Order (mod p)

**tarheelborn** · March 26th 2010, 12:24 PM

Show that if a has order 3 (mod p), then a + 1 has order 6 (mod p).

**tonio** · March 26th 2010, 01:52 PM

Originally Posted by tarheelborn

Show that if a has order 3 (mod p), then a + 1 has order 6 (mod p).

$(a+1)^6=[(a+1)^3]^2=[a^3+3a^2+3a+1]^2=(2+3a(a+1))^2=$ $4+12a(a+1)+9a^2(a+1)^2=$ $4+12a^2+12a+9a^4+18a^3+9a^2=$ $9a+18+21a^2+12a+4=21a^2+21a+22=<br /> 21(a^2+a+1)+1=0+1=1$ .

Now check that $(a+1)^3=-1$ and deduce what you want (remember that $a^3-1=(a-1)(a^2+a+1)$ ...

)

Tonio

Ps. The proof that $(a+1)^k\neq 1\,\,\forall\, 1\leq k\leq 5$ is pretty interesting in itself.

**tarheelborn** · March 31st 2010, 01:35 PM

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.

**tarheelborn** · April 9th 2010, 11:13 AM

How do I know that (a+1) doesn't have order < 6?

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