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Explanation about reduced residue system theorem

Theorem 3-8 from Leveque's Elementary Theorem of Numbers:

$$\varphi(m)=m\prod_{p|m}(1-1/p)$$ where the notation indicates product over all distinct primes which divide $m$.

Proof: By reduced residue system theorem 3-7, if

$$m=\prod_{i=1}^{r}p_i^{a_i}$$

then

$$\varphi(m)=\prod_{i=1}^{r}\varphi(p_i^{a_i})$$

But we can easily evaluate $\varphi(p^a)$ directly: all the positive integer not exceeding $p^a$ are prime to $p^a$ except the multiples of p, and there are just $p^{a-1}$ of these, so that

$$\varphi(p_i^{a_i})=p_i^{a_i}-p_i^{a_1-1}=p_i^{a_i}(1-1/p_i)$$

Thus

$$\varphi(m)=\prod_{i=1}^{r}p_i^{a_i}(1-1/p_i)=\prod_{i=1}^{r}p_i^{a_i}*\prod_{i=1}^{r}(1-1/p_i)$$

$$=m\prod_{p|m}(1-1/p)$$

This proof lacks a lot of explanation as well. I don't even understand what is going on here, What is $a_i$. Why is it taking such weird arguments?. How does m follow from theorem 3-7?. The proof is very short , so I hope someone can give me a better insight.

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daniel
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  • http://math.stackexchange.com/questions/999563/very-elementary-proof-of-that-eulers-totient-function-is-multiplicative – ajotatxe Jun 27 '16 at 04:52
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    So any integer can be written as the product of one or more primes, but why do we need $a_i$? – TheMathNoob Jun 27 '16 at 05:14
  • how do we know that there are exactly $p^{a-1}$ of these and how would that ensure that no two pairs from $p^{a_i}-p_i^{a_1-1}$ are congruent modulo $p^a$? – daniel Jun 27 '16 at 05:27
  • I just need a little explanation. I almost got the proof – daniel Jun 27 '16 at 05:38
  • Let's proceed towards explaining the proof step by step. But first tell me, (ignoring "By reduced residue system theorem 3-7") do you understand the implication $m=\displaystyle\prod_{i=1}^r p_{i}^{a_i}\implies \varphi(m)=\displaystyle\prod_{i=1}^r \varphi\left(p_{i}^{a_i}\right)$? –  Jun 27 '16 at 05:57
  • That's true if $(p_i^{a_i},.......,p_r^{a_r})=1$ – daniel Jun 27 '16 at 05:58
  • Here $p_{i}$'s are all primes and $a_i$'s are the highest positive exponent of $p_i$ that divides $m$. Do you see how then $\gcd(p_1^{a_1},p_2^{a_2},\ldots,p_r^{a_r})=1$? –  Jun 27 '16 at 06:01
  • I got the intuition, but I can't provide a proof, so no, so we have to prove if a and b are prime numbers then (a,b)=1, prove (a^n,b^n)=1 for any n – daniel Jun 27 '16 at 06:03
  • If not then suppose that the $\gcd$ is $d>1$. Then there exists a prime $p$ such that $p\mid d$ (why?). Then $p\mid p_i^{a_i}$ for all $i$. Which implies $p\mid p_i$ for all $i$ (why?). This implies $p=p_i$ for all $i$ (why?) and that's a contradiction (why?). –  Jun 27 '16 at 06:06
  • But $m=\displaystyle\prod_{i=1}^r p_i^{a_i}\implies \varphi(m)=\displaystyle\prod_{i=1}^r \varphi\left(p_i^{a_i}\right)$ isn't true for the reason you gave. It is true because the $\gcd(p_i^{a_i}, p_j^{a_j})=1$ for $i\ne j$ and for all $i,j$. –  Jun 27 '16 at 06:13
  • "Then there exists a prime p such that p|d". Why does p have to be prime? – daniel Jun 27 '16 at 06:18
  • Because we want to apply Euclid's Theorem that if $p\mid ab$ then $p\mid a $ or $p\mid b$ (where $p$ is a prime). –  Jun 27 '16 at 06:20
  • Let us continue this discussion in chat. –  Jun 27 '16 at 06:20

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