hcju/proofwikiqa huggingface dataset:
:400px From the General Equation of Straight Line in Plane, $\mathcal L$ can be expressed in the form: :$(1): \quad \alpha_1 x + \alpha_2 y = \beta$ where $\alpha_1, \alpha_2, \beta \in \R$ are given, and not both $\alpha_1, \alpha_2$ are zero. Substituting for the two points whose coordinates we know about: {{begin-eqn}} {{eqn | lo= x = a, y = 0:       | l = \alpha_1 \times a + \alpha_2 \times 0       | r = \beta       | c =  }} {{eqn | ll= \leadsto       | l = \alpha_1       | r = \dfrac \beta a       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | lo= x = 0, y = b:       | l = \alpha_1 \times 0 + \alpha_2 \times b       | r = \beta       | c =  }} {{eqn | ll= \leadsto       | l = \alpha_2       | r = \dfrac \beta b       | c =  }} {{end-eqn}} We know that $\beta \ne 0$ because none of $a, b, \alpha_1, \alpha_2$ are equal to $0$. Hence: {{begin-eqn}} {{eqn | l = \dfrac \beta a x + \dfrac \beta b y       | r = \beta       | c = substituting for $\alpha_1$ and $\alpha_2$ in $(1)$ }} {{eqn | ll= \leadsto       | l = \dfrac x a + \dfrac y a       | r = 1       | c = dividing both sides by $\beta$ }} {{end-eqn}} {{qed}}
:400px Let $A$ be the $x$-intercept of $\mathcal L$. Let $B$ be the $y$-intercept of $\mathcal L$. Let $A = \tuple {a, 0}$ and $B = \tuple {0, b}$. From the Equation of Straight Line in Plane: Two-Intercept Form, $\mathcal L$ can be expressed in the form: :$(1): \quad \dfrac x a + \dfrac y a = 1$ Then: {{begin-eqn}} {{eqn | l = p       | r = a \cos \alpha       | c = {{Defof|Cosine of Angle}} }} {{eqn | ll= \leadsto       | l = a       | r = \dfrac p {\cos \alpha}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = p       | r = b \sin \alpha       | c = {{Defof|Sine of Angle}} }} {{eqn | ll= \leadsto       | l = b       | r = \dfrac p {\sin \alpha}       | c =  }} {{end-eqn}} Substituting for $a$ and $b$ in $(1)$: {{begin-eqn}} {{eqn | l = \dfrac x a + \dfrac y a       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {x \cos \alpha} p + \dfrac {y \sin \alpha} p       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = x \cos \alpha + y \sin \alpha       | r = p       | c =  }} {{end-eqn}} {{Qed}}
We have that $\LL$ has the equation: :$(1): \quad a x + b y + c$ 500px Let a perpendicular be dropped from $P$ to $\LL$ at $Q$. The perpendicular distance $d$ that we are to find is then $PQ$. In order to simplify the algebra that will inevitably follow, we are to make a transformation as follows. Let $\MM$ be constructed parallel to $\LL$. Construct a perpendicular from $\MM$ to pass through the origin. Let this perpendicular intersect $\MM$ at $R$ and $\LL$ at $S$. We have that $PQSR$ is a rectangle, and so $RS = PQ$. It remains to establish the length of $RS$. We can manipulate $(1)$ into slope-intercept form as: :$y = -\dfrac a b x + \dfrac c b$ Thus the slope of $\LL$ is $-\dfrac a b$. From Slope of Orthogonal Curves, the  slope of $RS$ is then $\dfrac b a$. The next step is to find the coordinates of $R$ and $S$. From Equation of Straight Line in Plane: Point-Slope Form, the equation of $\MM$ can be given as: :$y - y_0 = -\dfrac a b \paren {x - x_0}$ or: :$(2): \quad y = \dfrac {-a x + a x_0 + b y_0} b$ From Equation of Straight Line in Plane: Slope-Intercept Form, the equation of $RS$ can be given as: :$(3): \quad y = \dfrac b a x$ $\MM$ and $RS$ intersect where these are equal: :$\dfrac b a x = \dfrac {-a x + a x_0 + b y_0} b$ which gives us: :$x = \dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2}$ Substituting back for $y$ in $3$, we find that: :$R = \tuple {\dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2}, \dfrac {b \paren {a x_0 + b y_0} } {a^2 + b^2} }$ Now to find the coordinates of $S$, which is the intersection of $\LL$ and $RS$. We can express $\LL$ as: :$y = -\dfrac {a x + c} b$ and so: :$\dfrac b a x = -\dfrac {a x + c} b$ which leads to: :$x = -\dfrac {a c} {a^2 + b^2}$ Substituting back for $y$ in $3$, we get (after algebra): :$S = \tuple {\dfrac {-a c} {a^2 + b^2}, \dfrac {-b c} {a^2 + b^2} }$ It remains to find the length $d$ of $RS$. From the Distance Formula: {{begin-eqn}} {{eqn | l = d       | r = \sqrt {\paren {\dfrac {-a c} {a^2 + b^2} - \dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2} }^2 + \paren {\dfrac {-b c} {a^2 + b^2} - \dfrac {b \paren {a x_0 + b y_0} } {a^2 + b^2} }^2 }       | c =  }} {{eqn | r = \sqrt {\dfrac {\paren {-a \paren {a x_0 + b y_0 + c} }^2 + \paren {-b \paren {a x_0 + b y_0 + c} }^2} {\paren {a^2 + b^2}^2 } }       | c =  }} {{eqn | r = \sqrt {\dfrac {\paren {a^2 + b^2} \paren {a x_0 + b y_0 + c}^2} {\paren {a^2 + b^2}^2 } }       | c =  }} {{eqn | r = \sqrt {\dfrac {\paren {a x_0 + b y_0 + c}^2} {a^2 + b^2} }       | c =  }} {{eqn | r = \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }       | c = as length is positive }} {{end-eqn}} {{qed}}
Let $p_1 = \tuple {x_1, y_1}$ and $p_2 = \tuple {x_2, y_2}$ be on $\mathcal L$. Thus: {{begin-eqn}} {{eqn | l = a x_1 - b y_1       | r = c }} {{eqn | l = a x_2 - b y_2       | r = c       | c =  }} {{end-eqn}} From Solution of Linear Diophantine Equation, it is necessary and sufficient that: :$\gcd \set {a, b} \divides c$ for such lattice points to exist. Also from Solution of Linear Diophantine Equation, all lattice points on $\mathcal L$ are solutions to the equation: :$\forall k \in \Z: x = x_1 + \dfrac b m k, y = y_1 - \dfrac a m k$ where $m = \gcd \set {a, b}$. So we have: {{begin-eqn}} {{eqn | l = x_2       | r = x_1 + \dfrac b m k }} {{eqn | l = y_2       | r = y_1 - \dfrac a m k       | c =  }} {{end-eqn}} for some $k \in \Z$ such that $k \ne 0$. The distance $d$ between $p_1$ and $p_2$ is given by the Distance Formula: {{begin-eqn}} {{eqn | l = d       | r = \sqrt {\paren {x_1 - \paren {x_1 + \dfrac b m k} }^2 + \paren {y_1 - \paren {y_1 - \dfrac a m k} }^2}       | c =  }} {{eqn | r = \sqrt {\paren {\dfrac {b k} m}^2 + \paren {\dfrac {a k} m}^2}       | c =  }} {{eqn | r = \sqrt {\dfrac {k^2 \paren {a^2 + b^2} } {m^2} }       | c =  }} {{eqn | r = k \dfrac {\sqrt {a^2 + b^2} } m       | c =  }} {{end-eqn}} This is a minimum when $k = 1$. Hence the result. {{qed}}
