hcju/proofwikiqa huggingface dataset:
We have: {{begin-eqn}} {{eqn | n = 1       | o =        | r = \left({x^{n + k} - y^{n + k} }\right) \left({x^{m - k} - y^{m - k} }\right) - \left({x^n - y^n}\right) \left({x^m - y^m}\right)       | c =  }} {{eqn | r = x^{n + m} + y^{n + m} - x^{m - k} y^{n + k} - x^{n + k} y^{m - k} - x^{n + m} - y^{n + m} + x^n y^m + x^m y^n       | c =  }} {{eqn | r = x^n y^m + x^m y^n - x^{m - k} y^{n + k} - x^{n + k} y^{m - k}       | c =  }} {{eqn | r = x^n y^n \left({y^{m - n} + x^{m - n} - x^{m - n - k} y^k - x^k y^{m - n - k} }\right)       | c =  }} {{eqn | r = x^n y^n \left({x^{m - n - k} \left({x^k - y^k}\right) - y^{m - n - k} \left({x^k - y^k}\right)}\right)       | c =  }} {{eqn | n = 2       | r = \left({x y}\right)^n \left({x^{m - n - k} - y^{m - n - k} }\right) \left({x^k - y^k}\right)       | c =  }} {{end-eqn}} Now substitute: {{begin-eqn}} {{eqn | l = x       | r = \phi       | c =  }} {{eqn | l = y       | r = \hat \phi       | c =  }} {{end-eqn}} where: : $\phi$ denotes the golden mean : $\hat \phi = 1 - \phi$ first into $(2)$: {{begin-eqn}} {{eqn | o =        | r = \left({\phi \hat \phi}\right)^n \left({\phi^{m - n - k} - \hat \phi^{m - n - k} }\right) \left({\phi^k - \hat \phi^k}\right)       | c =  }} {{eqn | r = \left({-1}\right)^n \left({\phi^{m - n - k} - \hat \phi^{m - n - k} }\right) \left({\phi^k - \hat \phi^k}\right)       | c = Reciprocal Form of One Minus Golden Mean: $\hat \phi = -\dfrac 1 \phi$ }} {{eqn | r = \left({-1}\right)^n F_{m - n - k} F_k \times \left({\sqrt 5}\right)^2       | c = Euler-Binet Formula }} {{end-eqn}} and then into $(1)$: {{begin-eqn}} {{eqn | o =        | r = \left({\phi^{n + k} - \hat \phi^{n + k} }\right) \left({\phi^{m - k} - \hat \phi^{m - k} }\right) - \left({\phi^n - \hat \phi^n}\right) \left({\phi^m - \hat \phi^m}\right)       | c =  }} {{eqn | r = F_{n + k} F_{m - k} - F_n F_m \times \left({\sqrt 5}\right)^2       | c = Euler-Binet Formula }} {{end-eqn}} $(1) = (2)$ and hence the result. {{qed}} {{Namedfor|Steven Vajda}}
{{begin-eqn}} {{eqn | l = {F_n}^2 + {F_{n + 1} }^2       | r = F_{n + 1} F_{n - 1} - \paren {-1}^n + F_{n + 2} F_n - \paren {-1}^{n + 1}       | c = Cassini's Identity }} {{eqn | r = F_n F_{n + 2} + F_{n - 1} F_{n + 1}       | c =  }} {{eqn | r = F_{n + n + 1}       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = F_{2 n + 1}       | c =  }} {{end-eqn}} {{qed}}
{{tidy}} Let $X$ be a non-empty set. Let $X$ contain at least two elements; otherwise, $X$ can be trivially well-ordered. We use the Hausdorff Maximal Principle to construct a well-ordering on $X$. We note that there exists at least the following strict well-orderings on subsets of $X$:  For any singleton $\{x\}$, $x$ is vacuously well-ordered with respect to every other element in the subset, of which there are none. Any doubleton $\left\{ { y,z } \right\}$ can be strictly well-ordered by defining $x < y$ and $y \not < x$. Let $\mathcal A$ be the set of all ordered pairs $\left({A,<}\right)$ such that $A$ is a subset of $X$ and $<$ is a strict well-ordering of $A$. Define $\prec$ as: :$\left({A,<}\right) \prec \left({A',<'}\right)$ {{iff}}  :$\left({A,<}\right)$ equals an initial