hcju/proofwikiqa huggingface dataset:
Let $R = \left\{{a, a^2, a^3, \ldots, a^{\phi \left({n}\right)}}\right\}$. Each element of $R$ is coprime to $n$ as $a \perp n$. Suppose $a^r \equiv a^s \pmod n$ where $1 \le r \le s \le \phi \left({n}\right)$. Then $a^{r-s} \equiv 1 \pmod n$. From the definition of primitive root, the multiplicative order of $a$ modulo $n$ is $\phi \left({n}\right)$. So from Integer to Power of Multiple of Order $\phi \left({n}\right)$ divides $r - s$ and so $r = s$. So no two elements are congruent modulo $n$. So as $R$ contains $\phi \left({n}\right)$ integers none of which are congruent modulo $n$ to any of the others, $R$ is a reduced residue system of $n$. {{qed}} Category:Number Theory h2qpitvvr5kf61e0ol177hrmyz96fyn
Let $a$ have multiplicative order $c$ modulo $n$. Consider $a^k$ and let $d = \gcd \left\{{c, k}\right\}$. Let $c = d c'$ and $k = d k'$ where $\gcd \left\{{c', k'}\right\} = 1$ from Integers Divided by GCD are Coprime. We want to show that the multiplicative order $a^k$ modulo $n$ is $c'$. Let the order $a^k$ modulo $n$ be $r$. Then: {{begin-eqn}} {{eqn | l = \left({a^k}\right)^{c'}       | r = \left({a^{d k'} }\right)^{c/d}       | c =  }} {{eqn | r = a^{k' c}       | c =  }} {{eqn | r = \left({a^c}\right)^{k'}       | c =  }} {{eqn | o = \equiv       | r = \left({1}\right)^{k'}       | rr= \pmod n       | c =  }} {{eqn | o = \equiv       | r = 1       | rr= \pmod n       | c =  }} {{end-eqn}} So, by Integer to Power of Multiple of Order, $c'$ is a multiple of $r$, that is, $r \mathrel \backslash c'$. On the other hand, $a^{kr} = \left({a^k}\right)^{r} \equiv 1 \pmod n$, and so $kr$ is a multiple of $c$. Substituting for $k$ and $c$, we see that $dk'r$ is a multiple of $dc'$ which shows $c'$ divides $k' r$. But from Euclid's Lemma (which applies because $\gcd \left\{{c', k'}\right\} = 1$), we have that $c'$ divides $r$, or $c' \mathrel \backslash r$. So, as $c' \mathrel \backslash r$ and $r \mathrel \backslash c'$, from Divisor Relation on Positive Integers is Partial Ordering, it follows that $c' = r$. {{qed}} === Proof of Corollary === Let $a$ be a primitive root of $n$. Then $R = \left\{{a, a^2, \ldots, a^{\phi \left({n}\right)}}\right\}$ is a reduced residue system for $n$, and so all primitive roots are contained in this set. The multiplicative order $a^k$ modulo $n$ is $\dfrac {\phi \left({n}\right)} {\gcd \left\{{\phi \left({n}\right), k}\right\}}$. Hence $a^k$ will be a primitive root of $n$ exactly when $\gcd \left\{{\phi \left({n}\right), k}\right\} = 1$, i.e. when $\phi \left({n}\right) \perp k$. So the primitive roots are the integers $a^k$, where $k$ is in the set $\left\{{1, 2, \ldots, \phi \left({n}\right)}\right\}$. By definition of $\phi$, there are $\phi \left({\phi \left({n}\right)}\right)$ such exponents $k$. Hence there are $\phi \left({\phi \left({n}\right)}\right)$ primitive roots of $n$. {{qed}} Category:Number Theory 60bfumrxh2fpnxzjslzfz11ayuwaxf6
