If $\,\gcd (a,0)=1,\,$ what can $a$ possibly be? I feel like a could be any number, but $0$ is divisible by any number,so they won't be mutually exclusive. I'm not sure, maybe this is not related, but it just confused me.
 A: Correction: Every integer (every number in fact) divides zero. $$a\mid 0\, \text{ for all }\,a \in \mathbb R$$
$$\gcd(a, 0) = a\,\text{ for all }\, a \in \mathbb Z$$ So, if $\,\gcd(a, 0) = 1,\,$ what must $a$ be?
A: Not sure why this was revived.  But it is confusing.  My advice to anyone confused is to take a deep breath and go back to definitions.
Some confusing concepts that make sense if you think about them but which might at first seem counter intuitive:
1) Every number is a divisor of $0$.
Pf: Let $n$ be any number.  $n*0 = 0$.  So there exists and integer $k$ (namely $k = 0$) so that $n*k = 0$.  That is the definition being a divisor.
2) If $n>0$   then the largest divisor of $n$ is $n$ itself.
Pf:  If $m > n > 0$ then $m*k > n*k \ge n$ if $k \ge 1$.  $m*k \le 0$ if $k \le 0$.  So there is not integer $k$ so that $m*k = n$.  So nothing larger than $n$ can be a divisor of $n$.  And as $n = n*1$, $n$ is a divisor of $n$.  So $n$ is the largest divisor of $n$.
3) $m$ is a divisor of $n$ if and only if $-m$ is a divisor of $n$ if and only if $m$ is a divisor of $-n$ if and ony if $-m$ is a divisor of $-n$.
Pf: $m*k =n$ if and only if $(-m)*(-k) = n$.... etc.
This means if $n \ne 0$ then $|n| \ge n$ and $|n|$ is the largest divisor of $n$.
4) $\gcd(0,0)$ is undefined but if $n \ne 0$ then $\gcd(0,n) = |n|$.
Pf: All numbers divide $0$ so the greatest number that divides $0$ is undefined.  So $\gcd(0,0)$.
All number divide $0$ so the common divisors of $0$ and $n$ are simply the divisors of $n$.  And the largest divisor of $n$ is $|n|$.
......
So....
So if $\gcd(a,0) =1$ then $\gcd(a,0) = |a| = 1$ and $a = \pm 1$.
Oh... some more concepts:
A)Every number is a divisor of $0$ but zero is not a divisor of any number except $0$.
B) If $a \ne 0$ then $\gcd(1,a) = 1$ and $\gcd(a,a) = |a|$.
A: If $|a|>1$ then $|a|$ is a common divisor of $a$ and of $0$,   so  $\gcd (a,0)\geq |a|>1.$
If $a=0$ then every $n\in \mathbb N$ is a divisor of $a$ and of $0$, so $\gcd (a,0)$ does not exist. 
If $a=\pm 1$ then the greatest divisor of $a$ is $1,$ and $1$ also divides $0$, so $\gcd(\pm 1,0)=1. $
