Product of Sigma and Euler Phi Functions

Theorem

Let $n$ be an integer such that $n \ge 2$, with prime decomposition $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$.

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

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

Then:

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

Proof

We have:

• From Euler Phi Function of an Integer:

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

• From Sigma of an Integer:

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

So:

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

Taking a general factor of this product:

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

So:

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

We notice that:

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

and the result follows.

$\blacksquare$