Help with proof involving Eisenstein's criterion Let $a\in\mathbb {Z}[X]$ and suppose that $2a\in\mathbb {Z}[X]$ is Eisenstein with respect to a prime $p\in\mathbb {Z}$. 
How can I prove that $a$ is an Eisenstein polynomial with respect to $p$?
Any help would be greatly appreciated!
 A: Hint 1 Show that $p \neq 2$. To do this, use $p\nmid 2a_n$.
Hint 2: If $p|2a_k$ and $p \neq 2$ deduce that $p|a_k$.
Hint 3: If $p^2 \nmid 2a_0$ show that $p^2 \nmid a_0$. 
A: $$\begin{align} 
&\ \ \ \ \overbrace{p\nmid 2a_n,\ \ \ \ p\mid 2a_{n-1},\ldots, p\mid 2a_0,\ p^2\nmid 2a_0}^{\textstyle 2a(x)\ {\rm is}\ p\text{-Eisenstein}}\\[.1em]
\Longrightarrow\  &p\nmid 2,\, \underbrace{p\nmid a_n,\ p\mid\ a_{n-1},\,\ldots,\, p\mid a_0,\ \ p^2\nmid a_0}_{\textstyle a(x)\ {\rm is}\ p\text{-Eisenstein}}\\[.5em]
&\ \ \ \text{since }\,p\nmid 2,\ p\mid 2a_i\,\Rightarrow\, p\mid a_i\ \ \rm by\ Euclid
\end{align}\qquad$$
More conceptually recall  the key idea in Eisenstein is the $\,a(x) \equiv cx^n\pmod{\!p}\,$ is (an associate) of a prime power $\,x^n\,$ and prime products factor uniquely. But this form is preserved by unit scalings, i.e.
$$\begin{align}\bmod p\!:\,\ 2a(x) &\equiv c\ x^n,\ \ c\not\equiv 0,\ \ p^2\nmid 2a(0)\\[.3em]
\Rightarrow\, 2\not\equiv 0,\ a(x)&\equiv \bar c\, x^n,\,\ \bar c\not\equiv 0,\ \ p^2\nmid a(0),\,\ \bar c = c/2\end{align}\qquad$$
A: If 
$2a(x) \in \Bbb Z[x] \tag 1$
is Eisenstein with respect to $p \in \Bbb P$, then $p \ne 2$ since Eisenstein forbids $p \mid 2a_n$, where
$a(x) = \displaystyle \sum_0^n a_i x^i \in \Bbb Z[x]; \tag 2$
since
$p \not \mid 2a_n, \tag 3$
we immediately have
$p \not \mid a_n; \tag 4$
with $p \ne 2$ and
$p \mid 2a_i, \; 0 \le i < n, \tag 5$
we may infer
$p \mid a_i, \; 0 \le i < n; \tag 6$
this follows from the well-known property of primes:  
$p \mid ab \Longleftrightarrow [p \mid a] \vee [p \mid b]; \tag 7$
and if $p \ne 2$ and
$p^2 \mid 2a_0, \tag 8$
then
$\exists r \in \Bbb Z, \; p^2r = 2a_0, \tag 9$
whence
$p \mid 2a_0 \Longrightarrow p \mid a_0 \Longrightarrow \exists s \in \Bbb Z, \; ps = a_0, \tag{10}$
from which
$p^2r = 2ps; \tag{11}$
thus
$pr = 2s \Longrightarrow p \mid s \Longrightarrow \exists t \in \Bbb Z, \; s = pt; \tag{12}$
now in light of (10) we may write
$p^2t = a_0 \Longrightarrow p^2 \mid a_0. \tag{13}$
(4), (6) and (13) together show that $a(x)$ is $p$-Eisenstein, that is, Eisenstein with respect to $p \in \Bbb P$, as desired.
