Question 192
Let $a$ and $b$ be positive integers that are co-prime and let $p$ be a prime number. Prove that
$$
\gcd(ab, a^2 + pb^2) = \begin{cases} 1 & \text{if } p \nmid a \\ p & \text{if } p \mid a. \end{cases}
$$
$$
\gcd(ab, a^2 + pb^2) = \begin{cases} 1 & \text{if } p \nmid a \\ p & \text{if } p \mid a. \end{cases}
$$