# Greatest common divisor

1. Nov 17, 2008

### kathrynag

1. The problem statement, all variables and given/known data
Suppsoe a, b$$\in$$natural numbers, and d=GCD(a,b). Then d^2=GCD(a^2,b^2). I need to find where the proof goes wrong.

2. Relevant equations

3. The attempt at a solution
By hypothesis, we have that d divides a and d divides b, so there are integres s and t with a=ds and b=dt. Then a^2=d^2s^2 and so d^2 divides a^2. Similarily d^2 divides b^2. Thus d^2 is a common divisor of a^2 and b^2, as desired.

2. Nov 17, 2008

### HallsofIvy

Staff Emeritus
Re: d^2=GCD(a^2,b^2)

No, not "as desired". What you "desire" is the greatest common divisor. Saying that d^2 is a common divisor does not mean it is the GREATEST common divisor.

3. Nov 17, 2008

### kathrynag

Re: d^2=GCD(a^2,b^2)

Ok, that made a lot more sense now that I see that!