Teorema:

Seja

$$n \in \mathbb{N}$$

, então

$$ 3^{3n} \equiv 1 \text{ mod 13} $$

.

Demonstração:

Seja

$$n \in \mathbb{N} $$

, tem-se

$$3^{3n} = (3^3)^n = 27^n = (2 \cdot 13 + 1)^n$$

Pelo Binómio de Newton {% cite Andre2000 -l 40 %} , tem-se: $$

\begin{align*} (2 \cdot 13 + 1)^n &= \sum^n*{k = 0} \binom{n}{k} (2 \cdot 13)^{n - k} \cdot 1^k\newline &= \sum^{n - 1}{k = 0} \binom{n}{k}(2 \cdot 13)^{n - k} \cdot 1^k + 1 \equiv 1 \text{ mod 13} \end{align}

$$

q.e.d.

Bibliografia #

{% bibliography –cited %}

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EDITED:

• 2023-06-02 23:52:30 +0000
  • Change location of EDITED
• 2023-06-02 18:47:00 +0000
  • Fix alignment of q.e.d.
• 2023-06-02 17:21:31 +0000
  • Added Bibliography $$