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Nov 21, 2019 at 23:58 vote accept John Greenwood
Nov 21, 2019 at 23:58 vote accept John Greenwood
Nov 21, 2019 at 23:58
Nov 21, 2019 at 19:44 answer added Robert Israel timeline score: 1
Nov 21, 2019 at 13:10 comment added Geoff Robinson The second condition seems to be equivalent to $\left( \sum_{i=1}^{n} a_{i} \right)^{2} = \sum_{i=1}^{n}a_{i}^{2}$. In combination with the first euation, this seems to imply that the number of $i$ for which $a_{i}$ is odd is divisibe by $4$.
Nov 21, 2019 at 7:34 history edited John Greenwood CC BY-SA 4.0
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Nov 21, 2019 at 7:33 comment added John Greenwood @darijgrinberg oops yes, thank you
Nov 21, 2019 at 7:14 comment added darij grinberg The first sum should start at $k=1$, right?
Nov 21, 2019 at 6:58 history edited Fedor Petrov CC BY-SA 4.0
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Nov 21, 2019 at 6:28 history asked John Greenwood CC BY-SA 4.0