Timeline for How to deduce an equation from this 3 Diophantine equations with 5 variables?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 28, 2017 at 6:20 | history | edited | DavitS |
edited tags
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| Aug 9, 2017 at 12:30 | answer | added | individ | timeline score: -2 | |
| Aug 3, 2017 at 9:29 | vote | accept | DavitS | ||
| Aug 3, 2017 at 7:05 | answer | added | Yaakov Baruch | timeline score: 8 | |
| Aug 2, 2017 at 16:34 | comment | added | DavitS | @YaakovBaruch but I am not asking to find all solutions, I just wanna show one of following, either $x=y$ or $m = n$. Or I wanna show that $x+y!=0mod3$ | |
| Aug 2, 2017 at 16:05 | comment | added | Yaakov Baruch | @DavitSargsyan: I would suggest changing the title, by replacing "deduce an equation from" with "solve", since what you are asking is tantamount to finding all the solutions. | |
| Aug 2, 2017 at 12:34 | comment | added | DavitS | @KonstantinosKanakoglou no it's not a contest problem. I need this in my research project. The story is very long to discuss in this post. | |
| Aug 2, 2017 at 11:58 | comment | added | Konstantinos Kanakoglou | May i ask about the origin of this problem? Seems like some contest problem. | |
| Aug 2, 2017 at 9:42 | history | edited | DavitS | CC BY-SA 3.0 |
added 135 characters in body
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| Aug 2, 2017 at 9:27 | comment | added | DavitS | @zen all the variables can only be natural numbers, so $k$ cannot be $0$ | |
| Aug 2, 2017 at 9:25 | comment | added | zen | If k=0 is allowed it can not be shown that x=y or m=n has to hold. consider for example k=0 then m=x and n=y and all natural numbers are allowed for either of these pairs | |
| Aug 2, 2017 at 9:18 | comment | added | DavitS | I also can deduce that if I show that $x + y$ is not divisible by $3$, i.e. $x + y != 0 mod 3$ | |
| Aug 2, 2017 at 9:12 | comment | added | Dirk | You might use this to transform your system into another system with variables, say, $a,b,c,d,e$, such that $a=b=c=d=e \geq 2$ are the only solutions of your new system. Maybe then showing that all unknowns in the new system have to be the same is easier than the problem you face now (just a thought, might or might not work). | |
| Aug 2, 2017 at 9:07 | comment | added | DavitS | It seems that you are right, I just want to show that either $m = n$, or $x = y$ | |
| Aug 2, 2017 at 8:57 | comment | added | Dirk | The solutions seem to be of the form $[m,n,k,x,y] = [a,a,4a+2,3a+1,3a+1]$ for $a$ a natural number and $a \geq 2$. | |
| Aug 2, 2017 at 6:54 | history | edited | Dima Pasechnik | CC BY-SA 3.0 |
adjusted the title to carry more meaning and updated tags
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| Aug 2, 2017 at 6:48 | review | First posts | |||
| Aug 2, 2017 at 7:02 | |||||
| Aug 2, 2017 at 6:47 | history | asked | DavitS | CC BY-SA 3.0 |