Evaluating the greatest common divisor of pairs $(a, a^2)$, $(a, a+1)$, $(a, a+2)$

Question

I have a homework question which i'm struggling with, i would be interested in what method i should use to solve the following problems:

> Let a be a positive integer. Evaluate the following:
> (i) gcd(a,a^2)
> (ii) gcd(a,a+1)
> (iii) gcd(a,a+2)
> (you do not need the Euclidean Algorithm to answer these questions.)

Im not looking for the answers to these questions rather the approach i should take to solving them and problems like them? I've searched and while i can find links to solving gcd's they all seem to be of the style gcd(2,4) etc.

Thanks for the help (i know its probably very simple.)

Answer 1

Hint: Pick some numbers for $a$ such as $a=5$. Now, what would divide both 5 and 25? Or for the second one, what would divide both 5 and 6? How about 5 and 7 for the third one? Also, for the third one, what if $a=6$ and we tried to find the gcd of 6 and 8?

Answer 2

i.) We need the largest integer $k$ such that $k \; | \;a$ and $k \; | \; a^2 = a(a)$ (the $|$ symbol means "divides".) Clearly, the largest integer is $k = a$.

Answer 3

Hint $\ \gcd(a,\,b) = \gcd(a,\,b-a) = \gcd(a,\,b-2a) = \,\cdots\, = \gcd(a,\,b\ {\rm mod}\ a)\ $ tackles all three.

Answer 4

i) let d=gcd(a,a^2) then ther exist integers x and y such that ax + a^2 y=d =>a(x+a^2 y)=d in other words a|d and we have d|a by definition which forces a=d .

ii) d|a+1-a => d|1 so d=1

!!!) d|a+2-a => d|2 so d=1 or d=2