Let $n$ be an integer such that $n \ge 2$. Let the prime decomposition of $n$ be: :$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$ Then from Tau Function from Prime Decomposition we have that: :$\displaystyle \map \tau n = \prod_{i \mathop = 1}^r \paren {k_i + 1}$ Let $n$ be square-free. Then by definition: :$\forall i: 1 \le i \le r: k_i = 1$ So: :$\displaystyle \map \tau n = \prod_{i \mathop = 1}^r \paren {1 + 1} = 2^r$ The statement about the converse is proved by counterexample: Let $n = p^3$ where $p$ is prime. Then $n$ is not square-free as $p^2 \divides n$. However: :$\map \tau n = 3 + 1 = 2^2$ {{qed}} Category:Tau Function Category:Square-Free Integers e8f6riw52gp0qtz7a0bcofvvejsrld6
Let $n$ be an integer such that $n \ge 2$. Let the prime decomposition of $n$ be: :$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$ Then from Tau Function from Prime Decomposition we have that: : $\displaystyle \tau \left({n}\right) = \prod_{i \mathop = 1}^r \left({k_i + 1}\right)$ === Sufficient Condition === Let $\tau \left({n}\right)$ be odd. Then all factors of $\displaystyle \prod_{i \mathop = 1}^r \left({k_i + 1}\right)$ are odd (and of course $\ge 3$). Therefore all factors of $\displaystyle \prod_{i \mathop = 1}^r \left({k_i}\right)$ are even. Thus $n = p_1^{2 s_1} p_2^{2 s_2} \ldots p_r^{2 s_r}$ for $r_i = k_i / 2$ for all $i$. Hence $n = \left({p_1^{s_1} p_2^{s_2} \ldots p_r^{s_r}}\right)^2$ and therefore is square. {{qed|lemma}} === Necessary Condition === Now suppose $n$ is square. The above argument reverses, and we see that all factors of $\displaystyle \prod_{i \mathop = 1}^r \left({k_i + 1}\right)$ are odd. Hence $\tau \left({n}\right)$ is itself odd. {{qed}} Category:Tau Function Category:Square Numbers scw54galnfy6hw3o2xigr6k775s2ei8
Let $n$ be an integer such that $n \ge 2$. Let the prime decomposition of $n$ be: :$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$ Then from Sigma Function of Integer we have that: : $\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ That is: : $\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \left({1 + p_i + p_i^2 + \ldots + p_i^{k_i}}\right)$ Let $\sigma \left({n}\right)$ be odd. Then all factors of $\displaystyle \prod_{i \mathop = 1}^r \left({1 + p_i + p_i^2 + \ldots + p_i^{k_i}}\right)$ are odd (and of course $\ge 3$). For $1 + p_i + p_i^2 + \ldots + p_i^{k_i}$ to be odd, one of two conditions must hold: : $p_i$ is even (so that all terms of $1 + p_i + p_i^2 + \ldots + p_i^{k_i}$ are even except the $1$); : $k_i$ is even (so that $1 + p_i + p_i^2 + \ldots + p_i^{k_i}$ has an odd number of odd terms). In the first case, that means $p_i^{k_i}$ is a power of $2$. In the second case, that means $p_i^{k_i}$ is a square. The result follows. The argument reverses. {{qed}} Category:Square Numbers Category:Sigma Function 1r15nrfitzxoo9w1hh0p2kez3vvryz5
