Residue of Fibonacci Number Modulo Fibonacci Number/Lemma

Theorem
Let $F_n$ denote the $n$th Fibonacci number.

Let $m$ be a non-negative integer.

Then:
 * $F_{m n + 1} \equiv \paren {\begin{cases} F_1 & : m \bmod 4 = 0 \\

F_{n - 1} & : m \bmod 4 = 1 \\ \paren {-1}^n F_1 & : m \bmod 4 = 2 \\ \paren {-1}^n F_{n - 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$

Proof
We prove this by induction on $m$.

For all $m \in \N$, let $\map P m$ be the proposition:
 * $F_{m n + 1} \equiv \paren {\begin{cases} F_1 & : m \bmod 4 = 0 \\

F_{n - 1} & : m \bmod 4 = 1 \\ \paren {-1}^n F_1 & : m \bmod 4 = 2 \\ \paren {-1}^n F_{n - 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$

Basis for the Induction
$\map P 0$ is the case:


 * $F_1 \equiv F_1 \pmod {F_n}$

So $\map P 0$ is true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P m$ is true then it logically follows that $\map P {m + 1}$ is true.

So this is our induction hypothesis:
 * $F_{m n + 1} \equiv \paren {\begin{cases} F_1 & : m \bmod 4 = 0 \\

F_{n - 1} & : m \bmod 4 = 1 \\ \paren {-1}^n F_1 & : m \bmod 4 = 2 \\ \paren {-1}^n F_{n - 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$

Then we need to show:

Induction Step
This is our induction step:

So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.