2022-09-14, 18:21 | #1 |
Mar 2018
2·269 Posts |
How to prove that only Powers of 6^(6+35j) are of the form648+23004s
This Is a Little more difficult:
How to prove that Powers 6^(6+35j) for j nonnegative integer are of the form 648+23004s and only that powers . I mean there Is no orher 6^g of the form 648+23004s, only Powers 6^(6+35j) Last fiddled with by enzocreti on 2022-09-14 at 19:20 |
2022-09-15, 13:49 | #2 |
Feb 2017
Nowhere
13730_{8} Posts |
If a congruence holds (mod n), it holds (mod d) for every divisor d of n. In particular, in the case at hand, as you more or less pointed out in this post,
if 6^e == 648 (mod 23004), then 6^e == 648 (mod 71), or 6^e == 9 (mod 71). Since 71 is a prime, the values of e are easily characterized. |
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