The quadratic residues of $p$ are the integers which result from the evaluation of the squares: : $1^2, 2^2, \ldots, \left({p - 1}\right)^2$ modulo $p$ But: :$r^2 = \left({-r}\right)^2$ and so these $p - 1$ integers fall into congruent pairs modulo $p$, namely: {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = \left({p - 1}\right)^2       | rr= \pmod p }} {{eqn | l = 2^2       | o = \equiv       | r = \left({p - 2}\right)^2       | rr= \pmod p }} {{eqn | o = \ldots }} {{eqn | l = \left({\frac {p - 1} 2}\right)^2       | o = \equiv       | r = \left({\frac {p + 1} 2}\right)^2       | rr= \pmod p       | c = Note: we require $p$ to be odd here. }} {{end-eqn}} Therefore each quadratic residue of $p$ is congruent modulo $p$ to one of the $\dfrac {p-1} 2$ integers $1^2, 2^2, \ldots, \left({\dfrac {p-1} 2}\right)^2$. Note that as $r^2 \not \equiv 0 \pmod p$ for $1 \le r < p$, the integer $0$ is not among these. All we need to do now is show that no two of these integers are congruent modulo $p$. So, suppose that $r^2 \equiv s^2 \pmod p$ for some $1 \le r \le s \le \dfrac {p-1} 2$. What we are going to do is prove that $r = s$. Now $r^2 \equiv s^2 \pmod p$ means that $p$ is a divisor of $r^2 - s^2 = \left({r + s}\right) \left({r - s}\right)$. From Euclid's Lemma either: :$p \mathrel \backslash \left({r + s}\right)$ or: :$p \mathrel \backslash \left({r - s}\right)$ $p \mathrel \backslash \left({r + s}\right)$ is impossible as $2 \le r + s \le p - 1$. Take $p \mathrel \backslash \left({r - s}\right)$. As $0 \le r - s < \dfrac {p-1} 2$, that can happen only when: :$r - s = 0$ or: :$r = s$ So there must be exactly $\dfrac {p-1} 2$ quadratic residues. That means there must also be exactly $\dfrac {p-1} 2$ quadratic non-residues. {{qed}} Category:Prime Numbers Category:Quadratic Residues gu8df47d8cj4ooq8wlpmxutqwv31zsi
First note that the square roots of $1$ are $1, -1 \pmod p$. Also, we have that $a^{p - 1} \equiv 1 \pmod p$ by Fermat's Little Theorem. Combining these two observations, we find: :$a^{\frac {p - 1} 2} \equiv 1 \text{ or } -1 \pmod p$ The theorem is therefore equivalent to stating that $a$ is a quadratic residue modulo $p$ {{iff}} $a^{\frac{p - 1} 2} \equiv 1 \pmod p$. Namely, considering the above, we see this also implies that all quadratic non-residues will be congruent to $-1 \pmod p$. We prove each direction of the equivalent statement separately: === Sufficient Condition === Assume $a$ is a quadratic residue modulo $p$. We pick $k$ such that $k^2 \equiv a \pmod p$. Then by Congruence of Powers and Fermat's Little Theorem: :$a^{\frac{p-1} 2} \equiv k^{p-1} \equiv 1 \pmod p$ === Necessary Condition === Now assume $a^{\frac{p - 1} 2} \equiv 1 \pmod p$. Then let $y$ be a primitive root modulo $p$, so that $a$ can be written as $y^j$. In particular: :$y^{j \frac {p - 1} 2} \equiv 1 \pmod p$ From the definition of $y$, it has order $p-1$. It follows that $p - 1 \divides j \dfrac {p - 1} 2$ from Element to Power of Multiple of Order is Identity. We conclude that $j$ is necessarily an even integer, and denote $j' = \dfrac j 2$. Let $k$ be such that $k \equiv y^{j'} \pmod p$. By construction, we have: :$k^2 \equiv y^{2 j'} \equiv y^j \equiv a \pmod p$ Hence $a$ is a quadratic residue modulo $p$. {{qed}}
