# identity of Euler's totient function

Let $$p$$ be a prime and $$\varphi(n)$$ be Euler's totient function and $$m,n$$ be natural numbers. Then $$\varphi(p^k)=\frac{p^{k+1}-1}{p-1}$$ and $$\varphi(mn)=\varphi(m)(n)$$ and $$\varphi(p)=p-1$$.But this $$\varphi(p^k)=\varphi(p)^k=(p-1)^k\neq \frac{p^{k+1}-1}{p-1}=\varphi(p^k)$$ bothers me. Where is the mistake here?

• usually Euler's totient function is denoted $\phi(n)$ Commented Feb 26 at 22:30
• Did you mean $\sigma(p^k)=p^k-p^{k-1}=p^{k-1}(p-1)=p^k\left(1-\dfrac1p\right)$? Commented Feb 26 at 22:37
• That's not the Euler's totient function. That's the "sum of divisors" function Commented Feb 26 at 22:40
• sum of divisors function $\sigma(p)$ should be $p\color{blue}+1$ Commented Feb 26 at 22:45
• $\phi (mn)$ only equals $\phi(m)\phi(n)$ if $\gcd(m,n) =1$. As $\gcd(p,p)\ne 1$ we do not have $\phi(p^k)=(\phi p)^k$. And $\phi(p^k)= (p-1)p^{k-1}\ne \frac {p^{k+1}-1}{p-1}$. Commented Feb 26 at 23:18

$$\sigma(mn)$$ is a multiplicative function, which means that $$\sigma(mn)=\sigma(m)\sigma(n)$$ when $$m$$ and $$n$$ are relatively prime.
A function $$f$$ is totally multiplicative when $$f(mn)=f(m)f(n)$$ for all $$m$$ and $$n$$. As shown by your example, $$\sigma$$ is not totally multiplicative.