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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!

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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 $(m_2,x_0)=1.$

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