EulerPhi(3+6I)=20 why ?can some one give me detailed explanation , I would be greatful
Remembering that is...
$\displaystyle \varphi(n) = n\ \prod_{p|n} (1-\frac{1}{p})$ (1)
... first we search the n for which is $\displaystyle \varphi(n)=20$. Searching the primes $\displaystyle p_{i}$ so that $\displaystyle (p_{i}-1)|20$ we find $\displaystyle p_{1}=2$, $\displaystyle p_{2}=3$, $\displaystyle p_{3}=5$ and $\displaystyle p_{4}=11$, so that n must contain one or more of these primes and necessarly $\displaystyle n>20$. Taking into account that we find that the possible values of n are $\displaystyle n=25$, $\displaystyle n=33$, $\displaystyle n=50$ and $\displaystyle n=66$. Among these n, the only for which is $\displaystyle 3+6\ i = n$ is $\displaystyle n=33$ so that is $\displaystyle i=5$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
That doesn't make sense. I have never heard of the phi function extended to $\displaystyle \mathbb{Z}[i]$. How do you define it, $\displaystyle \varphi(a+bi)$ is the number of coprime elements of $\displaystyle \mathbb{Z}[i]$ whose modulus is less than $\displaystyle |a+bi|=a^2+b^2$?