127.0.0.1

3^{3n} = 1 mod 13

tags: [mathematics][discrete-mathematics]

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 %}


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