# GCD proof in Number Theory

1. May 31, 2006

### melmath

Here is my problem:

Prove or disprove: If gcd(m, n) = d, then the gcd(a, mn) = gcd(a,m) * gcd(a.n)/d.

I can seem to get it started, sort of, but it just does not seem to get anywhere. I know by definition d | m and d | n. Then arbitrary integers x and y can be used such that m = xd and n = yd. I can then have gcd(a, mn) = E. Then E | a and E | mn. I can do the same as I did for the other. But the last part of gcd(a,m) * gcd(a.n)/d is giving me a really rough time getting it to know how tie it all in. I have a lot of integers and letters and I think I am very lost at this point. Please help!

melmath

Last edited: May 31, 2006
2. May 31, 2006

### AKG

Do you know the fundamental theorem of arithmetic which says that every number can be written uniquely as a product of primes? So we can write:

$$m = \prod _{i=1}^{\infty}p_i^{\mu _i},\ n = \prod _{i=1}^{\infty}p_i^{\nu _i}$$

where pi is the ith prime, and the $\mu _i,\, \nu _i$ are eventually zero since m and n are finite numbers. Using the notation above, how would you express gcd(m,n)?

Last edited: May 31, 2006
3. Jun 1, 2006

### matt grime

x and y are not aribtrary integers. x is m/d and y is n/d.

AKG's way is very useful, but it is also good to do it from the defining characteristics: the gcd of A and B is a the largest divisor of A and B. There is also another way of thinking about it: gcd(m,n) is the integer d such that m=dx, n=dy for some x,y and gcd(x,y)=1