=== Lemma === {{:Residue of Fibonacci Number Modulo Fibonacci Number/Lemma}}{{qed|lemma}} We prove the result for all $r$ by induction on $r$: For all $r \in \N$, let $\map P r$ be the proposition: :$F_{m n + r} \equiv \paren {\begin{cases} F_r & : m \bmod 4 = 0 \\ \paren {-1}^{r + 1} F_{n - r} & : m \bmod 4 = 1 \\ \paren {-1}^n F_r & : m \bmod 4 = 2 \\ \paren {-1}^{r + 1 + n} F_{n - r} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ === Basis for the Induction === $\map P 0$ is the case: :$F_{m n} \equiv \paren {\begin{cases} F_0 & : m \bmod 4 = 0 \\ \paren {-1} F_n & : m \bmod 4 = 1 \\ \paren {-1}^n F_0 & : m \bmod 4 = 2 \\ \paren {-1}^{1 + n} F_n & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ ::$\; \, \equiv 0 \pmod {F_n}$ This fact follows from Divisibility of Fibonacci Number. $\map P 1$ is the case: :$F_{m n + 1} \equiv \paren {\begin{cases} F_1 & : m \bmod 4 = 0 \\ F_{n - 1} & : m \bmod 4 = 1 \\ \paren {-1}^n F_1 & : m \bmod 4 = 2 \\ \paren {-1}^n F_{n - 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ This is our lemma. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P {k - 1}$ and $\map P {k - 2}$ are true, where $k \ge 2$, then it logically follows that $\map P k$ is true. So this is our induction hypotheses: :$F_{m n + k - 1} \equiv \paren {\begin{cases} F_{k - 1} & : m \bmod 4 = 0 \\ \paren {-1}^k F_{n - k + 1} & : m \bmod 4 = 1 \\ \paren {-1}^n F_{k - 1} & : m \bmod 4 = 2 \\ \paren {-1}^{k + n} F_{n - k + 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ :$F_{m n + k - 2} \equiv \paren {\begin{cases} F_{k - 2} & : m \bmod 4 = 0 \\ \paren {-1}^{k + 1} F_{n - k + 2} & : m \bmod 4 = 1 \\ \paren {-1}^n F_{k - 2} & : m \bmod 4 = 2 \\ \paren {-1}^{k + 1 + n} F_{n - k + 2} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ Then we need to show: :$F_{m n + k} \equiv \paren {\begin{cases} F_k & : m \bmod 4 = 0 \\ \paren {-1}^{k + 1} F_{n - k} & : m \bmod 4 = 1 \\ \paren {-1}^n F_k & : m \bmod 4 = 2 \\ \paren {-1}^{k + 1 + n} F_{n - k} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = F_{m n + k}       | r = F_{m n + k - 1} + F_{m n + k - 2}       | c = {{Defof|Fibonacci Numbers}} }} {{eqn | o = \equiv       | r = \paren {\begin{cases} F_{k - 1} + F_{k - 2} & : m \bmod 4 = 0 \\             \paren {-1}^k F_{n - k + 1} + \paren {-1}^{k + 1} F_{n - k + 2} & : m \bmod 4 = 1 \\             \paren {-1}^n F_{k - 1} + \paren {-1}^n F_{k - 2} & : m \bmod 4 = 2 \\             \paren {-1}^{k + n} F_{n - k + 1} + \paren {-1}^{k + 1 + n} F_{n - k + 2} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}       | c = Induction hypotheses }} {{eqn | o = \equiv       | r = \paren {\begin{cases} F_{k - 1} + F_{k - 2} & : m \bmod 4 = 0 \\             \paren {-1}^{k + 1} \paren {F_{n - k + 2} - F_{n - k + 1} } & : m \bmod 4 = 1 \\             \paren {-1}^n \paren {F_{k - 1} + F_{k - 2} } & : m \bmod 4 = 2 \\             \paren {-1}^{k + 1 + n} \paren {F_{n - k + 2} - F_{n - k + 1} } & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}       | c =  }} {{eqn | o = \equiv       | r = \paren {\begin{cases} F_k & : m \bmod 4 = 0 \\             \paren {-1}^{k + 1} F_{n - k} & : m \bmod 4 = 1 \\             \paren {-1}^n F_k & : m \bmod 4 = 2 \\             \paren {-1}^{k + 1 + n} F_{n - k} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}       | c = {{Defof|Fibonacci Numbers}} }} {{end-eqn}} So $\map P {k - 2} \land \map P {k - 1} \implies \map P k$ and the result follows by the Principle of Mathematical Induction. Therefore: :$F_{m n + r} \equiv \paren {\begin{cases} F_r & : m \bmod 4 = 0 \\ \paren {-1}^{r + 1} F_{n - r} & : m \bmod 4 = 1 \\ \paren {-1}^n F_r & : m \bmod 4 = 2 \\ \paren {-1}^{r + 1 + n} F_{n - r} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$ {{qed}}