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: {{begin-eqn}} {{eqn | l = F_{\paren {m + 1} n + 1}       | o = \equiv       | r = \paren {\begin{cases} F_1 & : \paren {m + 1} \bmod 4 = 0 \\             F_{n - 1} & : \paren {m + 1} \bmod 4 = 1 \\             \paren {-1}^n F_1 & : \paren {m + 1} \bmod 4 = 2 \\             \paren {-1}^n F_{n - 1} & : \paren {m + 1} \bmod 4 = 3 \end{cases} } \pmod {F_n} }} {{eqn | o = \equiv       | r = \paren {\begin{cases} F_{n - 1} & : m \bmod 4 = 0 \\             \paren {-1}^n & : m \bmod 4 = 1 \\             \paren {-1}^n F_{n - 1} & : m \bmod 4 = 2 \\             1 & : m \bmod 4 = 3 \end{cases} } \pmod {F_n} }} {{end-eqn}} === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = F_{\paren {m + 1} n + 1}       | r = F_{m n + r - 1} F_n + F_{m n + r} F_{n + 1}       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | o = \equiv       | r = F_{m n + r} F_{n + 1} \quad \pmod {F_n}       | c =  }} {{eqn | o = \equiv       | r = \paren {\begin{cases} F_1 F_{n + 1} & : m \bmod 4 = 0 \\             F_{n - 1} F_{n + 1} & : m \bmod 4 = 1 \\             \paren {-1}^n F_1 F_{n + 1} & : m \bmod 4 = 2 \\             \paren {-1}^n F_{n - 1} F_{n + 1} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}       | c = induction hypothesis }} {{eqn | o = \equiv       | r = \paren {\begin{cases} F_{n + 1} & : m \bmod 4 = 0 \\             F_{n - 1} F_{n + 1} - F_n^2 & : m \bmod 4 = 1 \\             \paren {-1}^n F_{n + 1} & : m \bmod 4 = 2 \\             \paren {-1}^n \paren {F_{n - 1} F_{n + 1} - F_n^2} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}       | c =  }} {{eqn | o = \equiv       | r = \paren {\begin{cases} F_n + F_{n - 1} & : m \bmod 4 = 0 \\             F_{n - 1} F_{n + 1} - F_n^2 & : m \bmod 4 = 1 \\             \paren {-1}^n \paren {F_n + F_{n - 1} } & : m \bmod 4 = 2 \\             \paren {-1}^n \paren {F_{n - 1} F_{n + 1} - F_n^2} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}       | c = {{Defof|Fibonacci Numbers}} }} {{eqn | o = \equiv       | r = \paren {\begin{cases} F_n + F_{n - 1} & : m \bmod 4 = 0 \\             \paren {-1}^n & : m \bmod 4 = 1 \\             \paren {-1}^n \paren {F_n + F_{n - 1} } & : m \bmod 4 = 2 \\             \paren {-1}^n \paren {-1}^n & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}       | c = Cassini's Identity }} {{eqn | o = \equiv       | r = \paren {\begin{cases} F_{n - 1} & : m \bmod 4 = 0 \\             \paren {-1}^n & : m \bmod 4 = 1 \\             \paren {-1}^n F_{n - 1} & : m \bmod 4 = 2 \\             1 & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}       | c =  }} {{end-eqn}} So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}} Category:Proofs by Induction Category:Residue of Fibonacci Number Modulo Fibonacci Number 5pvan7n5g8ac5azn1lhli07uuh3jwxt