segment of $\left({A',<'}\right)$. Let $\mathcal B$ be a set of ordered pairs in $\mathcal A$ such that $\mathcal B$ is ordered by $\prec$. Let $B'$ be the union of the sets $B$ for all $\left({B,<}\right) \in \mathcal B$. Let $<'$ be the union of the relations $<$ for all $\left({B,<}\right)$. By the lemma, $\left({B',<'}\right)$ is strictly well-ordered set. By the Hausdorff Maximal Principle, there exists a maximal chain defined by $<'$ imposed on a maximal subset of $X$. We claim that this maximal chain is imposed on the entirety of $X$. {{AimForCont}} that $\left({B',<'}\right)$ is a maximal chain but $B' \ne X$. Then consider $B' \cup \{ x \}$., with $x \in X \setminus B'$, with the extension of $<'$ such that: :$\forall a \in B': a <' x$.  Then $B' = S_x$, the initial segment in $B' \cup \{x\}$ determined by $x$. But by the definition of $\prec$, we then have: :$ \left({B'\cup \{x\},\text{extension of }<'}\right) \prec \left({B',<'}\right)$ But then by the definition of $B'$, we would have $B' \cup \{x\} \subseteq B'$, contrary to the assumption that $x \in X \setminus B'$. From this contradiction infer that $B' = X$. That is, all of $X$ can be well-ordered. {{qed}}
Let $u = \cos 72 \degrees, v = \cos 36 \degrees$. {{begin-eqn}} {{eqn | l = u       | r = 2 v^2 - 1       | c = Double Angle Formula for Cosine: Corollary 1 }} {{eqn | l = \cos 36 \degrees       | r = 1 - 2 \sin^2 18 \degrees       | c = Double Angle Formula for Cosine: Corollary 2 }} {{eqn | r = 1 - 2 u^2       | c = Cosine of Complement equals Sine }} {{eqn | ll= \leadsto       | l = u + v       | r = 2 \paren {v^2 - u^2}       | c =  }} {{eqn | r = 2 \paren {v + u} \paren {v - u}       | c =  }} {{eqn | ll= \leadsto       | l = 1       | r = 2 \paren {v - u}       | c =  }} {{eqn | r = 2 v - 4 v^2 + 2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {2 v}^2       | r = 2 v + 1       | c =  }} {{eqn | ll= \leadsto       | l = 2 v       | r = \phi       | c = Square of Golden Mean equals One plus Golden Mean }} {{end-eqn}} {{qed}}
From Complex Algebra: $z^4 - 3z^2 + 1 = 0$, the roots of $z^4 - 3z^2 + 1 = 0$ are: :$2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$ Then: {{begin-eqn}} {{eqn | l = z^4 - 3z^2 + 1       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = z^2       | r = \dfrac {3 \pm \sqrt {\paren {-3}^2 - 4 \times 1} } 2       | c = Quadratic Formula }} {{eqn | r = \dfrac 3 2 \pm \dfrac {\sqrt 5} 2       | c =  }} {{end-eqn}} Then let $z = a + \sqrt b$: {{begin-eqn}} {{eqn | l = \paren {a + \sqrt b}^2       | r = \dfrac 3 2 + \dfrac {\sqrt 5} 2       | c =  }} {{eqn | ll= \leadsto       | l = a^2 + b + 2 a \sqrt b       | r = \dfrac 3 2 + \dfrac {\sqrt 5} 2       | c =  }} {{eqn | ll= \leadsto       | l = a^2 + b       | r = \dfrac 3 2       | c =  }} {{eqn | l = 2 a \sqrt b       | r = \dfrac {\sqrt 5} 2       | c =  }} {{eqn | ll= \leadsto       | l = a       | r = \dfrac {\sqrt 5} {4 \sqrt b}       | c =  }} {{eqn | ll= \leadsto       | l = a^2       | r = \dfrac 5 {16 b}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 5 {16 b} + b       | r = \dfrac 3 2       | c =  }} {{eqn | ll= \leadsto       | l = 5 + 16 b^2       | r = 24 b       | c =  }} {{eqn | ll= \leadsto       | l = 16 b^2 - 24 b + 5       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = b       | r = \dfrac {24 \pm \sqrt {24^2 - 4 \times 5 \times 16} } {32}       | c =  }} {{eqn | r = \dfrac {2^3 \times 3 \pm \sqrt {\paren {2^3 \times 3}^2 - 2^6 \times 5} } {2^5}       | c =  }} {{eqn | r = \dfrac 3 4 \pm \dfrac {2^3 \sqrt {3^2 - 5} } {2^5}       | c =  }} {{eqn | r = \dfrac 3 4 \pm \dfrac {\sqrt {9 - 5} } 4       | c =  }} {{eqn | r = \dfrac 3 4 \pm \dfrac 2 4       | c =  }} {{eqn | r = \dfrac {3 \pm 2} 4       | c =  }} {{eqn | r = \dfrac 5 4 \text { or } \dfrac 1 4       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt b       | r = \dfrac {\sqrt 5} 2 \text { or } \dfrac 1 2       | c =  }} {{eqn | ll= \leadsto       | l = a       | r = \dfrac 3 2 - \dfrac {\sqrt 5} 2 \text { or } \dfrac 3 2 - \dfrac 1 2       | c =  }} {{end-eqn}} {{finish|Getting nowhere. Getting bored with it. Shouldn't try to do maths when "tired".}}
{{begin-eqn}} {{eqn | l = \sin 36^\circ       | r = \sqrt {1 - \cos^2 36^\circ}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \sqrt {1 - \frac {\phi^2} 4}       | c = Cosine of $36^\circ$ }} {{eqn | r = \frac 1 2 \sqrt {4 - \phi^2}       | c =  }} {{eqn | r = \frac 1 2 \sqrt {4 - \paren {\frac 1 2 + \frac {\sqrt 5} 2}^2}       | c = {{Defof|Golden Mean|index = 2}} }} {{eqn | r = \frac 1 2 \sqrt {4 - \paren {\frac 1 4 + \frac 5 4 + 2 \cdot \frac 1 2 \cdot \frac {\sqrt 5} 2} }       | c = Square of Sum }} {{eqn | r = \frac 1 2 \sqrt {4 - \paren {\frac 3 2 + \frac {\sqrt 5} 2} }       | c =  }} {{eqn | r = \frac 1 2 \sqrt {\frac 5 2 - \frac {\sqrt 5} 2}       | c =  }} {{eqn | r = \sqrt {\frac 5 8 - \frac {\sqrt 5} 8}       | c =  }} {{end-eqn}} We also have: {{begin-eqn}} {{eqn | l = \frac {\sqrt {\sqrt 5 / \phi} } 2       | r = \frac 1 2 \sqrt {\sqrt 5 \paren {\phi - 1} }       | c = {{Defof|Golden Mean|index = 3}}  }} {{eqn | r = \frac 1 2 \sqrt {\sqrt 5 \paren {\frac 1 2 + \frac {\sqrt 5} 2 - 1} }       | c = {{Defof|Golden Mean|index = 2}}  }} {{eqn | r = \frac 1 2 \sqrt {\frac 5 2 - \frac {\sqrt 5} 2} }} {{eqn | r = \sqrt {\frac 5 8 - \frac {\sqrt 5} 8} }} {{end-eqn}} {{qed}}
Multiplying the summation by $x^2 + x - 1$: {{begin-eqn}} {{eqn | r = \sum_{k \mathop = 0}^n F_k x^k \left({x^2 + x - 1}\right)       | o =        | c =  }} {{eqn | r = F_0 x^2 + F_1 x^3 + F_2 x^4 + \cdots + F_{n - 2} x^n + F_{n - 1} x^{n + 1} + F_n x^{n + 2}       | c =  }} {{eqn | o =        | ro= +       | r = F_0 x + F_1 x^2 + F_2 x^3 + \cdots + F_{n - 2} x^{n - 1} + F_{n - 1} x^n + F_n x^{n + 1}       | c =  }} {{eqn | o =        | ro= -       | r = F_0 - F_1 x - F_2 x^2 - \cdots - F_{n - 2} x^{n - 2} - F_{n - 1} x^{n - 1} - F_n x^n       | c =  }} {{eqn | r = F_0 + \left({F_0 - F_1}\right) x + \left({F_{n - 1} + F_n}\right) x^{n + 1} + F_n x^{n + 2}       | c = as, in general, $\left({F_{j - 2} + F_{j - 1} - F_j}\right) x^j = 0$ by {{Defof|Fibonacci Number}} }} {{eqn | r = 0 + \left({-1}\right) x + F_{n + 1} x^{n + 1} + F_n x^{n + 2}       | c = {{Defof|Fibonacci Number}}: $F_0 = 0, F_1 = 1$ }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^n F_k x^k       | r = \dfrac {F_{n + 1} x^{n + 1} + F_n x^{n + 2} - x} {x^2 + x - 1}       | c =  }} {{end-eqn}} If the denominator is $0$, then $x = \dfrac 1 \phi$ or $x = \dfrac 1 {\hat \phi}$ and the numerator is $0$ also. Thus we can differentiate the numerator and denominator {{WRT|Differentiation}} $x$ and use L'Hôpital's Rule: {{begin-eqn}} {{eqn | r = \dfrac {\dfrac \d {\d x} \left({F_{n + 1} x^{n + 1} + F_n x^{n + 2} - x}\right)} {\dfrac \d {\d x} \left({x^2 + x - 1}\right)}       | o =        | c =  }} {{eqn | r = \dfrac {\left({n + 1}\right) F_{n + 1} x^n + \left({n + 2}\right) F_n x^{n + 1} - 1} {2 x + 1}       | c = Power Rule for Derivatives }} {{end-eqn}} Hence the result. {{qed}}
From Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k: :$(1): \quad: \displaystyle \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$ Letting $t = 2$ in $(1)$: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \binom n k {F_2}^k {F_{2 - 1} }^{n - k} F_{m + k}       | r = \sum_{k \mathop \ge 0} \binom n k 1^k 1^{n - k} F_{m + k}       | c = {{Defof|Fibonacci Number}}: $F_1 = 1, F_2 = 1$ }} {{eqn | r = \sum_{k \mathop \ge 0} \binom n k F_{m + k}       | c =  }} {{eqn | r = F_{m + 2 n}       | c = from $(1)$ }} {{end-eqn}} {{qed}}
The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\forall m, t \in \N: \displaystyle \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$ $\map P 0$ is the case: :$\displaystyle \binom 0 0 {F_t}^0 {F_{t - 1} }^0 F_{m + 0} = F_m$ Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \binom 1 0 {F_t}^0 {F_{t - 1} }^1 F_{m + 0} + \binom 1 1 {F_t}^1 {F_{t - 1} }^0 F_{m + 1}       | r = F_{t - 1} F_m + F_t F_{m + 1}       | c =  }} {{eqn | r = F_{m + t}       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P q$ is true, where $q \ge 1$, then it logically follows that $\map P {q + 1}$ is true. So this is the induction hypothesis: :$\forall m, t \in \N: \displaystyle \sum_{k \mathop \ge 0} \binom q k {F_t}^k {F_{t - 1} }^{q - k} F_{m + k} = F_{m + t q}$ from which it is to be shown that: :$\forall m, t \in \N: \displaystyle \sum_{k \mathop \ge 0} \binom {q + 1} k {F_t}^k {F_{t - 1} }^{q + 1 - k} F_{m + k} = F_{m + t \paren {q + 1} }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = F_{m + t \paren {q + 1} }       | r = F_{m + t q} F_{t - 1} + F_{m + t q + 1} F_t       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = \sum_{k \mathop \ge 0} \binom q k {F_t}^k {F_{t - 1} }^{q - k} F_{m + k} F_{t - 1} + \sum_{k \mathop \ge 0} \binom q k {F_t}^k {F_{t - 1} }^{q - k} F_{m + 1 + k} F_t       | c = Induction Hypothesis }} {{eqn | r = \sum_{k \mathop \ge 0} \binom q k {F_t}^k {F_{t - 1} }^{q + 1 - k} F_{m + k} + \sum_{k \mathop \ge 0} \binom q k {F_t}^{k + 1} {F_{t - 1} }^{q - k} F_{m + 1 + k}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \binom q k {F_t}^k {F_{t - 1} }^{q + 1 - k} F_{m + k} + \sum_{k \mathop \ge 1} \binom q {k - 1} {F_t}^k {F_{t - 1} }^{q - \paren {k - 1} } F_{m + k}       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_{k \mathop \ge 0} \binom q k {F_t}^k {F_{t - 1} }^{q + 1 - k} F_{m + k} + \sum_{k \mathop \ge 0} \binom q {k - 1} {F_t}^k {F_{t - 1} }^{q - k + 1} F_{m + k}       | c = $\dbinom q {-1} = 0$ }} {{eqn | r = \sum_{k \mathop \ge 0} \paren {\binom q k + \binom q {k - 1} } {F_t}^k {F_{t - 1} }^{q + 1 - k} F_{m + k}       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} \binom {q + 1} k {F_t}^k {F_{t - 1} }^{q + 1 - k} F_{m + k}       | c = Pascal's Rule }} {{end-eqn}} So $\map P q \implies \map P {q + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n, t \in \N: \displaystyle \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$ {{qed}}
