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Let $n=\\var{n}$.

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Find:

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$\\mu(n) =\\;\\;$[[0]]

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$\\tau(n)=\\;\\;$[[1]]

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$\\sigma(n)=\\;\\;$[[2]]

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$\\phi(n)=\\;\\;$[[3]]

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Definitions.

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$\\mu(n)$ gives information on the prime decomposition of $n$ (see formula below).

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$\\tau(n)$ gives the number of divisors of $n$.

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$\\sigma(n)$ is the sum of all divisors of $n$

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$\\phi(n)$ is the number of natural integers $\\lt n$ which are coprime to $n$.

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All are multiplicative functions i.e. $f(n)$ is multiplicative if $n_1,\\;n_2$ are coprime then:
\\[ f(n_1n_2)=f(n_1)f(n_2)\\]

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More Information and Formulae.

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Recall that if $n=p_1^{\\beta_1}\\times p_2^{\\beta_2}\\times \\cdots \\times p_m^{\\beta_m},\\;\\;\\beta_j \\ge 1,\\;\\forall j$ is the prime factorization of $n$ then:
\\[\\begin{eqnarray*}\\mu(n)&=&0, \\mbox{ if }\\exists j, p_j \\ge 2\\\\\n\n&=&(-1)^m, \\mbox{ if }\\beta_j=1,\\;\\;\\forall j\\\\\n\n\\\\\n\n\\tau(n)&=&(\\beta_1+1)\\times (\\beta_2+1) \\times\\cdots \\times (\\beta_m+1)\\\\\n\n\\\\\n\n\\sigma(n)&=& \\frac{(p_1^{\\beta_1+1}-1)}{p_1-1}\\times \\frac{(p_2^{\\beta_2+1}-1)}{p_2-1} \\times \\cdots \\times \\frac{(p_m^{\\beta_m+1}-1)}{p_m-1}\\\\\n\n\\\\\n\n\\phi(n) &=& (p_1^{\\beta_1} - p_1^{\\beta_1-1})\\times (p_2^{\\beta_2} - p_2^{\\beta_2-1})\\times\\cdots \\times (p_m^{\\beta_m} - p_m^{\\beta_m-1})\n\n\\end{eqnarray*}\n\n\\]

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Number Theory. 

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Given $n \\in \\mathbb{N}$ find $\\mu(n),\\;\\tau(n),\\;\\sigma(n),\\;\\phi(n).$

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The factorization into a product of powers of prime factors is given by:

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$\\var{n} = \\simplify[unitFactor,unitPower,zeroPower]{{p[0]}^{c[0]}*{p[1]}^{c[1]}*{p[2]}^{c[2]}*{p[3]}^{c[3]}*{p[4]}^{c[4]}}$

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Finding $\\mu(\\var{n})$

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There are an {eorodd} number $\\var{notzero}$ of distinct prime factors, {mess} since the maximum power of any prime in this factorization is {d[4]} we see that

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$\\mu(\\var{n})=\\;\\var{mu}$

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Finding $\\tau(\\var{n})$

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Recall that if $n=p_1^{\\beta_1}\\times p_2^{\\beta_2}\\times \\cdots \\times p_m^{\\beta_m}$ is the prime factorization of $n$, then the number of divisors is given by:

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$\\tau(n)=(\\beta_1+1)\\times (\\beta_2+1) \\times\\cdots \\times (\\beta_m+1)$

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Hence in this case we have:

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$\\tau(\\var{n})=\\tau( \\simplify[unitFactor,unitPower,zeroPower]{{p[0]}^{c[0]}*{p[1]}^{c[1]}*{p[2]}^{c[2]}*{p[3]}^{c[3]}*{p[4]}^{c[4]}})=\\simplify[!basic,unitFactor,zeroTerm]{({c[0]}+1)*({c[1]}+1)*({c[2]}+1)*({c[3]}+1)*({c[4]}+1)}=\\var{ta}$

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Finding $\\sigma(\\var{n})$

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The sum over all divisors is given by :

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\\[\\sigma(n)= \\frac{(p_1^{\\beta_1+1}-1)}{p_1-1}\\times \\frac{(p_2^{\\beta_2+1}-1)}{p_2-1} \\times \\cdots \\times \\frac{(p_m^{\\beta_m+1}-1)}{p_m-1}\\]

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In this case we have:

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\\[\\begin{eqnarray*}\n\n\\sigma(\\var{n})&=&\\sigma( \\simplify[unitFactor,unitPower,zeroPower]{{p[0]}^{c[0]}*{p[1]}^{c[1]}*{p[2]}^{c[2]}*{p[3]}^{c[3]}*{p[4]}^{c[4]}})\\\\ \n\n&=&\\simplify[unitFactor]{{siga[0]}*{siga[1]}*{siga[2]}*{siga[3]}*{siga[4]}}\\\\\n\n&=&\\var{sig}\n\n\\end{eqnarray*}\n\n\\]

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Finding $\\phi(\\var{n})$

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This is the Euler totient function and $\\phi(n)$ it counts up the number of integers less than $n$ which are coprime to $n$.

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The formula we use for this is given by:

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\\[\\phi(n) = (p_1^{\\beta_1} - p_1^{\\beta_1-1})\\times (p_2^{\\beta_2} - p_2^{\\beta_2-1})\\times\\cdots \\times (p_m^{\\beta_m} - p_m^{\\beta_m-1})\\]

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$\\phi(\\var{n})=\\phi( \\simplify[unitFactor,unitPower,zeroPower]{{p[0]}^{c[0]}*{p[1]}^{c[1]}*{p[2]}^{c[2]}*{p[3]}^{c[3]}*{p[4]}^{c[4]}})=\\simplify[unitFactor]{{phia[0]}*{phia[1]}*{phia[2]}*{phia[3]}*{phia[4]}}=\\var{phi}$

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