Timeline for Sum Equals Product
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 13, 2010 at 7:41 | comment | added | dvitek | Ah sorry, I didn't take a look at the book, which of course would have clarified the issue at hand. Now to figure out how to solve this new problem... | |
| Aug 12, 2010 at 23:50 | comment | added | Gerry Myerson | For further information on what's in Guy D24, see my answer to question 35375, mathoverflow.net/questions/35375/… | |
| Aug 12, 2010 at 13:06 | comment | added | Gerry Myerson | drvitek, yes, but there are occasionally other solutions. E.g., 1, 1, 2, 2, 2, which is an extension of your construction. Guy asks not for values of $n$ but of $k$ (in the notation of your answer), and the problem is not so easily solved. Have a look at the book. | |
| Aug 12, 2010 at 1:02 | comment | added | dvitek | Gerry - if the problem is to solve the problem with only the unequal restriction removed, then aside from the trivial solutions $\\{n\\}$ we have solutions precisely for composite $n$. It is trivial to show that primes do not work, and for any composite $n$ we may take two non-trivial divisors $1 < j,k < jk = n$ and add the appropriate number of 1's. We have $j+k \le jk$, as $j, k \ge 2$ - this is just the inequality $(j-1)(k-1) \ge 1$. | |
| Aug 11, 2010 at 5:25 | vote | accept | Falco | ||
| Aug 10, 2010 at 23:32 | comment | added | Gerry Myerson | The Guy reference is to problem D24, pages 299-301 in the 3rd edition. Guy asks for solutions in positive integers, but does not require them to be distinct. | |
| Aug 10, 2010 at 18:34 | comment | added | Gerhard Paseman | Richard Guy mentions the more general case in his book Unsolved Problems in Number Theory. In particular, k (the number of summands)can assume only certain values. Gerhard "Ask Me About System Design" Paseman, 2010.08.10 | |
| Aug 10, 2010 at 17:54 | answer | added | dvitek | timeline score: 18 | |
| Aug 10, 2010 at 17:38 | comment | added | Kevin Buzzard | There is an algorithm! Here it is. Loop through all the finite sets of positive integers, and for each one compute the sum and the product, and output "yes" if they are equal, and "no" if not. | |
| Aug 10, 2010 at 17:33 | history | asked | Falco | CC BY-SA 2.5 |