Timeline for On finding all the positive integral solutions of $x^2+y^2=z^2+1$
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| May 6, 2015 at 22:15 | answer | added | KennyColnago | timeline score: 2 | |
| May 5, 2015 at 20:29 | vote | accept | Jr Antalan | ||
| May 5, 2015 at 19:44 | answer | added | george2079 | timeline score: 2 | |
| May 5, 2015 at 17:55 | answer | added | Michael E2 | timeline score: 5 | |
| May 5, 2015 at 14:56 | answer | added | djp | timeline score: 2 | |
| May 5, 2015 at 12:54 | vote | accept | Jr Antalan | ||
| May 5, 2015 at 20:29 | |||||
| May 5, 2015 at 12:54 | comment | added | Jr Antalan | To all: Thanks a lot. I got it. | |
| May 5, 2015 at 12:24 | comment | added | Michael E2 |
@Guesswhoitis. Many of the cases disappear, but you get m_goldberg's solution plus an unhelpful bit, (x | y | z) \[Element] Integers && ((x == 1 && y >= 1 && z == Abs[y]) || (x >= 2 && y >= 1 && z == Sqrt[-1 + x^2 + y^2])).
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| May 5, 2015 at 12:06 | comment | added | m_goldberg |
It seems obvious that triples of the form {1, n, n} and {n, 1, n} are solutions for any integer n.
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| May 5, 2015 at 11:59 | answer | added | shrx | timeline score: 5 | |
| May 5, 2015 at 11:53 | comment | added | J. M.'s missing motivation♦ |
@Michael, ah, now that's rather unhelpful output. What happens if you add a Positive[] restriction to all the variables? (How tricky these Diophantines are!)
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| May 5, 2015 at 11:48 | comment | added | Michael E2 |
@Guesswhoitis. Reduce does not parametrize the solutions in this case: i.sstatic.net/vAVfL.png
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| May 5, 2015 at 11:11 | comment | added | J. M.'s missing motivation♦ |
I was telling you to post the output. This is because Reduce[] will sometimes post parametrized solutions, if you're lucky.
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| May 5, 2015 at 11:01 | comment | added | Jr Antalan | It only gives me equations for x, y and z. @Guesswhoitis. | |
| May 5, 2015 at 10:56 | comment | added | J. M.'s missing motivation♦ |
@Jr, what does Reduce[] give you?
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| May 5, 2015 at 10:55 | history | edited | J. M.'s missing motivation♦ | CC BY-SA 3.0 |
added 4 characters in body; edited tags; edited title
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| May 5, 2015 at 10:14 | comment | added | Jr Antalan | Unfortunately, it does not give a result in terms of triple (x,y,z). | |
| May 5, 2015 at 10:12 | comment | added | Jr Antalan | Thanks @Pickett will try it now. | |
| May 5, 2015 at 10:08 | comment | added | C. E.♦ |
You can pick values according to Reduce[x^2 + y^2 == z^2 + 1, {x, y, z}, Integers].
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| May 5, 2015 at 10:07 | history | edited | C. E.♦ | CC BY-SA 3.0 |
deleted 1 character in body
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| May 5, 2015 at 10:02 | comment | added | Jr Antalan | thanks for your comment @Pickett. I want to find many of them using Mathematica. | |
| May 5, 2015 at 10:01 | comment | added | C. E.♦ | There are an infinite number of points that satisfy that equation. You can't list all of them. What are you trying to do? | |
| May 5, 2015 at 9:57 | review | First posts | |||
| May 5, 2015 at 10:24 | |||||
| May 5, 2015 at 9:57 | history | asked | Jr Antalan | CC BY-SA 3.0 |
