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$\def\mod{\mathrm{\;mod\;}}\def\pq{\{p_1,p_2,p_3,p_4\}}$$\def\dq{\{d_1,d_2,d_3,d_4\}}$ This question concerns quartets of consecutive primes $p_1 < p_2 < p_3 < p_4$ such that $p_4 - p_1 < 10$, and the associated quartets $\dq$ consisting of the last decimal digit of each of the $p_i$.

For example, if $\pq = \{11,13,17,19\}$, then $\dq = \{1,3,7,9\}$.

Assuming my code is correct (i.e. no bugs), then among the primes $< 10^8$ there are only three such quartets of primes whose associated quartets of last digits is not $\{1,3,7,9\}$. These three quartets are:

$$ \begin{array}{l} \{2, 3, 5, 7\} \\ \{3, 5, 7, 11\} \\ \{5, 7, 11, 13\} \end{array} $$

Are these really the only ones? If so, how does one prove it?

To state the question differently, let $x = 10\,n$ for some integer $n \geq 0$. Consider the following (overlapping) quartets of odd positive integers:

$$ \begin{array}{llllllll} \{& x+3, & x+7, & x+9, & x+11 & & &\} \\ \{& & x+7, & x+9, & x+11, & x+13 & &\} \\ \{& & & x+9, & x+11, & x+13, & x+17 &\} \end{array} $$

Can any one such quartet consist solely of prime numbers?

I have not been able to find one, but it's not immediately obvious to me why.

FWIW, at least in the range $[0, 10^8]$, there seems to be no shortage of positive integers $x = 10\,n$ such that the quartet $\{x+1, x+3, x+7, x+9\}$ consists solely of primes. I found $4767$ such $x$, the smallest and largest ones being being $x = 10$ and $x = 99982240$.

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    $\begingroup$ Except $2$ and $5$, the last digit of a prime must be one of $1,3,7,9$. If four consecutive primes are less than $10$ apart, none of these last digits can occur more than once, so the set must be $\{1,3,7,9\}$. Now it remains to see that the smallest of the four primes must be of the form $10k+1$. You can eliminate the other cases by considering remainders modulo $3$. $\endgroup$ – Daniel Fischer Mar 16 '14 at 15:26
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    $\begingroup$ After 3 and 5, all primes must be a multiple of 6 +/- 1. $\endgroup$ – Fred Kline Mar 16 '14 at 15:34
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    $\begingroup$ If you can show there are infinitely many of these prime quartets, you will have proved the Twin Prime Conjecture. Good luck! $\endgroup$ – Fred Kline Mar 16 '14 at 15:43
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If you consider the additions to $x$ modulo $3$ within

$\begin{array}{llllllll} \{& x+3, & x+7, & x+9, & x+11 & & &\} \\ \{& & x+7, & x+9, & x+11, & x+13 & &\} \\ \{& & & x+9, & x+11, & x+13, & x+17 &\} \end{array}$

You get

$\begin{array}{llllllll} \{& 0, & 1, & 0, & 2 & & &\} \\ \{& & 1, & 0, & 2, & 1 & &\} \\ \{& & & 0, & 2, & 1, & 2 &\} \end{array}$

Since each row contains a $0$, a $1$ and a $2$, whatever $x$ is modulo $3$ each row will have a number divisible by $3$. So it either contains $3$ itself, or contains a composite number. The only way it can be $3$ is the first row, with $x = 0$ but that would make $x + 9 = 9$, so there's always a composite number.

Note however ${1,3,7,9} \equiv {1,0,1,0} \mod 3$, so the argument doesn't apply to that case.

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  • $\begingroup$ If I understand it correctly, your argument looks incomplete to me, because, as far as I can tell, it covers only the case where $x$ is congruent to $0$ modulo $3$, whereas $x$ could also be congruent to $1$ (e.g. $x=10$) or to $2$ (e.g. $x = 20$) modulo $3$. $\endgroup$ – kjo Mar 16 '14 at 16:49
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    $\begingroup$ No because each row has a 0, a 1 and a 2, whatever $x$ is $\mod 3$ when you add it there will still be a 0, a 1 and a 2. If $x \equiv 1 \mod 3$ then the first row is 1,2,1,0 the second 2,1,0,1 and the third 1,0,2,0 - always a 0. $\endgroup$ – Neil W Mar 16 '14 at 22:06
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    $\begingroup$ I see. Thanks for the clarifiication! $\endgroup$ – kjo Mar 16 '14 at 22:44
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It's not clear if you care about the order of the last-digits, but if you did not then the problem seems very trivial. So I will assume that you do want the order of the last-digits to respect the order of the consecutive primes.

Then Daniel Fischer's comment does this for you. If a last digit is $5$, then the entire prime is $5$, and we can identify $\{2,3,5,7\}, \{3,5,7,11\}$ and $\{5,7,11,13\}$ as solutions. Beyond this, we can rule out $5$ (and $2$) as a last digit.

Do you know that there are no triplet primes except $\{3,5,7\}$? Writing sets in ascending order, each of $\{\ldots3,\color{red}{\ldots7,\ldots9,\ldots1}\}, \{\color{red}{\ldots7,\ldots9,\ldots1},\ldots3\}, \{\color{red}{\ldots9,\ldots1,\ldots3},\ldots7\}$ that have diameter under $10$ would have to have triplet primes (indicated in red). So the only option left is $\{\ldots1,\ldots3,\ldots7,\ldots9\}$.

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