1.
first make numbers from 1 to mn in a rectangular table with m rows and n columns as below
1 m+1 2m+1 ...... (n-1)m+1
2 m+2 2m+2 ...... (n-1)m+2
3 m+3 2m+3 ...... (n-1)m+3
.
.
.
m 2m 3m ....... nm
here, the numbers in rth row are of the form km+r ( k changes from 0 to m-1 )
If two numbers are relative prime then gcd of those numbers is 1.
So if gcd(r,m) =1 , then every entry in rth row is realtive prime to m.
so gcd(km+r,m)=1 ( as mentioned above (m,n are coprime) )
We have shown that there are ϕ(m) rows in the table which contain numbers relatively prime to mn, and each of those contain exactly ϕ(n) such numbers.
So there are, in total, ϕ(m)ϕ(n) numbers in the table which are relatively prime to mn. This proves the theorem.
2. yes, the converse of Euler's Toient theorm is also true.
Suppose that for some B, aB=1 mod n . Then we have some integers k,m and an equation of the form aBk+mn=1. This means that aB and n are coprime. Then a and n must be coprime.
Exercise 2. Let φ denote the Euler totient function. (i) Prove that for all positive integers...
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oblemns for Solution: 1. Recall that Euler's phi function (or called Euler's totient function) φ(n) is defined as the number of integers m in the range 1< m<n such that m and n are relatively prime, i.e., gcd(rn, n) l. Find a formula for φ(n), n 2. (Hint: Factor n as the product of prime powers. i.e., n-TiỀ, where pi's are distinct primes and ei 〉 1, i, where p;'s are distinct primes and e > 1 t. oblemns for...
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blems for Solution: Recall that Euler's phi function (or called Euler's totient function) o(n) is defined as the number of integers m in the range 1 S m S n such that m and n are relatively prime, ie, gcd(mn) = 1. Find a formula for (n), n 2. (Hint: Factor n as the product of prime powers, ie., n llis] pr., where p's are distinct primes and c, 1, blems for Solution: Recall that Euler's phi function (or called...
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