Let $c a \equiv c b \pmod n$. Then we have that $c a - c b = k n$ for some $k \in \Z$ by definition of congruence. Now $d = \gcd \set {c, n}$, so from Integers Divided by GCD are Coprime we have: :$\exists r, s \in Z: r \perp s: c = d r, n = d s$ So we substitute for $c$ and $n$ in $c a - c b = k n$: :$d r a - d r b = k d s$ which leads us to: :$r \paren {a - b} = k s$ So $s \divides \paren {a - b}$ and as $r \perp s$, from Euclid's Lemma $s \divides \paren {a - b}$. So $a \equiv b \pmod s$ where $s = \dfrac n d$. Now suppose $a \equiv b \pmod {n / d}$ where $d = \gcd \set {c, n}$. Then: :$\exists k \in \Z: a - b = k \dfrac n d$ Hence: :$c a - c b = \dfrac {k c} d n$ As $d = \gcd \set {c, n}$ we have $d \divides c$ and so $\dfrac c d \in \Z$. So: :$c a \equiv c b \pmod n$ {{qed}} Category:Modulo Multiplication Category:Modulo Arithmetic Category:Cancellability of Congruences pky8rvdw5ydnippe4xss52lr97tegog
We assume that both $a$ and $b$ are non-zero, otherwise the solution is trivial. The first part of the problem is a direct restatement of Set of Integer Combinations equals Set of Multiples of GCD: The set of all integer combinations of $a$ and $b$ is precisely the set of integer multiples of the GCD of $a$ and $b$: :$\gcd \set {a, b} \divides c \iff \exists x, y \in \Z: c = x a + y b$ Now, suppose that $x', y'$ is any solution of the equation. Then we have: :$a' x_0 + b' y_0 = c'$ and $a' x' + b' y' = c'$ Substituting for $c'$ and rearranging: :$a' \paren {x' - x_0} = b' \paren {y_0 - y'}$ So: :$a' \divides b' \paren {y_0 - y'}$ Since $\gcd \set {a', b'} = 1$, from Euclid's Lemma we have: :$a' \divides \paren {y_0 - y'}$. So $y_0 - y' = a' k$ for some $k \in \Z$. Substituting into the above gives $x' - x_0 = b' k$ and so: :$x' = x_0 + b' k, y' = y_0 - a'k$ for some $k \in \Z$ which is what we claimed. Substitution again gives that the integers: :$x_0 + b' k, y_0 - a' k$ constitute a solution of $a' x + b' y = c'$ for any $k \in \Z$. {{qed}}
The only divisors of $p$ are $1$ and $p$ itself by definition. $\gcd \set {n, p} = p$ precisely when $p$ divides $n$. Hence the result. {{qed}} Category:Greatest Common Divisor Category:Prime Numbers 1ohxvpud1h23z1w5q5zd1t1dqu30upr
=== Sufficient Condition === Let $p^n \divides a^n$. We have $p \divides p^n$ as $p^n = p \paren {p^{n - 1} }$. From the fact that Divisor Relation is Transitive, we have that $p \divides a^n$. {{qed|lemma}} === Necessary Condition === Let $p \divides a^n$. Using Euclid's Lemma for Prime Divisors with $a_1 = a_2 = \cdots = a_n = a$ we have that: :$p \divides a^n \implies p \divides a$ Hence $a = p r$ for some $r \in \Z$. Raising both sides of this equation to the power $n$ we get that: :$a^n = p^n r^n$ So: :$p^n \divides a^n$ {{qed}} Category:Prime Numbers ale5gk38c1x2y1oa7bbkdad464qtabv
We are given that $\exists k \in \Z: z = k m$. Thus: {{begin-eqn}} {{eqn | l = a       | o = \equiv       | r = b       | rr= \pmod z       | c =  }} {{eqn | ll= \implies       | l = \exists k' \in \Z: a       | r = b + k' z       | c = Definition of Congruence }} {{eqn | ll= \implies       | l = a       | r = b + k' k m       | c =  }} {{eqn | l = a       | o = \equiv       | r = b       | rr= \pmod m       | c = Definition of congruence: $k' k$ is an integer }} {{end-eqn}} {{qed}}
