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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.

Lord_Farin
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Sherloek holmes
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4 Answers4

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$3^5=243=11(22)+1\equiv 1\mod 11$

So if $3^{(n+1)}\equiv 1 \mod 11$ we must have $5|(n+1)$(as $5$ is the smallest integer such that $3^5\equiv 1\mod 5$)

Abhra Abir Kundu
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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, $m\ge 1$

lab bhattacharjee
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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

Amr
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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]

eccstartup
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    It is my strong opinion that numerical evidence should not be presented as a "proof" that something holds for all natural numbers $n$. Therefore, I consider your effort to fail in answering the question. – Lord_Farin Jun 15 '13 at 15:12
  • @Lord_Farin , Could you please help me solve $12737*x \equiv 1\pmod {1000000}$ IN what you mean by a "proof"? I was helping to give evidence for Sherloek holmes, hoping that he can have a knowledge of the structure of the solution. – eccstartup Jun 15 '13 at 23:37