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Old 2018-03-05, 02:23   #14
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Jul 2009

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We get for free that if they aren't then the divisor that opposes the square are 7 mod 8. Doesn't help much though.edit: okay it limits the number of arrangements to n for 2n+1 7 mod 8 divisors.

Last fiddled with by science_man_88 on 2018-03-05 at 03:17
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