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).

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- February 17th 2007, 04:53 PMIdeasmanGCD = 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). - February 17th 2007, 05:37 PMSoroban
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)

. . 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

Quote:

2) .GCD(a², b)

. . . .and: . b . = .p·n

Therefore: .GCD(a², b) .= .p

Quote:

3) .GCD(a³, b²)

We have: .a³ .= .p³·m³

. . . and: . b² .= .p²·n²

Therefore: .GCD(a³, b²) .= .p²

- February 19th 2007, 02:48 PMIdeasman
Are you sure that these are unique?

- February 19th 2007, 04:33 PMPlato
- February 19th 2007, 05:25 PMThePerfectHacker
You cannot define gcd(a,b) without making it unique, otherwise the function gcd: N x N --> N is not well-defined.