Timeline for Find four natural numbers with a certain property [closed]
Current License: CC BY-SA 4.0
16 events
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| Sep 4, 2023 at 2:47 | comment | added | user44191 | @AndreaLanzillo It's the only solution if you assume $X, Y, Z, T > 1$; otherwise, $(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 2, 3)$, and $(1, 2, 3, 7)$ all work. I haven't checked extremely closely, but I'm pretty sure similar reasoning to the above will work. | |
| Sep 3, 2023 at 10:54 | comment | added | Andrea Lanzillo | Thanks everyone for your help, you made my knowledge more complete. Now I'm just trying to demonstrate thate (2,3,7,43) is di only solution | |
| Sep 2, 2023 at 6:58 | comment | added | user44191 | In fact, the solution below is the only one where no variable is $1$. It's not hard to check that for $(3, 4, 5, 6)$ the sum is less than $1$, so the smallest variable must be $2$; then with the other variables being odd, the sum for $(2, 5, 7, 9)$ is less than $1$, so the second smallest must be $3$; the sum for $(2, 3, 5)$ is too large, so the third smallest must be at least $7$. It also must be less than $18$, and manual checking shows that $(2, 3, 7, 43)$ is the only possibility. | |
| Sep 1, 2023 at 20:48 | comment | added | Michael Hardy | Can one find a set of natural numbers for which multiplying finitely many members and adding $1$ always yields a number divisible by some member of the set? Certainly the set of all primes does that. Does it have a proper subset that does that? In the posted answer, $\{\,2,3,7,43\,\}$ the product of any three, plus one, is divisible by the fourth AND the product of any two, plus one, is divisible by one of the other two. But $2\times7+1$ is divisible not only by $3$ but also by a number coprime to all four members of this set. And similarly $2\times43+1, \,\,\, 3\times43+1,$ and $3\times7+1.$ | |
| Sep 1, 2023 at 20:35 | history | closed |
Will Jagy GH from MO Benjamin Dickman Dave Benson Max Horn |
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| Sep 1, 2023 at 20:29 | comment | added | user44191 | @DenisShatrov That does limit it pretty heavily; the sum of five distinct integer reciprocals can only be an integer if either a) one of them is $1$, or b) the sum is exactly $1$, as $H_6 = \frac{59}{20} < 3$. | |
| Sep 1, 2023 at 20:01 | comment | added | Denis Shatrov | It might be possible to find all such quadruples. $XYZT \mid XYZ + XYT + XZT + YZT + 1$ and number $1/X + 1/Y + 1/Z + 1/T + 1/(XYZT)$ is a positive integer. | |
| Sep 1, 2023 at 19:14 | answer | added | Maarten Havinga | timeline score: 5 | |
| Sep 1, 2023 at 19:09 | comment | added | GH from MO | @user44191 Sorry, I misread the last word as "four". | |
| Sep 1, 2023 at 19:05 | comment | added | user44191 | @GHfromMO I may be missing something, but I don't see how that satisfies the question; $XYZ + 1 = 232$, which is not divisible by $T = 15$. | |
| Sep 1, 2023 at 18:59 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Sep 1, 2023 at 18:46 | review | Close votes | |||
| Sep 1, 2023 at 20:35 | |||||
| Sep 1, 2023 at 18:42 | comment | added | Maarten Havinga | True, but the question is interesting even so | |
| Sep 1, 2023 at 18:35 | comment | added | user44191 | Is there a reason for the mismatch between question title and content, and for the metric-geometry and classical-analysis-and-odes tags (and probably the algebraic-geometry and prime-numbers tags as well)? | |
| S Sep 1, 2023 at 18:10 | review | First questions | |||
| Sep 1, 2023 at 19:25 | |||||
| S Sep 1, 2023 at 18:10 | history | asked | Andrea Lanzillo | CC BY-SA 4.0 |