Follows directly from the definition of Modulo Addition: {{begin-eqn}} {{eqn | l = a       | o = \equiv       | r = b       | rr= \pmod z       | c = given }} {{eqn | l = c       | o = \equiv       | r = c       | rr= \pmod z       | c = Congruence Modulo Real Number is Equivalence Relation }} {{eqn | ll= \leadsto       | l = a + c       | o = \equiv       | r = b + c       | rr= \pmod z       | c = {{Defof|Modulo Addition}} }} {{end-eqn}} {{qed}} Category:Modulo Addition mcnjhhb8e1wychegup4ic439u6s1c55
By definition of congruence modulo $n$: :$a = b + k n$ Dividing through by $d$ (which you can do because $d$ divides all three terms), we get: :$\dfrac a d = \dfrac b d + k \dfrac n d$ from which the result follows directly. {{qed}}
From Congruence by Product of Moduli, we have that: :$\dfrac a d \equiv \dfrac b d \pmod {n / d} \iff d \dfrac a d \equiv d \dfrac b d \pmod {n / d} \iff a \equiv b \pmod n$ {{qed}}
As: :$10 \equiv -1 \pmod {11}$ we have: :$10^r \equiv \paren {-1}^r \pmod {11}$ from Congruence of Powers. Thus: :$N \equiv a_0 + \paren {-1} a_1 + \paren {-1}^2 a_2 + \cdots + \paren {-1}^n a_n \pmod {11}$ from the definition of Modulo Addition. The result follows. {{qed}}
Let $P \left({x}\right) = c_m x^m + c_{m-1} x^{m-1} + \cdots + c_1 x + c_0$. Since $a \equiv b \pmod n$, from Congruence of Product and Congruence of Powers, we have $c_r a^r \equiv c_r b^r \pmod n$ for each $r \in \Z: r \ge 1$. From Modulo Addition we then have: {{begin-eqn}} {{eqn | l=P \left({a}\right)       | r=c_m a^m + c_{m-1} a^{m-1} + \cdots + c_1 a + c_0       | c= }} {{eqn | o=\equiv       | r=c_m b^m + c_{m-1} b^{m-1} + \cdots + c_1 b + c_0       | rr=\pmod n       | c= }} {{eqn | o=\equiv       | r=P \left({b}\right)       | rr=\pmod n       | c= }} {{end-eqn}} In particular, $P \left({a}\right) \equiv 0 \iff P \left({b}\right) \equiv 0 \pmod n$. That is, $a$ is a solution to the polynomial congruence $P \left({x}\right) \equiv 0 \pmod n$ iff $b$ is also. {{qed}} Category:Modulo Arithmetic fui1xeaxw55wryejdhmemm9s1if7185
Let $\displaystyle N = \prod_{i \mathop = 1}^r n_i$ and $N_k = \dfrac N {n_k}$ for $k = 1, 2, \ldots, r$. Because $n_i \perp n_k$ for each $i \ne k$, we have: :$\gcd \set {N_k, n_k} = \gcd \set {n_1 n_2 \cdots n_{k - 1} n_{k + 1} \cdots n_r, n_k} = 1$ So by Solution of Linear Congruence, the linear congruence $N_k x \equiv 1 \pmod {n_k}$ has a unique solution modulo $n_k$. Let this solution be $x_k$. So $N_k x_k \equiv 1 \pmod {n_k}$ We are going to show that $x_0 = b_1 N_1 x_1 + b_2 N_2 x_2 + \cdots + b_r N_r x_r$ satisfies each of the congruences. So, we evaluate $x_0$ modulo $n_k$ for each $k = 1, 2, 3, \ldots, r$. We start by noting that: :$i \ne k \implies n_k \divides N_i$ so: :$N_i \equiv 0 \pmod {n_k}$ So for each $k = 1, 2, \ldots, r$ we have $x_0 \equiv b_k N_k x_k$ because each of the remaining terms in the sum is congruent modulo $n_k$ to $0$. Finally, since $x_k$ was found such that $N_k x_k \equiv 1 \pmod {n_k}$ we have $x_0 \equiv b_k \pmod {n_k}$ as we claimed. All we need to do now is show that this solution we have discovered is unique modulo $N$. So, suppose that $x'$ is a second solution of the system. That is: :$x_0 \equiv x' \equiv b_k \pmod {n_k}$ for each $k = 1, 2, \ldots, r$. So each $n_k$ divides $x' - x_0$. But because each of the moduli are pairwise coprime, we have: :$\displaystyle \prod_{i \mathop = 1}^r n_i \divides x' - x_0$ That is: :$\displaystyle x' \equiv x_0 \pmod {\prod_{i \mathop = 1}^r n_i}$ as we wanted to show. {{qed}}
We take the case where $r = 2$. Suppose $x \in \Z$ satisfies both: {{begin-eqn}} {{eqn | l = a_1 x       | o = \equiv       | r = b_1       | rr= \pmod {n_1}       | c =  }} {{eqn | l = a_2 x       | o = \equiv       | r = b_2       | rr= \pmod {n_2}       | c =  }} {{end-eqn}} That is, $\exists r, s \in \Z$ such that: {{begin-eqn}} {{eqn | l = x - b_1       | r = n_1 r       | c =  }} {{eqn | l = x - b_2       | r = n_2 r       | c =  }} {{end-eqn}} Eliminating $x$, we get: :$b_2 - b_1 = n_1 r - n_2 s$ The {{RHS}} is an integer combination of $n_1$ and $n_2$ and so is a multiple of $\gcd \left\{{n_1, n_2}\right\}$. Thus $\gcd \set {n_1, n_2}$ divides $b_2 - b_1$, so this is a necessary condition for the system to have a solution. To show sufficiency, we reverse the argument. Suppose $\exists k \in \Z: b_2 - b_1 = k \gcd \set {n_1, n_2}$. We know that $\exists u, v \in \Z: \gcd \set {n_1, n_2} = u n_1 + v n_2$ from Bézout's Identity. Eliminating $\gcd \set {n_1, n_2}$, we have: :$b_1 + k u n_1 = b_2 - k v n_2$. Then: :$b_1 + k u n_1 = b_1 + \paren {k u} n_1 \equiv b_1 \pmod {n_1}$ :$b_1 + k u n_1 = b_2 + \paren {k v} n_2 \equiv b_2 \pmod {n_2}$ So $b_1 + k u n_1$ satisfies both congruences and so simultaneous solutions do exist. Now to show uniqueness. Suppose $x_1$ and $x_2$ are both solutions. That is: :$x_1 \equiv x_2 \equiv b_1 \pmod n_1$ :$x_1 \equiv x_2 \equiv b_2 \pmod n_2$ Then from Intersection of Congruence Classes the result follows. {{qed}} The result for $r > 2$ follows by a tedious induction proof. {{finish}} Category:Modulo Arithmetic iwfmpb9elqjtmw11en8efgwdoqi8lk3
Consider the numbers in the set $S = \set {2 \times 1, 2 \times 2, 2 \times 3, \dots, 2 \times \dfrac {p - 1} 2} = \set {2, 4, 6, \dots, p - 1}$. From Gauss's Lemma: :$\paren {\dfrac 2 p} = \paren {-1}^n$ where $n$ is the number of elements in $S$ whose least positive residue modulo $p$ is greater than $\dfrac p 2$. As they are, the elements of $S$ are already least positive residues of $p$ (as they are all less than $p$). What we need to do is count how many are greater than $\dfrac p 2$. We see that: :$2 k > \dfrac p 2 \iff k > \dfrac p 4$ So the first $\floor {\dfrac p 4}$ elements of $S$ are not greater than $\dfrac p 2$, where $\floor {\dfrac p 4} $ is the floor function of $\dfrac p 4$. The rest of the elements of $S$ ''are'' greater than $\dfrac p 2$. So we have: :$n = \dfrac {p - 1} 2 - \floor {\dfrac p 4}$ Consider the four possible residue classes modulo $8$ of the odd prime $p$. $p = 8 k + 1$: {{begin-eqn}} {{eqn | l = p       | r = 8 k + 1       | c =  }} {{eqn | ll= \leadsto       | l = n       | r = 4 k - \floor {2 k + \frac 1 4}       | c =  }} {{eqn | r = 4 k - 2 k       | c =  }} {{eqn | r = 2k       | c =  }} {{end-eqn}} $p = 8 k + 3$: {{begin-eqn}} {{eqn | l = p       | r = 8 k + 3       | c =  }} {{eqn | ll= \leadsto       | l = n       | r = 4 k + 1 - \floor {2 k + \frac 3 4}       | c =  }} {{eqn | r = 4 k + 1 - 2 k       | c =  }} {{eqn | r = 2 k + 1       | c =  }} {{end-eqn}} $p = 8 k + 5$: {{begin-eqn}} {{eqn | l = p       | r = 8 k + 5       | c =  }} {{eqn | ll= \leadsto       | l = n       | r = 4 k + 2 - \floor {2 k + \frac 5 4}       | c =  }} {{eqn | r = 4 k + 2 - \paren {2 k + 1}       | c =  }} {{eqn | r = 2 k + 1       | c =  }} {{end-eqn}} $p = 8 k + 7$: {{begin-eqn}} {{eqn | l = p       | r = 8 k + 7       | c =  }} {{eqn | ll= \leadsto       | l = n       | r = 4 k + 3 - \floor {2 k + \frac 7 4}       | c =  }} {{eqn | r = 4 k + 3 - \paren {2 k + 1}       | c =  }} {{eqn | r = 2 k + 2       | c =  }} {{end-eqn}} We see that $n$ is even when $p = 8 k + 1$ or $p = 8 k + 7$ and odd in the other two cases. So from Gauss's Lemma, we have: {{begin-eqn}} {{eqn | l = \paren {\dfrac 2 p}       | r = \paren {-1}^n = 1       | c = when $p = 8 k + 1$ or $p = 8 k + 7$ }} {{eqn | l = \paren {\dfrac 2 p}       | r = \paren {-1}^n = -1       | c = when $p = 8 k + 3$ or $p = 8 k + 5$ }} {{end-eqn}} As $7 \equiv -1 \pmod 8$ and $5 \equiv -3 \pmod 8$ the result follows. {{qed}}
From Euler's Criterion, and the definition of the Legendre symbol, we have that: :$\paren {\dfrac a p} \equiv a^{\paren {p - 1} / 2} \pmod p$ The result follows by putting $a = -1$. {{qed}}
We have that both series are absolutely convergent. Thus by Manipulation of Absolutely Convergent Series, it is permitted to expand the product as: :$\displaystyle \left({\sum_{i \mathop \in A} a_i }\right) \left({\sum_{j \mathop \in B} b_j }\right) = \sum_{i \mathop \in A}  \left({a_i \sum_{j \mathop \in B} b_j}\right)$. But since $a_i$ is a constant, it may be brought into the summation, to obtain: :$\displaystyle \sum_{i \mathop \in A} \sum_{j \mathop \in B} a_i b_j$ Hence the result. {{qed}} Category:Number Theory Category:Analysis ooo7xwe9uanll0mib3s4nut70cc87c7
Let: :$\mathbf m = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ :$\mathbf n = -w \mathbf 1 + x \mathbf i + y \mathbf j + z \mathbf k$ be two quaternions. Then: {{begin-eqn}} {{eqn | l = \left\vert \mathbf m \right\vert \left\vert \mathbf n \right\vert       | r = \left\vert \mathbf m \mathbf n \right\vert       | c = Quaternion Modulus of Product of Quaternions }} {{eqn | ll= \implies       | l = \left({a^2 + b^2 + c^2 + d^2}\right) \left({\left({-w}\right)^2 + x^2 + y^2 + z^2}\right)       | r = \left\vert \mathbf m \mathbf n \right\vert       | c = {{Defof|Quaternion Modulus}} }} {{eqn | r = \left({-a w - b x - c y - d z}\right)^2       | c = {{Defof|Quaternion Multiplication}} }} {{eqn | o =        | ro= +       | r = \left({a x - b w + c z - d y}\right)^2       | c =  }} {{eqn | o =        | ro= +       | r = \left({a y - b z - c w + d x}\right)^2       | c =  }} {{eqn | o =        | ro= +       | r = \left({a z + b y - c x - d w}\right)^2       | c =  }} {{eqn | ll= \implies       | l = \left({a^2 + b^2 + c^2 + d^2}\right) \left({w^2 + x^2 + y^2 + z^2}\right)       | r = \left({a w + b x + c y + d z}\right)^2       | c = simplifying }} {{eqn | o =        | ro= +       | r = \left({a x - b w + c z - d y}\right)^2       | c =  }} {{eqn | o =        | ro= +       | r = \left({a y - b z - c w + d x}\right)^2       | c =  }} {{eqn | o =        | ro= +       | r = \left({a z + b y - c x - d w}\right)^2       | c =  }} {{end-eqn}} {{qed}}
It suffices to demonstrate the theorem for rectangular regions in the $x y$-plane. The Riemann-sum nature of the double integral will then guarantee the proof of the theorem for arbitrary regions, because a Riemann-sum is technically a summation of the areas of arbitrarily small rectangles.  As the proof is for a rectangle, the proof will work for arbitrary regions, which can be approximated by collections of ever smaller rectangles. {{handwaving|To do this formally involves the Monotone Convergence Theorem (Measure Theory) for $U$ and approximating $\Gamma$ by means of a "piecewise rectification". Its arduous, yet necessary.}} Let $R = \set {\tuple {x, y}: a \le x \le b, c \le y \le d}$ be a rectangular region. Let the boundary $C$ of $R$ be oriented counterclockwise. We break the boundary into $4$ pieces: :$C_1$, which runs from $\tuple {a, c}$ to $\tuple {b, c}$ :$C_2$, which runs from $\tuple {b, c}$ to $\tuple {b, d}$ :$C_3$, which runs from $\tuple {b, d}$ to $\tuple {a, d}$ :$C_4$, which runs from $\tuple {a, d}$ to $\tuple {a, c}$ Then: {{begin-eqn}} {{eqn | l = \iint_R \frac {\partial B} {\partial x} \rd x \rd y       | r = \int_c^d \int_a^b \frac {\partial B} {\partial x} \rd x \rd y       | c =  }} {{eqn | r = \int_c^d \paren {\map B {b, y} - \map B {a, y} } \rd y       | c =  }} {{eqn | r = \int_c^d  \map B {b, y} \rd y + \int_d^c \map B {a, y} \rd y       | c =  }} {{eqn | r = \int_{C_2} B \rd y + \int_{C_4} B \rd y       | c =  }} {{end-eqn}} We note that $y$ is constant along $C_1$ and $C_3$. So: :$\displaystyle \int_{C_1} B \rd y = \int_{C_3} B \rd y = 0$ Hence: {{begin-eqn}} {{eqn | l = \int_{C_2} B \rd y + \int_{C_4} B \rd y       | r = \int_{C_1} B \rd y + \int_{C_2} B \rd y + \int_{C_3} B \rd y + \int_{C_4} B \rd y       | c =  }} {{eqn | r = \oint_C B \rd y       | c =  }} {{end-eqn}} A similar argument demonstrates that: :$\displaystyle \iint_R \frac {\partial A} {\partial y} \rd x \rd y = -\oint_C A \d x$ and hence: :$\displaystyle \oint_C \paren {A \rd x + B \rd y} = \iint_R \paren {\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} } \rd x \rd y$ {{qed}}