Do we conclude from these relations that $ny-hx \mid x(nx-h)$? We have the following relations $$p^i \mid ny-hx \\ (ny-hx)q=(nx-h)n^f \\ p^i \mid x(nx-h)$$ where $p$ is a prime, $x, y \in \mathbb{Z}$, $n>1$, $|h|<n$, $hx\geq 0$, $i>0$. 
Do we conclude from these relations the following? $$ny-hx \mid x(nx-h)$$ 
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EDIT: 
I am looking at the following proof: 


 
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At the part "By the Chinese Remainder Theorem ... $ny-hx \mid x(nx-h)$." 
I haven't understood how from the relations $(7)$ and $(5)$ we conclude that $ny-hx \mid x(nx-h)$. 
Could you explain it to me? 
 A: First of all, the part of the proof you don't understand is (correct me if I'm wrong).


*

*We can prove easily that, by the Chinese remainder theorem there exists an $h\mod n$ such that $h=h(p) \mod p$ for every prime $p$ dividing $n$. (this one is clear ?)

*$(7)$  For every prime $p$, if $h=h(p) \mod p$ then $$ p^i| ny-hx \implies p^i|x$$

*$(5)$ For all $k$ such that $|k|<n$ we have :
$$ny-kx|_n nx-k $$



Question : How can we deduce from $1.,2.$ and $3.$ that $ny-hx|x(nx-h)$ ?

1. Sketch or hint
As we can see, from $1.$ we can apply $2.$ for every prime $p$ which divides $n$ hence we would say that "the commune part between $n$ and $ny-hx$ must divide $x$" and from $3$ we can deduce that "the other part of $ny-hx$ must divide $nx-h$". These two observations both implies that $ny-hx$ divides $x(nx-h)$
2. Solution :  In order to justify $1$, let $n=q_1^{\alpha_1}\cdots q_k^{\alpha_k}$ the factorization of $n$. let $a_i=0$ if $ny$ and $x$ are divisible by the same powers of $q_i$ and $a_i=1$ otherwise, this means that $a_i=h(q_i)$ for every $i=1,\dots,k$. the Chinese remainder theorem implies that there exist an $h<n$ such that:
$$\begin{align}h&\equiv a_1\mod q_1\\ h &\equiv a_2 \mod q_2 \\  h &\equiv \cdots \mod \cdots \\h &\equiv a_k\mod q_k\end{align} $$.
Now we can write :
$$ny-hx =q_1^{r_1}\cdots q_k^{r_k} a \quad \gcd(a,n)=1\tag {*}$$
for every $i=1,cdot,k$ we have $h=h(p_i)\mod p_i$ so we can apply $2$: because $q_i^{r_i}$ divides $ny-hx$ hence $q_i^{r_i}$ divides $x$ this is true for every $i=1,cdot,k$ then :
$$q_1^{r_1}\cdots q_k^{r_k}|x \tag{**}$$
But we have also $h<n$ so we can apply $3.$ hence there exists $f,q$ such that
$$(ny-hx)q=(nx-h)n^f $$
but $a|ny-hx$ hence $a|(nx-h)n^f$ and because $\gcd(a,n^f)=1$ we can deduce (using Euclid's lemma) that $a|nx-h \tag {***}$
Finally from ($*$),($**$) and ($***$) we can deduce the result.
Observation : In the beginning you wrote the problem without quantifiers and conditions which gives another sense to the assertions so every time you copy a problem you have to understand the relevant parts of implications and the result you may for example write:
Let $h,n,x,y$ be some integers such that :
$$\begin{align}\exists f,q \quad (ny-hx)q=(nx-h)n^f \\
  \forall i>0 \quad p^i| ny-hx \implies p^i|x\end{align}$$
for every prime $p$ dividing $n$ (which is equivalent to $h=h(p)\mod p$ because of the definition of $h$)
If you have written the problem this way there would not be such a misunderstanding.
