# Why there exists an integer which is the product of those prime-power factors $p^{\alpha}$ of 4n but also $p | x_0$ for the integral quadratric form?

The background follows from Niven's number theory textbook's proof of Theorem $$3.13$$

Theorem $$3.13$$ :Let $$n$$ and $$d$$ be given integers with $$n \neq 0$$. There exists a binary quadratic form of discriminant $$d$$ that represents $$n$$ properly if and only if the congruence $$x^2 \cong d (mod 4|n|)$$ has a solution.

...Conversely, suppose we have a proper representation $$f(x_0, y_0)$$ of $$n$$ by a form $$f(x, y) = ax^2 + bxy + cy^2 = n$$ with discriminant $$b^2 - 4ac = d$$. Since $$g.c.d.(x_0, y_0) = 1$$, we can choose integers $$m_1 , m_2$$ such that $$m_1m_2 = 4|n|$$, $$(m_1, y_0) = 1$$ and $$(m_2 , x_0) = 1$$. For example, take $$m_1$$ to be the product of those prime-power factors $$p^a$$ of $$4n$$ for which $$p|x_0$$, and then put $$m_2 = \frac{4|n|}{m1}$$.

I am quite confused about the bolded statement. First off, my understanding of the statement is to say that we can have a $$m_1=p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}=4f(x_0,y_0)$$ for which simultaneously $$p_1,p_2...p_n|x_0$$.
Is this understanding correct or not? If yes, then for example if $$x_0=3,y_0=5$$, and then we would have $$4n=3^{\alpha}5^{\beta}$$, which an obvious paradox?, If my interpretation is wrong, Then which particular step made me wrong? the understanding? or the example which I constructed is incorrect? Thank you for any advices!

The bolded statement means: if $$(x_0,y_0)=1$$ and $$4|n|=p_1^{\alpha_1}\dots p_r^{\alpha_r}$$ for distinct primes $$p_i$$ such that $$p_1,\dots,p_s$$ divide $$x_0$$ and $$p_{s+1},\dots,p_r$$ don't, then, letting $$m_1=p_1^{\alpha_1}\dots p_s^{\alpha_s}\quad\text{and}\quad m_2=p_{s+1}^{\alpha_{s+1}}\dots p_r^{\alpha_r},$$ you get $$m_1m_2=4|n|$$, $$(m_1,y_0)=1$$ and $$(m_2,x_0)=1.$$