We note that: :$D = \begin{vmatrix}   1 & b_{12} & \cdots & b_{1n} \\   0 & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\   0 & b_{n2} & \cdots & b_{nn} \end{vmatrix}$ is the transpose of: :$D^\intercal = \begin{vmatrix}   1 & 0 & \cdots & 0 \\   b_{12} & b_{22} & \cdots & b_{n2} \\ \vdots & \vdots & \ddots & \vdots \\   b_{1n} & b_{2n} & \cdots & b_{nn} \end{vmatrix}$ From Determinant with Unit Element in Otherwise Zero Row: :$D^\intercal = \begin{vmatrix}   b_{22} & \cdots & b_{n2} \\ \vdots & \ddots & \vdots \\   b_{2n} & \cdots & b_{nn} \end{vmatrix}$ The result follows by Determinant of Transpose. {{Qed}}
Let $X$ be a non-empty set. Let $X$ contain at least two elements; otherwise, any non-empty ordering on $X$ is trivially a maximal chain. By the Well-Ordering Theorem, $X$ can be well-ordered. Fix such a well-ordering. Let $\le$ be any ordering on $X$. Let $P\left({a,Y}\right)$ be the predicate: ''$a$ is $\le$-comparable to every $y \in Y$.'' (Here $a$ and $Y$ are bound variables.) That is, $P\left({a,Y}\right)$ holds if, for every $y \in Y$, $a \le y$ or $y \le a$. Define the mapping: :$\rho\left({f: S_x \to \mathcal P(X) }\right) = \begin{cases} f[S_x]  \cup \{ x \} & P(x,S_x) \\ f[S_x] & \text{ otherwise }  \end{cases}$ where $S_x$ is the initial segment in $X$ defined by $x$, and $f[\,\cdot\,]$ denotes an image set. Using the Principle of Recursive Definition for Well-Ordered Sets, we can use $\rho$ to uniquely define: :$h: X \to \mathcal P(X)$:  :$h(\alpha) = \rho\left({ h {\restriction_{S_\alpha}} }\right) = \begin{cases}  h[S_\alpha] \cup \{\alpha\} & P(\alpha,S_\alpha) \\ h[S_\alpha] & \text{ otherwise } \end{cases}$ Then $h(\alpha)$ is a $\le$-totally ordered set, by virtue of the construction of $h$ in terms of the predicate $P$. Similarly, $h[S_\alpha]$ is totally ordered, for the same reason. Then the union: :$Z = \displaystyle \bigcup_{\alpha \mathop \in X} h(\alpha)$ admits an ordered sum in terms of $\le$ imposed on each $h(\alpha)$. By Ordered Sum of Tosets is Totally Ordered Set, this is a totally ordered set. In particular, it is a chain of $\left({X,\le}\right)$. We claim this chain is maximal. To see this, consider $\displaystyle x_0 \notin \displaystyle \bigcup_{\alpha \mathop \in X} h(\alpha)$. Then $x_0$ is not $\le$-comparable to the elements of $Z$, because $\neg P\left({x,Z}\right)$. That is, there are no proper supersets of $Z$ that are totally ordered. Therefore the ordered sum on $Z$ is a maximal chain in $\left({X,\le}\right)$. {{qed}}
