Proof by induction: For all $n, r \in \N_{>0}$ where $n > r$, let $\map P {n, r}$ be the proposition: :${F_n}^2 - F_{n - r} F_{n + r} = \paren {-1}^{n - r} {F_r}^2$ === Basis for the Induction === $n = 1$ yields no suitable $r$, so we look at $n = 2$ instead, which only gives us $r = 1$. $\map P {2, 1}$ is true: :${F_2}^2 - F_3 F_1 = 1^2 - 2 \times 1 = -1 = -1 \times {F_1}^2$ $n = 3$ gives us only $r = 1$ and $r = 2$. $\map P {3, 1}$ is true: :${F_3}^2 - F_2 F_4 = 2^2 - 1 \times 3 = 1 = 1 \times {F_1}^2$ $\map P {3, 2}$ is true: :${F_3}^2 - F_1 F_5 = 2^2 - 1 \times 5 = -1 = -1 \times {F_2}^2$ This is our basis for the induction. === First Induction Hypothesis === Now we need to show that, if $\map P {n, r}$ is true for all $r$, where $n > 3$, then it logically follows that $\map P {n + 1, r}$ is true for all $r$. So this is our induction hypothesis: :$\forall r < n : {F_n}^2 - F_{n - r} F_{n + r} = \paren {-1}^{n - r} {F_r}^2$ Then we need to show: :$\forall r < n : {F_{n + 1} }^2 - F_{n - r + 1} F_{n + r + 1} = \paren {-1}^{n - r + 1} {F_r}^2$ === Induction Step === This is our induction step: It will again be a proof by induction. ==== Basis for the Induction ==== When $r = 1$: {{begin-eqn}} {{eqn | l = {F_{n + 1} }^2 - F_n F_{n + 2}       | r = {F_{n + 1} }^2 - F_n \paren {F_{n + 1} + F_n}       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = {F_{n + 1} }^2 - F_n F_{n + 1} - {F_n}^2       | c =  }} {{eqn | r = F_{n + 1} \paren {F_{n + 1} - F_n} - {F_n}^2       | c =  }} {{eqn | r = F_{n + 1} F_{n - 1} - {F_n}^2       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \paren {-1} \paren {F_n^2 - F_{n - 1} F_{n + 1} }       | c =  }} {{eqn | r = \paren {-1} \paren {-1}^{n - 1} {F_1}^2       | c = First induction hypothesis }} {{eqn | r = \paren {-1}^n {F_1}^2       | c =  }} {{end-eqn}} So $\map P {n + 1, 1}$ holds. This is our basis for the induction. ==== Second Induction Hypothesis ==== Now we need to show that, if $\map P {n + 1, r}$ is true, where $2 < r < n$, then it logically follows that $\map P {n + 1, r + 1}$ is true. So this is our second induction hypothesis: :${F_{n + 1} }^2 - F_{n - r + 1} F_{n + r + 1} = \paren {-1}^{n - r + 1} {F_r}^2$ Then we need to show: :${F_{n + 1} }^2 - F_{n - r} F_{n + r + 2} = \paren {-1}^{n - r} {F_{r + 1} }^2$ ==== Induction Step ==== This is our induction step: {{begin-eqn}} {{eqn | l = {F_{n + 1} }^2 - F_{n - r} F_{n + r + 2}       | r = {F_{n + 1} }^2 - F_{n - r + 1} F_{n + r + 1} + F_{n - r + 1} F_{n + r + 1} - F_{n - r} F_{n + r + 2}       | c =  }} {{eqn | r = \paren {-1}^{n - r + 1} {F_r}^2 + F_{n - r + 1} F_{n + r + 1} - F_{n - r} F_{n + r + 2}       | c = Second induction hypothesis }} {{eqn | r = \paren {-1}^{n - r + 1} {F_r}^2 + F_{n - r + 1} F_{n + r + 1} - F_{n - r} \paren {F_{n + r} + F_{n + r + 1} }       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \paren {-1}^{n - r + 1} {F_r}^2 + \paren {F_{n - r + 1} - F_{n - r} } F_{n + r + 1} - F_{n - r} F_{n + r}       | c =  }} {{eqn | r = \paren {-1}^{n - r + 1} {F_r}^2 + F_{n - r - 1} F_{n + r + 1} - F_{n - r} F_{n + r}       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \paren {-1}^{n - r + 1} {F_r}^2 - F_{n - r} F_{n + r} + F_{n - r - 1} F_{n + r + 1}       | c =  }} {{eqn | r = \paren {-1}^{n - r + 1} {F_r}^2 + {F_n}^2 - F_{n - r} F_{n + r} - {F_n}^2 + F_{n - r - 1} F_{n + r + 1}       | c =  }} {{eqn | r = \paren {-1}^{n - r + 1} {F_r}^2 + \paren {-1}^{n - r} {F_r}^2 - \paren {-1}^{n - r + 1} {F_{r + 1} }^2       | c = First induction hypothesis }} {{eqn | r = \paren {-1}^{n - r + 1} {F_r}^2 + \paren {-1}^{n - r} {F_r}^2 + \paren {-1}^{n - r} {F_{r + 1} }^2       | c =  }} {{eqn | r = \paren {-1}^{n - r} \paren {- {F_r}^2 + {F_r}^2} + \paren {-1}^{n - r} {F_{r + 1} }^2       | c =  }} {{eqn | r = \paren {-1}^{n - r} {F_{r + 1} }^2       | c =  }} {{end-eqn}} So $\map P {n + 1, r} \implies \map P {n + 1, r + 1}$ and the result follows by the Principle of Mathematical Induction. So $\map P {n, r} \implies \map P {n + 1, r}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$F_n^2 - F_{n - r} F_{n + r} = \paren {-1}^{n - r} {F_r}^2$ {{qed}}