2020-02-14, 15:19 | #1 |
Mar 2018
17×31 Posts |
cubes congruent to 2^n (mod 215)
1^3 is congruent to 2^n (mod 215) n=0
6^3 is congruent to 2^n (mod 215) n=0 7^3 is congruent to 2^n (mod 215) n=7 any other cubes congruent to 2^n (mod 215)? Last fiddled with by enzocreti on 2020-02-14 at 15:23 |
2020-02-14, 16:56 | #2 |
"Dylan"
Mar 2017
574_{10} Posts |
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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