I am reading a text on number theory, but I am confused about the following.
Let $a$ be neither $\pm 1$ nor a perfect square. Suppose $h$ is the largest positive integer such that $a$ is a perfect $h$th power. Let $k$ be a square-free number, and $k_1$ be $k/(h,k)$ where $(h,k)$ means the greatest common divisor of $h$ and $k$.
Then for prime $p\nmid a$, I have the simultaneous conditions $$ v^q \equiv a \pmod p \ \text{solvable and} \ p\equiv 1 \pmod q,$$ hold for every prime divisor $q$ of $k$. The texts say that the former congruence is always possible when $q|h$, which I can see why, but I am not sure how it leads us to infer equivalent simultaneous conditions $$ v^{k_1} \equiv a \pmod p \ \text{solvable and} \ p\equiv 1 \pmod k.$$
I tried to show $k_1|h$ or $k_1|q$ so that the former congruence will be true, but I failed to show that, so I am wondering if I miss some properties of $\gcd$ or any other ingredients. For the second condition, I think the equivalence just follows from the Chinese Remainder Theorem.
Could anyone help me with my confusion? Thank you in advance.