Product of Divisor Sum 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:
 * $$\sigma \left({n}\right) \phi \left({n}\right) = n^2 \prod_{1 \le i \le r} \left({1 - \frac 1 {p_i^{k_i + 1}}}\right)$$

Proof
We have:
 * From Euler Phi Function of an Integer: $$\phi \left({n}\right) = \prod_{1 \le i \le r} p_i^{k_i - 1} \left({p_i - 1}\right)$$;
 * From Sigma of an Integer: $$\sigma \left({n}\right) = \prod_{1 \le i \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$$.

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

Taking a general factor of this product:

$$ $$ $$

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

We notice that $$\prod_{1 \le i \le r} p_i^{2k_i} = \left({\prod_{1 \le i \le r} p_i^{k_i}}\right)^2 = n^2$$, and the result follows.