From Fibonacci Number by Power of 2: :$2^p F_p = 2 \displaystyle \sum_{k \mathop \in \Z} \dbinom p {2 k + 1} 5^k$ From Binomial Coefficient of Prime: :$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$ and from Fermat's Little Theorem: Corollary 1: :$2^p \equiv 2 \pmod p$ Hence: {{begin-eqn}} {{eqn | l = 2^p F_p       | o = \equiv       | r = 2 F_p       | rr= \pmod p       | c =  }} {{eqn | ll= \leadsto       | l = 2 \sum_{k \mathop \in \Z} \dbinom p {2 k + 1} 5^k       | o = \equiv       | r = 2 \times 5^n       | c = as all $\dbinom p k \equiv 0 \pmod p$ except $\dbinom p p$ }} {{eqn | ll= \leadsto       | l = 2^p F_p       | o = \equiv       | r = 2 \times 5^n       | rr= \pmod p       | c =  }} {{eqn | ll= \leadsto       | l = F_p       | o = \equiv       | r = 5^n       | rr= \pmod p       | c = because $p \perp 2$ }} {{end-eqn}} {{qed}}
First we consider the edge cases: For $p = 2$ we have that: :$F_{2 + 1} = F_3 = 2$ which is a multiple of $2$. :$F_{2 - 1} = F_1 = 1$ which is not a multiple of $2$. For $p = 3$ we have that: :$F_{3 + 1} = F_4 = 3$ which is a multiple of $3$. :$F_{3 - 1} = F_2 = 1$ which is not a multiple of $3$. Let $p$ be an odd prime. From Cassini's Identity: :$(1): \quad F_{p - 1} F_{p + 1} - {F_p}^2 = \left({-1}\right)^p = -1$ Then: {{begin-eqn}} {{eqn | l = F_p       | o = \equiv       | r = 5^{\left({p - 1}\right) / 2}       | rr= \pmod p       | c = Fibonacci Number with Prime Index 2n+1 is Congruent to 5^n Modulo p }} {{eqn | ll= \leadsto       | l = {F_p}^2       | o = \equiv       | r = 5^{p - 1}       | rr= \pmod p       | c =  }} {{eqn | ll= \leadsto       | l = {F_p}^2       | o = \equiv       | r = 1       | rr= \pmod p       | c = Fermat's Little Theorem, as long as $p \ne 5$ }} {{eqn | ll= \leadsto       | l = F_{p - 1} F_{p + 1}       | o = \equiv       | r = 0       | rr= \pmod p       | c = from $(1)$ above }} {{end-eqn}} Hence either $F_{p - 1}$ or $F_{p + 1}$ is a multiple of $p$. But as $F_{p + 1} = F_p + F_{p - 1}$, only one of them can be. Finally, note that if $p = 5$: :$F_{5 + 1} = F_6 = 8$ which is a multiple of $5$. :$F_{5 - 1} = F_4 = 3$ which is not a multiple of $5$. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$a_n = \begin{cases} \dfrac {A^n - B^n} {A - B} & : A \ne B \\ n A^{n - 1} & : A = B \end{cases}$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \dfrac {A^0 - B^0} {A - B}       | r = \dfrac {1 - 1} {A - B}       | c =  }} {{eqn | r = \dfrac 0 {A - B}       | c =  }} {{eqn | r = 0       | c =  }} {{eqn | r = a_0       | c =  }} {{end-eqn}} When $A = B$: {{begin-eqn}} {{eqn | l = 0 A^{-1}       | r = 0       | c =  }} {{eqn | r = a_0       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \dfrac {A^1 - B^1} {A - B}       | r = \dfrac {A - B} {A - B}       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = 0 \times A + 1       | c =  }} {{eqn | r = A a_0 + B^0       | c =  }} {{eqn | r = a_1       | c =  }} {{end-eqn}} When $A = B$: {{begin-eqn}} {{eqn | l = 1 \times  A^0       | r = 1       | c =  }} {{eqn | r = A \times 0 + 1       | c =  }} {{eqn | r = A \times a_0 + A^0       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. This is the induction hypothesis: :$a_k = \begin{cases} \dfrac {A^k - B^k} {A - B} & : A \ne B \\ k A^{k - 1} & : A = B \end{cases}$ from which it is to be shown that: :$a_{k + 1} = \begin{cases} \dfrac {A^{k + 1} - B^{k + 1} } {A - B} & : A \ne B \\ \left({k + 1}\right) A^k & : A = B \end{cases}$ === Induction