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.
Let: .a = p·m .and .b = p·nSuppose 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
We have: .a² .= .p²·m²2) .GCD(a², b)
. . . .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