Let $n \in \Z$ be of the form $2^2 p q$ where $p$ and $q$ are odd primes. Then: :$n = \paren {2 p} \times \paren {2 q} = 2 \times \paren {2 p q}$ A specific example that can be cited is $n = 60$: :$60 = 6 \times 10$ and: :$60 = 2 \times 30$. Each of $2, 6, 10, 30$ are even-times odd integers: {{begin-eqn}} {{eqn | l = 2       | r = 2 \times 1 }} {{eqn | l = 6       | r = 2 \times 3 }} {{eqn | l = 10       | r = 2 \times 5 }} {{eqn | l = 30       | r = 2 \times 15 }} {{end-eqn}} Every $n \in \Z$ which has a divisor in that same form $2^2 p q$ can similarly be decomposed non-uniquely into even-times odd integers. {{qed}}
{{AimForCont}} there exists $n \in \Z$ which can be decomposed into a power of $2$ and an odd integer in more than one way. That is: :$n = 2^a r$ and: :$n = 2^b s$ where: :$a, b \in \Z_{\ge 0}$ :$r$ and $s$ are odd integers. :either $a \ne b$ or $r \ne s$. Suppose $r = s$. Then: :$\dfrac n r = \dfrac n s = 2^a = 2^b$ which leads to: :$a = b$ which contradicts our supposition that either $r \ne s$ or $a \ne b$. Similarly, suppose $a = b$. Then: :$\dfrac n {2^a} = \dfrac n {2^b} = r = s$ which also contradicts our supposition that either $r \ne s$ or $a \ne b$.   Thus both $r \ne s$ and $a \ne b$. {{WLOG}}, let $a > b$. Then $2^{a - b} = 2^c$ where $c \ge 1$. Thus: :$\dfrac n {2^b} = 2^c r = s$ which means $s = 2 u$ for some $u \in \Z$. Thus $s$ is an even integer. This contradicts our supposition that $r$ and $s$ are both odd. All possibilities have been investigated: :$a = b, r \ne s$ :$a \ne b, r = s$ :$a \ne b, r \ne s$ All have been shown to lead to a contradiction. Thus by Proof by Contradiction it can be seen that an decomposition of $n$ into the product of a power of $2$ and an odd integer is unique. Hence the result. {{qed}}
From Integer is Expressible as Product of Primes, $n$ can be expressed as the product of one or more primes. Let $n = q_1 q_2 \dotsm q_s$ where $q_1, q_2, \ldots, q_s$ are all primes such that: :$(1): \quad n = q_1 \le q_2 \le \dotsb \le q_s$ From Expression for Integer as Product of Primes is Unique, the expression for $(1)$ is unique. By the Fundamental Theorem of Equivalence Relations, we can partition $\set {q_1, q_2, \dotsc, q_s}$ in $(1)$ according to equality. Thus the equivalence classes $\eqclass {q_j} =$ contain all repetitions of $q_j$. Hence the contribution of $q_j$ to $n$ is: :${q_j}^{k_j}$ where $k_j = \card {\eqclass {q_j} =}$, the cardinality of $\eqclass {q_j} =$. Renaming the representative elements of the various $\eqclass {q_r} =$ as $p_1, p_2, \ldots, p_r$, where $r$ is the number of equivalence classes. Hence: :$n = {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k^r}$ As $n = q_1 \le q_2 \le \dotsb \le q_s$ is a unique representation, so is $n = {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k^r}$. {{qed}}
In the prime decompositions $(1)$ and $(2)$, we have that: :$q_1 < q_2 < \dotsb < q_r$ and: :$s_1 < s_2 < \dotsb < s_u$ Hence we can define: {{begin-eqn}} {{eqn | l = E       | r = \set {q_1, q_2, \ldots, q_r} }} {{eqn | l = F       | r = \set {s_1, s_2, \ldots, s_u} }} {{end-eqn}} as all the $q_1, q_2, \dotsc, q_r$ are distinct, and all the $s_1, s_2, \dotsc, s_u$ are distinct. Then let: :$T = E \cup F$ and let the elements of $T$ be renamed as: :$T = \set {t_1, t_2, \ldots, t_v}$ where all the $t_1, t_2, \dotsc, t_v$ are distinct, and: :$t_1 < t_2 < \dotsb < t_v$ Let $\iota: E \to T$ be the inclusion mapping: :$\forall q_i \in E: \map \iota {q_i} = q_i$ Let $\iota: F \to