Modular arithmetic: How to solve $3^{n+1} \equiv 1 \pmod{11}$? Please, I can't solve this equation: 
$$3^{n+1}\equiv 1 \pmod{11}$$
for $n \in \mathbb{N}$.
So what should I do please? Thanks.
 A: $3^5=243=11(22)+1\equiv 1\mod 11$
So if $3^{(n+1)}\equiv 1 \mod 11$ we must have $5|(n+1)$(as $5$ is the smallest integer such that $3^5\equiv 1\mod 5$)
A: First of all, using Fermat's Little Theorem, $3^{10}\equiv1\pmod {11}$ as $(3,11)=1$
Now,as we know if $a^n\equiv1\pmod m$  multiplicative order $\text{ord}_ma$ must divide $n$ where $a,m$ are any integers
So, we need to check for the following powers of $3 : 1,2,5,10$
Now,  $3^2=9\equiv-2\pmod {11}$
$3^5=(3^2)^2\cdot3^1\equiv(-2)^2\cdot3\equiv1\pmod {11} \implies \text{ord}_{11}3=5$ 
So, $\text{ord}_{11}3=5$ must divide $(n+1)\implies n=5m-1$ where $m$ is any integer 
As $n\in \mathbb N,$ we need  $5m-1>0$ or $\ge0$ as "there seems to be no general agreement about whether to include $0$ in the set of natural numbers."
In either case, $m\ge 1$
A: Fermat's little theorem states that for any non zero integer $a$ and any prime $p$, we have:
$$a^{p-1}\equiv1\pmod p$$
Thus,:
$$3^{10}\equiv 1\pmod {11}$$
This will allow you to find some of the solutions
A: I use Haskell to get :
4,9,14,19,24,29,34,39,44,49,54,59,64,69,74,79,84,89,94,99...

[n | n <- [0..100], mod (3^(n+1)) 11 == 1]

