Function Fi Help: Simplifying the (m,n)=d Demonstration

[catala]

Hi,

I need help in the following demonstration:

If (m,n)=d then \Phi(mn)=\frac{d}{\Phi(d)}\Phi(m)\Phi(n)

Thank you very much for your support! :D

 

[PiAreSquared]

Ok, so first begin by letting m= p^{κ_{1}}_{1}...p^{κ_{r}}_{r} and n=p^{β_{1}}_{1}...p^{β_{r}}_{r}, where κ_{i},β_{i}≥ 0 \foralli. Then (m,n)= p^{δ_{1}}_{1}...p^{δ_{r}}_{r}, where δ_{i} = min{κ_{i},β_{i}}. Then use the fact that \Phi(w) = w∏^{i=1}_{r} (1-1/p_{i}), given that w = p^{κ_{1}}_{1}...p^{κ_{r}}_{r} to rewrite \Phi(mn). At this point, it isn't too difficult to come up with the desired result. If you rewrite the right side of what you are trying to prove, it helps to at least see where you should go next.

Also, as a side note, \Phi is spelled 'phi'