Number theory, prove that a prime number $p \mid 1$

Question

Consider a prime number $p > 1$ and $a \in \mathbb{Z}$ and $p < a$. We know $p \mid a$, then $a = p.b$ for $b \in \mathbb{N}$.

We also already know the congruence $a \equiv 1 (\text{mod } m)$ which implies $$ a = m.x + 1 $$

We need to prove that $p \mid 1$ (which in fact is a contradiction of a theorem).

I give you the proof and then let you know what my problem is.

Proof:

$a = p.b = m.x + 1 = p.(m/p). x +1$

and since if $p \mid x, p \mid y, x = y + z \Rightarrow p \mid z$

in our case $p \mid \overbrace{a}^{ = x}, p \mid \overbrace{p.(m/p).x}^{ = y}$ and $a = p.(m/p).x + \overbrace{1}^{ = z}$ so $p \mid 1$.

My Problem

I can't convince my self that $p \mid p.(m/p).x$. Because it is true only if $(m/p).x \in \mathbb{Z}$ and I can't prove that $(m/p).x$ is an integer.

Thanks in advance.

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UPDATE

Answer 1

In the 6th line of the text, $m$ is defined in such a way that $p_r|m$. Hence $\frac{m}{p_r} \in \mathbb{Z}$.