Consider the set $P_{n+1} = \{p_1, \dotsc, p_{n+1}\}$ of the first $n+1$ primes. Does there always exist a $p \in P_{n+1}$ and a partition $\{A, B\}$ of $P_{n+1} \setminus \{p\}$ (in other words, $A$ and $B$ are disjoint and nonempty subsets whose union is $P_{n+1} \setminus \{p\}$) such that $$ \prod_{q \in A} q - \prod_{r \in B} r = p? $$ For example for $n=2$, we have $5-2=3$ and $5-3=2$; for $n=3$ we have $2\cdot 5-3=7$ and $2\cdot 5-7=3$; and for $n=4$ we have $5\cdot 7-3\cdot 11=2$. After these five I couldn't find any more solutions. Are there more? I'm especially interested in cases where $p > 2$.
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Sorry, I can't understand what operation you are doing or what your question is. – MJD Mar 12 '15 at 19:42
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@MJD what parts you don't understand ? – Mario Gomez Mar 12 '15 at 19:47
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What does "joined together" mean? Are your examples a single example, or more than one? In your example, what are the two groups? – MJD Mar 12 '15 at 19:50
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7 joined together with 2,3,5 becomes 2,3,5,7 – Mario Gomez Mar 12 '15 at 19:53
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one example 2,3,5,7,11 divided into 2 groups becomes 2,3,7 and 5,11 – Mario Gomez Mar 12 '15 at 19:54
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I think you are then performing some operation on 2,3,7 and 5,11 to obtain 2,3,5,7,11,13, but I don't understand from your example what operation that is. – MJD Mar 12 '15 at 19:57
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@MJD it only involves 2 operations: multiplication and difference:/ – Mario Gomez Mar 12 '15 at 19:59
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@MJD I think Mario means: Consider the set $P_{n+1} = {p_1, \dotsc, p_{n+1}}$ of the first $n+1$ primes. Does there always exist a $p^* \in P_{n+1}$ and a partition $A \cup B$ of $P_{n+1} \setminus {p^}$ such that $\prod_{q \in A} q - \prod_{r \in B} r = p^$? – epimorphic Mar 12 '15 at 20:01
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@epimorphic yes thanks,is there a better sentence/language than'joined together with' – Mario Gomez Mar 12 '15 at 20:08
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Perhaps "Divide $n$ primes into two groups so that those $n$ primes, together with the difference between the respective products of all elements in each of the two groups, form the first $n+1$ primes" if you insist on using mostly words. Using more mathematical symbols would make things clearer though (as long as you understand them). – epimorphic Mar 12 '15 at 20:23
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I could, I suppose. Though it might be considered a major edit and I can't make edits unilaterally yet so ideally you should be around to approve the edit. Will you still be on the site for the next 10 minutes? – epimorphic Mar 12 '15 at 20:33
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@epimorphic please do what you can do :) bye.. – Mario Gomez Mar 12 '15 at 20:52
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1Where did this problem come from? Did you think of it yourself? – Barry Cipra Mar 12 '15 at 21:11
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For $n=4$ you also have $2\cdot11-3\cdot5=7$ and $7\cdot3-2\cdot5=11$.
I don't see any examples with $n=5$, but it's possible I missed one (or more); it'd be nice to have a proof there are none (for $n=5$) that doesn't simply amount to exhaustive search. The larger $n$ gets, the more exhausting the search gets!
Added 3/16/15: Karen C has shown I missed at least one example with $n=5$, namely $5\cdot11-2\cdot3\cdot7=13$. She has also found a example with $n=6$ (but opines confidence there are none with $n=7$).
Barry Cipra
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A second example for $n=6$ is $13\cdot17-2\cdot3\cdot5\cdot7=11$. Mathematica found nothing else up to $n=16$, and the smallest difference between products seems to grow quickly with $n$. For $n=16$, it was $236297$, if my program was correct. – Steve Kass Mar 16 '15 at 18:44
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I have solutions for n=5 and n=6. For n=5 we have 5.11 - 2.3.7 = 13 and for n=6 we have 2.7.13 - 3.5.11 =17. (I am doing somekind of smart strategies,based on the fact that we only need the product's difference to be a single prime,so A and B must be close in values). And finally I'm 99.99 % sure that there are no solutions for n=7.
Karen C
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Holy cow, I can't believe I missed that solution for $n=5$. I'm just glad I didn't state for a fact that there were none! – Barry Cipra Mar 16 '15 at 17:09