This is the question and solution:
Q: Prove that for any integer $a$, $2a + 1$ and $4a^2$ + 1 are relatively prime. A: Since $4a^2 + 1 = (2a − 1)(2a + 1) + 2$, any common divisor of $2a + 1$ and $4a^2 + 1$ must be a divisor of $2$. This means that $d=1$ or $d=2$. However, $2$ is not a common divisor as both $2a + 1$ and $4a^2 + 1$ are odd. Therefore, the greatest common divisor must be $1$.
Can someone explain how they knew that the common divisor of $2a + 1$ and $4a^2 + 1$ would be a divisor of $2$ (the bolded portion)? Is this some number theory that I just don't know? Also, this topic is $\gcd$ (greatest common divisor in discrete mathematics). Thanks in advance.