Thread: 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).

Hello, Ideasman!

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

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: .a² .= .p²·m²
. and: .b² .= .p²·n²

Therefore: .GCD(a², b²) .= .p²

2) .GCD(a², b)

We have: .a² .= .p²·m²
. . . .and: . b . = .p·n

Therefore: .GCD(a², b) .= .p

3) .GCD(a³, b²)

We have: .a³ .= .p³·m³
. . . and: . b² .= .p²·n²

Therefore: .GCD(a³, b²) .= .p²

Are you sure that these are unique?

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?

You cannot define gcd(a,b) without making it unique, otherwise the function gcd: N x N --> N is not well-defined.