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Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site #Formula_phi_mn_=phi_m_phi_n#,This lecture contains a proof of formula: phi(mn) = phi(m).phi(n), where phi is the Euler's phi function. We know that $\phi$ is a multiplicative function, which means that if $(n, m) = 1$, then $\phi(nm) = \phi(n) \phi(m)$. We will use this property to prove the given identity. Step 3/5 Apr 28, 2012. #2. math2011 said: Suppose m and n are relatively prime positive integers; show that. m Ï ( n) + n Ï ( m) â¡ 1 ( mod m n) where Ï is the Euler Totient function. I can only see that Ï ( m n) = Ï ( m) Ï ( n) because g c d ( m, n) = 1 . I am reading Peskin and Schroeder, chapter ten, and my Lagrangian is $$ \mathcal{L}=\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2-\frac{\lambda}{4!}z^2\phi Since n Ï (m) â¡ 0 (m o d n) n^{\phi(m)} \equiv 0 \pmod{n} n Ï (m) â¡ 0 (mod n), we can add this congruence to the above equation to obtain. m Ï (n) + n Ï (m) â¡ 1 (m o d n). \begin{aligned} m^{\phi(n)}+n^{\phi(m)}&\equiv 1&\pmod{n}. \end{aligned} m Ï (n) + n Ï (m) â¡ 1 (mod n). Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site I'm studying a topic of the Nikitin's book (see pages 101 and 105) which deals with nonadiabatic electronic transitions, considering the two-state