Show that if a has order 3 (mod p), then a + 1 has order 6 (mod p).
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Originally Posted by tarheelborn Show that if a has order 3 (mod p), then a + 1 has order 6 (mod p). .
Now check that and deduce what you want (remember that ... )
Ps. The proof that is pretty interesting in itself.
If a^3 == 1 (mod p) does it imply that (a+1)^3 == -1 (mod p)? I have gotten only as far as (a+1)^3 = 2 + 3a(a+1) but I can't see that it helps me any in getting to -1.
How do I know that (a+1) doesn't have order < 6?
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