Teorema:

Seja $\Psi: \mathbb{N} \to \mathbb{N}$ uma função definida por

$$ \Psi(n) = \text{mdc}(n + 1, n^2 + 1) $$

, então $$

\Psi(n) = \begin{cases} 1, n \equiv 0 \text{ mod 2} \newline 2, n \equiv 1 \text{ mod 2} \end{cases}

$$

Demonstração:

Seja $n \in \mathbb{N} $, tem-se

$$n^2 + 1 = n^2 - 1 + 2 = (n + 1)(n - 1) + 2 \equiv 2 \text{ mod } n + 1$$

Pelo lema de Euclides {% cite Martinez -L Lema -l Lema 1.4 %}, este resultado implica que

$$ \Psi(n) = \text{mdc}(n+1, n^2 + 1) = \text{mdc}(n + 1, 2).$$

Logo,

$$

\begin{align*} \forall n \in \mathbb{N}, \exists k \in \mathbb{N}: n + 1 &= \begin{cases} 2k + 1, n \equiv 0 \text{ mod 2} \newline 2k, n \equiv 1 \text{ mod 2} \end{cases} \newline \implies \Psi(n) &= \begin{cases} \text{mdc}(2k + 1, 2), n \equiv 0 \text{ mod 2}\newline \text{mdc}(2k, 2), n \equiv 1 \text{ mod 2} \end{cases} \newline &= \begin{cases} 1, n \equiv 0 \text{ mod 2}\newline 2, n \equiv 1 \text{ mod 2} \end{cases} \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 17:18:31 +0000

• Added Bibliography $$