Step === This is the induction step: First let $A \ne B$. {{begin-eqn}} {{eqn | l = a_{k + 1}       | r = A a_k + B^k       | c = by definition }} {{eqn | r = A \dfrac {A^k - B^k} {A - B} + B^k       | c = Induction Hypothesis }} {{eqn | r = \dfrac {A \left({A^k - B^k}\right) + \left({A - B}\right) B^k} {A - B}       | c =  }} {{eqn | r = \dfrac {A^{k + 1} - A B^k + A B^k - B^{k + 1} } {A - B}       | c =  }} {{eqn | r = \dfrac {A^{k + 1} - B^{k + 1} } {A - B}       | c =  }} {{end-eqn}} When $A = B$ we have for $k > 0$: {{begin-eqn}} {{eqn | l = a_{k + 1}       | r = A a_k + A^k       | c =  }} {{eqn | r = A \left({k A^{k - 1} }\right) + A^k       | c =  }} {{eqn | r = k A^k + A^k       | c =  }} {{eqn | r = \left({k + 1}\right) A^k       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and by the Principle of Mathematical Induction: :$a_n = \begin{cases} \dfrac {A^n - B^n} {A - B} & : A \ne B \\ n A^{n - 1} & : A = B \end{cases}$ {{qed}}
Recurrence Relation for Fibonomial Coefficients gives: :$\dbinom n k_\mathcal F = F_{k - 1} \dbinom {n - 1} k_\mathcal F + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\mathcal F$ where $F_{k - 1}$ etc. denote Fibonacci numbers. It follows that each Fibonomial coefficient is the sum of integers, and so an integer. {{qed}}
{{begin-eqn}} {{eqn | r = F_{k - 1} \dbinom {n - 1} k_\mathcal F + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\mathcal F       | o =        | c =  }} {{eqn | r = F_{k - 1} \dfrac {F_{n - 1} F_{n - 2} \cdots F_{n - k + 1} F_{n - k} } {F_k F_{k - 1} F_{k - 2} F_{k - 3} \cdots F_1} + F_{n - k + 1} \dfrac {F_{n - 1} F_{n - 2} \cdots F_{n - k + 1} } {F_{k - 1} F_{k - 2} F_{k - 3} \cdots F_1}       | c = {{Defof|Fibonomial Coefficient}} }} {{eqn | r = \dfrac {F_{k - 1} F_{n - 1} F_{n - 2} \cdots F_{n - k + 1} F_{n - k} + F_{n - k + 1} F_k F_{n - k + 1} F_{n - 1} F_{n - 2} \cdots F_{n - k + 1} } {F_k F_{k - 1} F_{k - 2} F_{k - 3} \cdots F_1}       | c =  }} {{eqn | r = F_{n - 1} F_{n - 2} \cdots F_{n - k + 1} \dfrac {F_{k - 1} F_{n - k} + F_{n - k + 1} F_k} {F_k F_{k - 1} F_{k - 2} F_{k - 3} \cdots F_1}       | c =  }} {{eqn | r = F_{n - 1} F_{n - 2} \cdots F_{n - k + 1} \dfrac {F_{\left({n - k}\right) + k} } {F_k F_{k - 1} F_{k - 2} F_{k - 3} \cdots F_1}       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers: $F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$ }} {{eqn | r = \dfrac {F_n F_{n - 1} F_{n - 2} \cdots F_{n - k + 1} } {F_k F_{k - 1} F_{k - 2} F_{k - 3} \cdots F_1}       | c =  }} {{eqn | r = \dbinom n k_\mathcal F       | c = {{Defof|Fibonomial Coefficient}} }} {{end-eqn}} {{qed}}
=== Lemma 1 === {{:Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1}} === Lemma 2 === {{:Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 2}} {{ProofWanted}}
{{begin-eqn}} {{eqn | l = F_{2 n} \phi       | r = \phi \dfrac {\phi^{2 n} - \hat \phi^{2 n} } {\sqrt 5}       | c = Euler-Binet Formula }} {{eqn | r = \dfrac {\phi^{2 n + 1} - \phi \hat \phi^{2 n} } {\sqrt 5}       | c =  }} {{eqn | r = \dfrac {\phi^{2 n + 1} + \hat \phi^{2 n - 1} } {\sqrt 5}       | c = Golden Mean by One Minus Golden Mean equals Minus 1 }} {{end-eqn}} {{finish|probably going in the wrong direction}}