If $a$ has order $3 \pmod p$, where $p$ is an odd prime, show that $a+1$ must have order $6 \pmod p$

Question

Since $3$ is the order of $a$ :
$a^3\equiv 1 \pmod p$

I am stuck after this. Not getting ideas. Any help ?

This is a fun one. What happens when you consider $a^3-1\equiv 0\pmod p$? — abiessu, Sep 10 '14 at 16:40
$$(a-1)(a^2+a+1) \equiv 0 \pmod p$$ but I'm not so good with quadratic congruences. Need little more help :3 — rrr, Sep 10 '14 at 16:42
http://math.stackexchange.com/questions/463010/show-that-if-a-has-order-3-bmod-p-then-a1-has-order-6-bmod-p — lab bhattacharjee, Sep 10 '14 at 16:44
thank you all so much for the very useful pointers, going through them :) — rrr, Sep 10 '14 at 16:50

score 2 · Accepted Answer · answered Sep 10 '14 at 16:46

2

We have that

$$a^3-1=(a-1)(a^2+a+1)=0\pmod p\stackrel{\text{since}\;a\neq 1}\iff a^2+a+1=0\pmod p\implies$$

$$a+1=-a^2=\pmod p$$

and from here

$$(a+1)^6=(-a^2)^6=a^{12}=(a^3)^4\pmod p\ldots$$

answered Sep 10 '14 at 16:46

Timbuc

1

Note that you can use \equiv to produce $\equiv$ for the modular equivalence symbol – abiessu Sep 10 '14 at 16:48
1

this looks simpler to follow :) thanks a lot !! – rrr Sep 10 '14 at 16:53

1 Answers1