# Math Help - gcd questions (fairly simple)

1. ## gcd questions (fairly simple)

I feel like these are stupid questions but I just want to double check myself. (also would i have to show some sort of proof/explanation for my answers?)

(a) Let a be a positive integer. What is the greatest common divisor of a and a^2? This would be a right? With the idea of if a is a prime number than obviously the only divisor is going to be 1 and itself.

(b) Let a be a positive integer. What is the greatest common divisor of a and a+2? Would this be 1 if a is odd and 2 if a is even?

(c) show that if a and be are integers with (a,b)=1, then (a+b,a-b)=1 or 2.

2. Originally Posted by alice8675309
I feel like these are stupid questions but I just want to double check myself. (also would i have to show some sort of proof/explanation for my answers?)

(a) Let a be a positive integer. What is the greatest common divisor of a and a^2? This would be a right? With the idea of if a is a prime number than obviously the only divisor is going to be 1 and itself.

(b) Let a be a positive integer. What is the greatest common divisor of a and a+2? Would this be 1 if a is odd and 2 if a is even?

(c) show that if a and be are integers with (a,b)=1, then (a+b,a-b)=1 or 2.
I don't think these are stupid questions; I myself would like to know the solutions of these for practice.

For (a), do you know the fact that every integer can be expressed as a product of primes, and the prime factorization of the gcf?

ie $a = \prod p^{\alpha_p}$

and that $gcf(a,b) = \prod p^{min \alpha_p , \beta_p}$ ?

In this case, $min{\alpha_p, \alpha_p^2} = \alpha_p$ which shows the gcf(a, a^2) = a.
--

For (b) and (c), by inspection, I would most likely recourse to integer sums. They look tricky though.

3. Originally Posted by alice8675309
I feel like these are stupid questions but I just want to double check myself. (also would i have to show some sort of proof/explanation for my answers?)

(a) Let a be a positive integer. What is the greatest common divisor of a and a^2? This would be a right? With the idea of if a is a prime number than obviously the only divisor is going to be 1 and itself.

(b) Let a be a positive integer. What is the greatest common divisor of a and a+2? Would this be 1 if a is odd and 2 if a is even?

(c) show that if a and be are integers with (a,b)=1, then (a+b,a-b)=1 or 2.
I agree with your answers for (a) and (b), here is some justification.

(a) a^2 = a*a so clearly a divides a^2, and also a=1*a so a divides a (which you probably can use without thinking by this point). It should also be obvious that any integer greater than a cannot divide a. (You can prove this rigorously if you wish, it's easy.) Therefore gcd(a,a^2)=a.

(b) You should know in general that if d | x and d | y then d | (x-y) and d | (x+y). So gcd(a,a+2) divides 2. So it is either 1 or 2. You can then easily conclude that if a is even then gcd is 2, else it is 1.

(c) I think this is fairly tricky. Let g = gcd(a+b,a-b). We know from what I wrote in (b) that g | 2a and g | 2b. Can you take it from here?

4. infact for (c) you can find out when would it be 1 or 2 and this depends on whether a,b are even/odd