If $a$ is of order $3$ mod a prime $p$, then ...

Question

The question says:

Prove that if $a$ is of order $3$ modulo a prime $p$, then $1+a+a^2\equiv 0 \pmod p$. Moreover, $a+1$ is of order $6$.

For the First Part:

The typical idea is to start with $a^3 \equiv 1 \pmod p \to a^3 -1 \equiv 0 \pmod p$. Factoring the term on the left hand side, the rest is straightforward.

However, I need to check the following idea:

$$1+a+a^2 \equiv a^3+a^2+a \equiv a(1+a+a^2)\equiv a^2(1+a+a^2)$$ $$\equiv a^3(1+a+a^2) \equiv 0 \pmod p$$

Since, by the hypothesis, $a^3$ cannot be zero modulo the prime $p$, the desired result holds.

If this idea holds true, it can be generalized to the following result:

if $a$ is of order $k$, then, modulo prime $p$, then $1+a+a^2+\cdots + a^{k-1}$ is divisible by $p$.

Is it??

For the Second Part:

I can see that:

$$1+a+a^2 \equiv 0 \to 1+a \equiv -a^2 \pmod p$$

However, does this implies anything? Given that $a$ is of order $3$, but what about $-a$??

Please Help, and Thanks in advance,,

https://math.stackexchange.com/questions/220493/order-of-elements-modulo-p — lab bhattacharjee, Nov 23 '18 at 07:16
@labbhattacharjee, I would like to see your comments about thoughts on the first part. — Maged Saeed, Nov 23 '18 at 08:12

DeepSea · Accepted Answer · 2018-11-23T04:42:47.673

2

For the second part, taking from the first part that $1+a = -a^2\pmod p\implies (1+a)^6= (-a^2)^6 = a^{12}\pmod p= (a^3)^4 =1^4\pmod p=1\pmod p$ . Thus $1+a$ is of order $6$ as claimed. And if $a$ is of order $3$ then $(-a)^3 = -a^3 = -1 = p-1\pmod p$. And from this we have: $(-a)^6 = ((-a)^3)^2 = (-1)^2 = 1\pmod p$. So $-a$ is of order $6$ as well.

edited Nov 23 '18 at 04:42

answered Nov 23 '18 at 04:25

DeepSea

78,689

Thanks, but how to be sure that $6$ is the least power congruent to $1 \pmod p$ for $(1+a)$?? – Maged Saeed Nov 23 '18 at 04:28
I need also to make sure my thoughts are correct for the first part. Thanks in advance. – Maged Saeed Nov 23 '18 at 04:29
1

@MagedSaeed: You can prove it yourself that if $k$ is such that $(a+1)^k = 1\pmod p \implies k = 6$. In fact, such $k$ must satisfies $k \mid 6 \implies k = 1,2,3$. And none of these can satisfy since $a$ is of order $3$. – DeepSea Nov 23 '18 at 04:35
Oh, thanks,, I see. – Maged Saeed Nov 23 '18 at 04:37

score 0 · Answer 2 · answered Feb 04 '19 at 14:39

0

Using geometrical sum formula we get

$$1+a+a^2 = \frac{a^3-1}{a-1} \equiv 0 \pmod p$$

And generally we get

$$1+a+a^2+\dots + a^{k-1}= \frac{a^k-1}{a-1} \equiv 0 \pmod p$$

answered Feb 04 '19 at 14:39

user183748292

375

If $a$ is of order $3$ mod a prime $p$, then ...

For the First Part:

For the Second Part:

2 Answers2