \Longrightarrow\quad p\ \ \le\ \ (-p)\,\ /_q\,\ d\ \ <\ \ q $$
This significantly reduces the number of options for $\ d\ $ when it is (easily!) applied to all primes $\ q>p\ $ at the same time (in the same computer program).

APPLICATIONS


Let prime $\ p>3\ $ be a younger sibling of prime $ q:=p+2.\ $ Then

$$ (-p)\,\ /_q\,\ d\,\ =\,\ -\!1\,\ \mbox{or}\ -\!2 $$
$\qquad$ (Thus, even one prime $\ q\ $ contributes to the computational savings).

Let $\ p:=11\ $. The (see above) $\ D=2\cdot 3\cdot 5\cdot 7=210,\ $ and let $\ d := r\cdot D\ $ be the respective difference of an arbitrary $11$-app. Then, by the above THEOREM 1, when $\ d\not\equiv 0 \mod 13\ $ (i.e.
$\ d\not\equiv 0 \mod 13)\ $ then

$$ 2\,\ /_{13}\ d\ \equiv\ -\!1\,\ \mbox{or}\ -\!2\quad \mod 13 $$ 
or
$$ d\ \equiv\ -\!1\,\ \mbox{or}\ -\!2\quad \mod 13 $$
Since $\,\ d = r\cdot D = r\cdot 210 \equiv 2\cdot r\,\ \mod 13,\ $ and allowing for the divisibility $13\,|\,r,\ $ we finally obtain,
$$ r\,\ \equiv\,\ 0\ \mbox{or}\ 6\ \mbox{or}\ 12 \mod 13 $$
This reduces the amount of computation $\frac 3{13}\ $ time.
Then, taking into account another prime, $\ q:=17,\ $ we reduce the computation time again $\ \frac 7{17}\ $ times, or for a total saving
$$ \frac 3{13}\cdot\frac 7{17}\ =\ \frac {21}{221}$$
times (more then ten times faster), etc.