T$ be the inclusion mapping: :$\forall s_i \in F: \map \iota {s_i} = s_i$ Then we have that: {{begin-eqn}} {{eqn | l = a       | r = \prod_{i \mathop = 1}^r {q_i}^{e_i}       | c =  }} {{eqn | r = \prod_{q_i \mathop \in E} {q_i}^{e_i} \times \prod_{t_i \mathop \in T \mathop \setminus E} {t_i}^0       | c =  }} {{eqn | r = \prod_{t_j \mathop \in T} {t_j}^{g_j}       | c = where $g_j = \begin {cases} e_i & : t_j = q_i \\ 0 & : t_j \notin E \end{cases}$ }} {{eqn | r = {t_1}^{g_1} {t_2}^{g_2} \dotsm {t_v}^{g_v}        | c = for some $g_1, g_2, \dotsc, g_v \in \Z_{\ge 0}$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = b       | r = \prod_{i \mathop = 1}^r {s_i}^{f_i}       | c =  }} {{eqn | r = \prod_{s_i \mathop \in F} {s_i}^{f_i} \times \prod_{t_i \mathop \in T \mathop \setminus F} {t_i}^0       | c =  }} {{eqn | r = \prod_{t_j \mathop \in T} {t_j}^{h_j}       | c = where $h_j = \begin {cases} f_i & : t_j = s_i \\ 0 & : t_j \notin F \end{cases}$ }} {{eqn | r = {t_1}^{h_1} {t_2}^{h_2} \dotsm {t_v}^{h_v}        | c = for some $h_1, h_2, \dotsc, h_v \in \Z_{\ge 0}$ }} {{end-eqn}} Thus $a$ and $b$ can be expressed as the product of powers of the same primes, on the understanding that one or more of the powers in either product may be zero. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition: :for all $a_i \in \set {a_1, a_2, \ldots, a_n}$: there exists a set $T = \set {t_1, t_2, \dotsc, t_v}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_v$ such that: ::$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$ === Basis for the Induction === $\map P 2$ is the case: there exist prime numbers $t_1 < t_2 < \dotsb < t_v$ such that: {{begin-eqn}} {{eqn | l = a_1       | r = \prod_{j \mathop = 1}^v {t_j}^{g_{1 j} } }} {{eqn | l = a_2       | r = \prod_{j \mathop = 1}^v {t_j}^{g_{2 j} } }} {{end-eqn}} This has been proved in Expression for Integers as Powers of Same Primes. Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :for all $a_i \in \set {a_1, a_2, \ldots, a_k}$: there exists a set $T = \set {t_1, t_2, \dotsc, t_v}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_v$ such that: ::$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$ from which it is to be shown that: :for all $a_i \in \set {a_1, a_2, \ldots, a_{k + 1} }$: there exists a set $T' = \set {t_1, t_2, \dotsc, t_w}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_w$ such that: ::$\displaystyle a_i = \prod_{j \mathop = 1}^w {t_j}^{g_{i j} }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = a_{k + 1}       | r = \prod_{\substack {q_i \mathop \divides a_{k + 1} \\ \text {$q_i$ is prime} } } {q_i}^{e_i}       | c = {{Defof|Prime Decomposition}} }} {{eqn | l = a_k       | r = \prod_{j \mathop = 1}^v {t_j}^{g_{k j} }       | c = by the induction hypothesis }} {{end-eqn}} Let $E = \set {q_i: q_i \divides a_{k + 1} , \text {$q_i$ is prime} }$. Then let: :$T' = E \cup T$ and let the elements of $T$ be renamed as: :$T' = \set {t_1, t_2, \ldots, t_w}$ where all the $t_1, t_2, \dotsc, t_w$ are distinct, and: :$t_1 < t_2 < \dotsb < t_w$ Then we have that: {{begin-eqn}} {{eqn | l = a_{k + 1}       | r = \prod_{q_i \mathop \in E} {q_i}^{e_i} \times \prod_{t_i \mathop \in T' \mathop \setminus