Euler Phi Function of Product with Prime

Theorem
Let $$p$$ be prime and $$n \in \Z: n \ge 1$$.

Then: $$ \phi \left({p n}\right) = \begin{cases} \left({p - 1}\right) \phi \left({n}\right) & : p \nmid n \\ p \phi \left({n}\right) & : p \backslash n \end{cases} $$

Thus for all $$n \ge 1$$ and for any prime $$p$$, we have that $$\phi \left({n}\right)$$ divides $$\phi \left({p n}\right)$$.

Corollary
If $$d \backslash n$$ then $$\phi \left({d}\right) \backslash \phi \left({n}\right)$$.

Proof

 * First suppose that $$p \nmid n$$.

Then by Prime Not Divisor then Coprime, $$p \perp n$$.

So by Euler Phi Function is Multiplicative, $$\phi \left({p n}\right) = \phi \left({p}\right) \phi \left({n}\right)$$.

It follows from Euler Phi Function of a Prime that $$\phi \left({p n}\right) = \left({p - 1}\right) \phi \left({n}\right)$$.


 * Now suppose that $$p \backslash n$$.

Then $$n = p^k m$$ for some $$k, m \in \Z: k, m \ge 1$$ such that $$p \perp m$$.

Then:

$$ $$ $$

At the same time:

$$ $$ $$ $$

Proof of Corollary
Let $$d \backslash n$$.

We can write $$n$$ as $$n = d p_1 p_2 p_3 \ldots p_r$$, where $$p_1, p_2, \ldots, p_r$$ are all the primes (not necessarily distinct) which divide $$n$$.

Thus, repeatedly using the above result:

$$ $$ $$ $$ $$

As the last expression is $$\phi \left({n}\right)$$, the result follows from Divides is Partial Ordering on Positive Integers.