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!

$$\begin{align} 
&\ \ \ \ \overbrace{p\nmid 2a_n,\ \ \ \ p\mid 2a_{n-1},\ldots, p\mid 2a_0,\ p^2\nmid 2a_0}^{\textstyle 2a(x)\ {\rm is}\ p\text{-Eisenstein}}\\[.1em]
\Longrightarrow\  &p\nmid 2,\, \underbrace{p\nmid a_n,\ p\mid\ a_{n-1},\,\ldots,\, p\mid a_0,\ \ p^2\nmid a_0}_{\textstyle a(x)\ {\rm is}\ p\text{-Eisenstein}}\\[.5em]
&\ \ \ \text{since }\,p\nmid 2,\ p\mid 2a_i\,\Rightarrow\, p\mid a_i\ \ \rm by\ Euclid
\end{align}\qquad$$
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.
$$\begin{align}\bmod p\!:\,\ 2a(x) &\equiv c\ x^n,\ \ c\not\equiv 0,\ \ p^2\nmid 2a(0)\\[.3em]
\Rightarrow\, 2\not\equiv 0,\ a(x)&\equiv \bar c\, x^n,\,\ \bar c\not\equiv 0,\ \ p^2\nmid a(0),\,\ \bar c = c/2\end{align}\qquad$$
