Suppose that every prime dividing n also divides m. Prove that φ(mn) = nφ(m).
This follows at once from the formula: if $\displaystyle n=p_1^{a_1}\cdot p_2^{a_2}\cdot\ldots\cdot p_r^{a_r}\,,\,\,then\,\,\,\phi(n)=n\prod^r_{i=1}\l eft(1-\frac{1}{p_i}\right)$ , $\displaystyle p_i,\,a_i\in\mathbb{N}\,,\,\,p_i$ primes. But you can also do a counting argument.
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