Why are these numbers coprime?

Question

I have encountered the following argument in a number theory class:

Let $p$ be an odd prime and let $n,x$ and $y$ be integers with $\gcd(x,y) = 1$, $n > 0$ and $p \not \mid n$. Suppose that $p \mid x^2+ny^2$, it then follows that $\gcd(y,p) = 1$.

I do not understand why $\gcd(y,p) = 1$ should hold here. Could you give me a hint?

Answer 1

Since $p$ is prime, either it is coprime to $y$ or divides $y$. But it can't divide both $x$ and $y$, since they are coprime to each other, and it can't divide $y$ but not $x$, since then it would not divide $x^2+ny^2$.

Answer 2

We have:

$$x^2+ny^2=kp$$

Or:

$$x^2=kp-ny^2$$

Due to definition of question:

$(x, y)=1$

If $(p, y)=a$, that is p and y have a common divisor like a, then $a|x$ which in turn means $a|y$ or x and y have a common divisor like a, this contradict the definition of question.