Hello, Ideasman!

I don't know what they mean by "EVERY possible value".

. . For a given set of numbers, there is exactlyoneGCD.

Let: .a = p·m .and .b = p·nSuppose GCD(a, b) = p, a prime,

then determine EVERY possible value of:

1) .GCD(a², b²)

. . wheremandnare 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