Product of Sigma and Euler Phi Functions

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Theorem

Let $n$ be an integer such that $n \ge 2$.

Let the prime decomposition of $n$ be:

$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$

Let $\sigma \left({n}\right)$ be the sigma function of $n$.

Let $\phi \left({n}\right)$ be the Euler phi function of $n$.


Then:

$\displaystyle \sigma \left({n}\right) \phi \left({n}\right) = n^2 \prod_{1 \mathop \le i \mathop \le r} \left({1 - \frac 1 {p_i^{k_i + 1}}}\right)$


Proof

From Euler Phi Function of Integer:

$\displaystyle \phi \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i - 1} \left({p_i - 1}\right)$

From Sigma Function of Integer:

$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$


So:

$\displaystyle \sigma \left({n}\right) \phi \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \left({\frac {p_i^{k_i + 1} - 1} {p_i - 1}}\right) p_i^{k_i - 1} \left({p_i - 1}\right)$


Taking a general factor of this product:

\(\displaystyle \left({\frac {p_i^{k_i + 1} - 1} {p_i - 1} }\right) p_i^{k_i - 1} \left({p_i - 1}\right)\) \(=\) \(\displaystyle \left({p_i^{k_i + 1} - 1}\right) p_i^{k_i - 1}\) $\quad$ cancelling $p_i - 1$ top and bottom $\quad$
\(\displaystyle \) \(=\) \(\displaystyle p_i^{2 k_i} - p_i^{k_i - 1}\) $\quad$ multiplying out the bracket $\quad$
\(\displaystyle \) \(=\) \(\displaystyle p_i^{2 k_i} \left({1 - \frac 1 {p_i^{k_i + 1} } }\right)\) $\quad$ extracting $p_i^{2 k_i}$ as a factor $\quad$


So:

$\displaystyle \sigma \left({n}\right) \phi \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} p_i^{2 k_i} \left({1 - \frac 1 {p_i^{k_i + 1} } }\right)$

Hence:

$\displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{2 k_i} = \left({\prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}}\right)^2 = n^2$

and the result follows.

$\blacksquare$