n is a number and n=(p1^c) * (p2^d).Here p1 and p2 are prime. Let a=p1^c and b=p2^d.
gcd(i,n)=
gcd(i,a) *
gcd(i,b)
How to prove this? Any explanation? Thanks in advance.
# | User | Rating |
---|---|---|
1 | tourist | 3845 |
2 | jiangly | 3707 |
3 | Benq | 3630 |
4 | orzdevinwang | 3573 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | jqdai0815 | 3532 |
8 | ecnerwala | 3501 |
9 | gyh20 | 3447 |
10 | Rebelz | 3409 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 171 |
2 | adamant | 164 |
3 | awoo | 163 |
4 | TheScrasse | 155 |
5 | nor | 153 |
6 | maroonrk | 152 |
6 | -is-this-fft- | 152 |
8 | Petr | 145 |
9 | pajenegod | 144 |
9 | orz | 144 |
n is a number and n=(p1^c) * (p2^d).Here p1 and p2 are prime. Let a=p1^c and b=p2^d.
gcd(i,n)=
gcd(i,a) *
gcd(i,b)
How to prove this? Any explanation? Thanks in advance.
Name |
---|
Each divisor of n looks like p1i × p2j. The sum on the right is
(φ(p1a) + p1 + p12 + p13 + ... + p1a) × (φ(p2b) + p2 + p22 + p23 + ... + p2b)
Using CRT you can make a bijection between terms on the right and and i on the left. For example, there are φ(n) values for which gcd(i, n) = 1, and there is φ(p1a)φ(p2b) = φ(n) on the right. Also, you can take φ(p1a) values which are coprime to p and then you can take only one of p2, p22, ... to make i for which gcd(i, n) = p2, or p22, etc.
Please explain briefy. I am too weak in math.