# GCD = p (prime)

• Feb 17th 2007, 03:53 PM
Ideasman
GCD = p (prime)
Suppose (a, b) = p, which is a prime, then determine EVERY possible value of:

1.) (a^2, b^2)

2.) (a^2, b)

3.) (a^3, b^2)

Note: (a, b) means gcd(a, b).
• Feb 17th 2007, 04:37 PM
Soroban
Hello, Ideasman!

I don't know what they mean by "EVERY possible value".
. . For a given set of numbers, there is exactly one GCD.

Quote:

Suppose GCD(a, b) = p, a prime,
then determine EVERY possible value of:

1) .GCD(a², b²)

Let: .a = p·m .and .b = p·n
. . where m and n are relatively prime (they have no common factors).

Then: . .= .p²·m²
. and: . .= .p²·n²

Therefore: .GCD(a², b²) .= .

Quote:

2) .GCD(a², b)
We have: . .= .p²·m²
. . . .and: . b . = .p·n

Therefore: .GCD(a², b) .= .p

Quote:

3) .GCD(a³, b²)

We have: . .= .p³·m³
. . . and: . .= .p²·n²

Therefore: .GCD(a³, b²) .= .

• Feb 19th 2007, 01:48 PM
Ideasman
Are you sure that these are unique?
• Feb 19th 2007, 03:33 PM
Plato
Quote:

Originally Posted by Ideasman
Are you sure that these are unique?

Why would you think it could be otherwise?
Do you understand the unique prime factorization theorem?
• Feb 19th 2007, 04:25 PM
ThePerfectHacker
You cannot define gcd(a,b) without making it unique, otherwise the function gcd: N x N --> N is not well-defined.