E} {t_i}^0       | c =  }} {{eqn | r = \prod_{t_j \mathop \in T'} {t_j}^{g_j}       | c = where $g_j = \begin {cases} e_i & : t_j = q_i \\ 0 & : t_j \notin E \end{cases}$ }} {{eqn | r = {t_1}^{g_1} {t_2}^{g_2} \dotsm {t_w}^{g_w}        | c = for some $g_1, g_2, \dotsc, g_w \in \Z_{\ge 0}$ }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore, for all $n \in \Z_{\ge 2}$: :for all $a_i \in \set {a_1, a_2, \ldots, a_n}$: there exists a set $T = \set {t_1, t_2, \dotsc, t_v}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_v$ such that: ::$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \map \gcd {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{i j}: \, i \in \N_n} }$ === Basis for the Induction === $\map P 2$ is the case: :$\displaystyle \gcd \set {a_1, a_2} = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{1 j}, e_{2 j} } }$ This is GCD from Prime Decomposition. Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \map \gcd {A_k} = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{i j}: \, i \in \N_k} }$ from which it is to be shown that: :$\displaystyle \map \gcd {A_{k + 1} } = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{i j}: \, i \in \N_{k + 1} } }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \map \gcd {A_{k + 1} }       | r = \map \gcd {A_k \cup a_{k + 1} }       | c =  }} {{eqn | r =        | c =  }} {{end-eqn}} {{finish|I think we need to prove $\gcd \set {a, b, c} {{=}} \gcd \set {\gcd \set {a, b}, c}$ }} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 2}: \displaystyle \map \gcd {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{i j}: \, i \in \N_n} }$
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$ === Basis for the Induction === $\map P 2$ is the case: :$\displaystyle \gcd \set {a_1, a_2} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{1 j}, e_{2 j} } }$ This is GCD from Prime Decomposition. Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \map \lcm {A_k} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_k} }$ from which it is to be shown that: :$\displaystyle \map \lcm {A_{k + 1} } = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_{k + 1} } }$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \map \lcm {A_{k + 1} }       | r = \map \lcm {A_k \cup a_{k + 1} }       | c =  }} {{eqn | r =        | c =  }} {{end-eqn}} {{finish|I think we need to prove $\lcm \set {a, b, c} {{=}} \lcm \set {\lcm \set {a, b}, c}$ }} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 2}: \displaystyle \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$ {{Qed}}
{{begin-eqn}} {{eqn | l = \lcm \set {a, b, c}       | r = \lcm \set {a, \lcm \set {b, c} } }} {{eqn | r = \frac {a \lcm \set {b, c} } {\gcd \set {a, \lcm \set {b, c} } }       | c = Product of GCD and LCM }} {{eqn | r = \frac {a b c} {\gcd \set {b, c} } \paren {\frac 1 {\gcd \set {a, \lcm \set {b, c} } } }       | c = Product of GCD and LCM }} {{eqn | r = \frac {a b c} {\gcd \set {b, c} } \paren {\frac 1 {\lcm \set {\gcd \set {a, b}, \gcd \set {a, c} } } }       | c = GCD and LCM Distribute Over Each Other }} {{eqn | r = \frac {a b c} {\gcd \set {b, c} } \paren {\frac {\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } } {\gcd \set {a, b} \gcd \set {a, c} } }       | c = Product of GCD and LCM }} {{eqn | r = \frac {a b c \gcd \set {a, b, c} } {d_1 d_2 d_3}       | c = by Lemma }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{s \mathop = 0}^n \binom n s \paren {-1}^s a_s       | r = \sum_{s \mathop = 0}^n \binom n s \paren {-1}^s \paren {\sum_{r \mathop = 0}^s \binom s r b_r}       | c =  }} {{eqn | r = \sum_{s \mathop = 0}^n \sum_{\substack {r \mathop \le 0 \mathop \le n \\ s \mathop \ge r} } \binom n s \binom s r \paren {-1}^s b_r       | c =  }} {{eqn | r = \sum_{r \mathop = 0}^n b_r \sum_{s \mathop = r}^n \dfrac {n!} {s! \, \paren{n - s}!} \dfrac {s!} {r! \, \paren{s - r}!} \paren {-1}^s       | c =  }} {{eqn | r = \sum_{r \mathop = 0}^n \dfrac {n!} {r! \, \paren{n - r}!} b_r \sum_{s \mathop = r}^n \dfrac {\paren{n - r}! \paren {-1}^s} {\paren{n - s}! \paren{s - r}!}       | c =  }} {{eqn | r = \sum_{r \mathop = 0}^n \binom n r b_r \sum_{s \mathop = r}^n \binom {n - r} {s - r} \paren {-1}^s       | c =  }} {{eqn | n = 1       | r = \sum_{r \mathop = 0}^n \binom n r b_r \sum_{s \mathop = 0}^{n - r} \binom {n - r} s \paren {-1}^{s + r}       | c =  }} {{end-eqn}} For $n - r > 1$ we can use the Binomial Theorem with $x = 1$ and $y = -1$: :$0 = \paren {1 - 1}^{n - r} = \displaystyle \sum_{s \mathop = 0}^{n \mathop - r} \binom {n - r} s \paren {-1}^s$ Hence all terms of $(1)$ vanish except for where $n - r$. That term is: :$\paren {-1}^n b_n$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$F_{2 n} = \displaystyle \sum_{k \mathop = 1}^n \dbinom n k F_k$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = F_0       | r = 0       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^0 \dbinom 0 k F_k       | c = vacuously }} {{end-eqn}} Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = F_2       | r = 1       | c =  }} {{eqn | r = \dbinom 1 1 F_1       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^1 \dbinom 1 k F_k       | c =  }} {{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 $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true. So this is the induction hypothesis: :$F_{2 m} = \displaystyle \sum_{k \mathop = 1}^m \dbinom m k F_k$ from which it is to be shown that: :$F_{2 \paren {m + 1} } = \displaystyle \sum_{k \mathop = 1}^{m + 1} \dbinom {m + 1} k F_k$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^{m + 1} \dbinom {m + 1} k F_k       | r = \sum_{k \mathop = 1}^m \dbinom {m + 1} k F_k + \dbinom {m + 1} {m + 1} F_{m + 1}       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^m \paren {\dbinom m {k - 1} + \dbinom m k} F_k + F_{m + 1}       | c = Pascal's Rule }} {{eqn | r = \sum_{k \mathop = 1}^m \dbinom m {k - 1} F_k  + F_{2 m} + F_{m + 1}       | c = Induction Hypothesis }} {{eqn | r = \sum_{k \mathop = 1}^m \dbinom m {k - 1} \paren {F_{k - 1} + F_{k - 2} } + F_{2 m} + F_{m + 1}       | c = {{Defof|Fibonacci Numbers}} }} {{end-eqn}} {{finish|Can't see where this is going -- anyone care to have a go at this?}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: F_{2 n} = \displaystyle \sum_{k \mathop = 1}^n \dbinom n k F_k$
