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Old 2020-02-14, 16:56   #2
Dylan14
 
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"Dylan"
Mar 2017

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Firstly, since n+215k is n mod 215 for any n and k in the integers, we know that (1+215k)^3 must be 1 mod 215 = 2^0 mod 215, and similarly for the others cases stated ((6+215k)^3 = 2^0 mod 215, and (7+215k)^3 = 2^7 mod 215).

So it suffices to check on the integers in the set [0, 214], then use the equivalence classes to quickly generate the rest.
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