Timeline for two's and three's survive in gcd of Lagrange
Current License: CC BY-SA 3.0
21 events
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| Jan 31, 2017 at 15:23 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Nov 30, 2016 at 16:31 | vote | accept | T. Amdeberhan | ||
| Nov 30, 2016 at 16:24 | comment | added | GH from MO | I think showing that even the exponents of $2$ and $3$ are bounded in the gcd can be seen similiarly from the papers mentioned in my post. The only difference is that one needs to study the conditions $d_i\mid a_i$, where $d_i$ are various small powers of $2$ and $3$. The mentioned papers contain this information in the generality needed, but I don't have time to collect it carefully. | |
| Nov 30, 2016 at 12:50 | history | edited | T. Amdeberhan |
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| Nov 30, 2016 at 5:15 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Nov 30, 2016 at 4:50 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Nov 29, 2016 at 18:37 | history | edited | GH from MO |
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| Nov 29, 2016 at 18:22 | comment | added | Gerhard Paseman | For clarity, it should be stated that using 4n+1 implies b is at least 3, and similarly b is at least 2 if 4n+2 is used. Gerhard "Simple Should Be Made Clear" Paseman, 2016.11.29. | |
| Nov 29, 2016 at 17:49 | answer | added | GH from MO | timeline score: 21 | |
| Nov 29, 2016 at 16:40 | comment | added | T. Amdeberhan | GCD over all decompositions. | |
| Nov 29, 2016 at 16:20 | comment | added | Gerhard Paseman | Surely this disappears with large n? Take any m which has two representations as sums of two squares, and for M a sufficiently large multiple of 4 also a sum of two squares that M+m has two representations with large gcd? Or are you taking gcd over all decompositions? Gerhard "Is Squaring Up Both Decomposition" Paseman, 2016.11.29. | |
| Nov 29, 2016 at 15:58 | comment | added | T. Amdeberhan | I've not engaged elliptic curves, but maybe they are. | |
| Nov 29, 2016 at 15:26 | comment | added | Sylvain JULIEN | Are there elliptic curves behind that ? | |
| Nov 29, 2016 at 14:59 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Nov 29, 2016 at 13:31 | comment | added | T. Amdeberhan | $n=4(30)+1=8^2+5^2+4^2+4^2=8^2+7^2+2^2+2^2=10^2+4^2+2^2+1^2$ only, no $3$'s. | |
| Nov 29, 2016 at 13:23 | comment | added | Farewell | @T.Amdeberhan Do you know of any example for $n \geq 26$ where none of $a_1,a_2,a_3,a_4$ is divisible by 3? | |
| Nov 29, 2016 at 12:45 | comment | added | T. Amdeberhan | Uhmm... interesting. | |
| Nov 29, 2016 at 12:38 | comment | added | Fedor Petrov | In other words, given a prime $p>3$ we want to find four squares non-divisible by $p$ which sum up to $4n+1$. This is possible at least modulo $p$ by standard reasons like Cauchy-Davenport. We may try to count the number of representations of $n$ as $n=x^2+y^2+z^2+p^2t^2$ (how to do it?) and prove that it is strictly less than the number of representations as a sum of four squares. Alternatively, we may try to modify the inductive proof of Lagrange theorem carefully avoiding divisibility by $p$. | |
| Nov 29, 2016 at 12:28 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Nov 29, 2016 at 12:14 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Nov 29, 2016 at 12:02 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |