Improve grammar
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@ -37,11 +37,11 @@ are integers, if not otherwise stated.
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\index{factor}
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\index{divisor}
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A number $a$ is a \key{factor} or \key{divisor} of a number $b$
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A number $a$ is called a \key{factor} or a \key{divisor} of a number $b$
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if $a$ divides $b$.
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If $a$ is a factor of $b$,
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we write $a \mid b$, and otherwise we write $a \nmid b$.
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For example, the factors of the number 24 are
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For example, the factors of 24 are
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1, 2, 3, 4, 6, 8, 12 and 24.
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\index{prime}
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@ -49,15 +49,14 @@ For example, the factors of the number 24 are
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A number $n>1$ is a \key{prime}
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if its only positive factors are 1 and $n$.
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For example, the numbers 7, 19 and 41 are primes.
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The number 35 is not a prime, because it can be
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divided into the factors $5 \cdot 7 = 35$.
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For example, 7, 19 and 41 are primes,
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but 35 is not a prime, because $5 \cdot 7 = 35$.
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For each number $n>1$, there is a unique
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\key{prime factorization}
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\[ n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k},\]
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where $p_1,p_2,\ldots,p_k$ are primes and
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where $p_1,p_2,\ldots,p_k$ are distinct primes and
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$\alpha_1,\alpha_2,\ldots,\alpha_k$ are positive numbers.
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For example, the prime factorization for the number 84 is
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For example, the prime factorization for 84 is
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\[84 = 2^2 \cdot 3^1 \cdot 7^1.\]
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The \key{number of factors} of a number $n$ is
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@ -66,7 +65,7 @@ because for each prime $p_i$, there are
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$\alpha_i+1$ ways to choose how many times
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it appears in the factor.
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For example, the number of factors
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of the number 84 is
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of 84 is
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$\tau(84)=3 \cdot 2 \cdot 2 = 12$.
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The factors are
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1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.
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@ -74,14 +73,14 @@ The factors are
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The \key{sum of factors} of $n$ is
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\[\sigma(n)=\prod_{i=1}^k (1+p_i+\ldots+p_i^{\alpha_i}) = \prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1},\]
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where the latter formula is based on the geometric progression formula.
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For example, the sum of factors of the number 84 is
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For example, the sum of factors of 84 is
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\[\sigma(84)=\frac{2^3-1}{2-1} \cdot \frac{3^2-1}{3-1} \cdot \frac{7^2-1}{7-1} = 7 \cdot 4 \cdot 8 = 224.\]
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The \key{product of factors} of $n$ is
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\[\mu(n)=n^{\tau(n)/2},\]
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because we can form $\tau(n)/2$ pairs from the factors,
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each with product $n$.
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For example, the factors of the number 84
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For example, the factors of 84
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produce the pairs
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$1 \cdot 84$, $2 \cdot 42$, $3 \cdot 28$, etc.,
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and the product of the factors is $\mu(84)=84^6=351298031616$.
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@ -91,7 +90,7 @@ and the product of the factors is $\mu(84)=84^6=351298031616$.
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A number $n$ is \key{perfect} if $n=\sigma(n)-n$,
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i.e., $n$ equals the sum of its factors
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between $1$ and $n-1$.
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For example, the number 28 is perfect,
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For example, 28 is a perfect number,
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because $28=1+2+4+7+14$.
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\subsubsection{Number of primes}
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@ -471,7 +470,7 @@ because $36 \bmod 17 = 2$ and $6^{-1} \bmod 17 = 3$.
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However, a modular inverse does not always exist.
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For example, if $x=2$ and $m=4$, the equation
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\[ x x^{-1} \bmod m = 1 \]
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cannot be solved, because all multiples of the number 2
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cannot be solved, because all multiples of 2
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are even and the remainder can never be 1 when $m=4$.
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It turns out that the value of $x^{-1} \bmod m$
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can be calculated exactly when $x$ and $m$ are coprime.
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@ -487,7 +486,7 @@ x^{-1} = x^{m-2}.
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\]
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For example, if $x=6$ and $m=17$, then
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\[x^{-1}=6^{17-2} \bmod 17 = 3.\]
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Using this formula, we can calculate the modular inverse
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Using this formula, we can calculate modular inverses
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efficiently using the modular exponentation algorithm.
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The above formula can be derived using Euler's theorem.
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@ -684,7 +683,7 @@ For example, $(3,4,5)$ is a Pythagorean triple.
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If $(a,b,c)$ is a Pythagorean triple,
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all triples of the form $(ka,kb,kc)$
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are also Pythagorean triples where $k>1$.
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A Pythagorean triple is \key{primitive} if
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A Pythagorean triple is \emph{primitive} if
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$a$, $b$ and $c$ are coprime,
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and all Pythagorean triples can be constructed
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from primitive triples using a multiplier $k$.
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