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?
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2$\begingroup$ usually Euler's totient function is denoted $\phi(n)$ $\endgroup$– J. W. TannerCommented Feb 26 at 22:30
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1$\begingroup$ Did you mean $\sigma(p^k)=p^k-p^{k-1}=p^{k-1}(p-1)=p^k\left(1-\dfrac1p\right)$? $\endgroup$– J. W. TannerCommented Feb 26 at 22:37
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$\begingroup$ That's not the Euler's totient function. That's the "sum of divisors" function $\endgroup$– jjagmathCommented Feb 26 at 22:40
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$\begingroup$ sum of divisors function $\sigma(p)$ should be $p\color{blue}+1$ $\endgroup$– J. W. TannerCommented Feb 26 at 22:45
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1$\begingroup$ $\phi (mn)$ 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 $\phi(p^k)= (p-1)p^{k-1}\ne \frac {p^{k+1}-1}{p-1}$. $\endgroup$– fleabloodCommented Feb 26 at 23:18
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1 Answer
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$\sigma(mn)$ 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.
