Theorem: Catalan's Identity

:${F_n}^2 - F_{n - r} F_{n + r} = \left({-1}\right)^{n - r} {F_r}^2$

Proof:
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}}