From Fermat's Little Theorem: Corollary 1: :$n^5 \equiv n \pmod 5$ Suppose $n \equiv 1 \pmod 2$. Then from Congruence of Powers: :$n^5 \equiv 1^5 \pmod 2$ and so: :$n^5 \equiv 1 \pmod 2$ Similarly, suppose $n \equiv 0 \pmod 2$. Then from Congruence of Powers: :$n^5 \equiv 0^5 \pmod 2$ and so: :$n^5 \equiv 0 \pmod 2$ Hence: :$n^5 \equiv n \pmod 2$ So we have: :$n^5 \equiv n \pmod {2 \times 5}$ and the result follows. {{Qed}}
Assume $p$ is composite, and $q$ is a prime such that $q \divides p$. Then both $p$ and $\paren {p - 1}!$ are divisible by $q$. If the congruence $\paren {p - 1}! \equiv -1 \pmod p$ were satisfied, we would have $\paren {p - 1}! \equiv -1 \pmod q$. However, this amounts to $0 \equiv -1 \pmod q$, a contradiction. Hence for $p$ composite, the congruence $\paren {p - 1}! \equiv -1 \pmod p$ cannot hold. {{qed}}
If $p = 2$ the result is obvious. Therefore we assume that $p$ is an odd prime. Let $p$ be prime. Consider $n \in \Z, 1 \le n < p$. As $p$ is prime, $n \perp p$. From Law of Inverses (Modulo Arithmetic), we have: :$\exists n' \in \Z, 1 \le n' < p: n n' \equiv 1 \pmod p$ By Solution of Linear Congruence, for each $n$ there is '''exactly one''' such $n'$, and $\paren {n'}' = n$. So, provided $n \ne n'$, we can pair any given $n$ from $1$ to $p$ with another $n'$ from $1$ to $p$. We are then left with the numbers such that $n = n'$. Then we have $n^2 \equiv 1 \pmod p$. Consider $n^2 - 1 = \paren {n + 1} \paren {n - 1}$ from Difference of Two Squares. So either $n + 1 \divides p$ or $n - 1 \divides p$. Observe that these cases do not occur simultaneously, as their difference is $2$, and $p$ is an odd prime. From Negative Number is Congruent to Modulus minus Number: :$p - 1 \equiv -1 \pmod p$ Hence $n = 1$ or $n = p - 1$. So, we have that $\paren {p - 1}!$ consists of numbers multiplied together as follows: :in pairs whose product is congruent to $1 \pmod p$ :the numbers $1$ and $p - 1$. The product of all these numbers is therefore congruent to $1 \times 1 \times \cdots \times 1 \times p - 1 \pmod p$ by modulo multiplication. From Negative Number is Congruent to Modulus minus Number: :$\paren {p - 1}! \equiv -1 \pmod p$ {{qed}}
If $p = 2$ the result is obvious. Therefore we assume that $p$ is an odd prime. Consider $p$ points on the circumference of a circle $C$ dividing it into $p$ equal arcs. By joining these points in a cycle, we can create polygons, some of them stellated. From Number of Different n-gons that can be Inscribed in Circle, the number of different such polygons is $\dfrac {\paren {p - 1}!} 2$. When you rotate these polygons through an angle of $\dfrac {2 \pi} p$, exactly $\dfrac {p - 1} 2$ are unaltered. These are the regular $p$-gons and regular stellated $p$-gons. That there are $\dfrac {p - 1} 2$ of them follows from Number of Regular Stellated Odd n-gons. The remaining $\dfrac {\paren {p - 1}!} 2 - \dfrac {p - 1} 2$ polygons can be partitioned into sets of $p$ elements: those which can be obtained from each other by rotation through multiples of $\dfrac {2 \pi} p$. {{Explain|Why are there always $p$ of them? See Fermat's Little Theorem/Corollary 1.}} The total number of such sets is then: :$\dfrac {\paren {p - 1}! - \paren {p - 1} } {2 p}$ Thus we have that: :$2 p \divides \paren {\paren {p - 1}! - \paren {p - 1} }$ That is: :$p \divides \paren {\paren {p - 1}! + 1}$ {{qed}}