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Math,Gcd
By sahasumit288, history, 8 years ago, In English

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.

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8 years ago, # |
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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.

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