Q:

2-Adic Height of Binomial Coefficients

Prove that out of all the terms in the summation 
$$\sum^{n-1} _{\substack{i=0 \\ {i \; \text{even}}}} \dbinom{n}{i} x^{n-1-i}y^{i} \; \text{ for n odd},$$
no two terms have the same $2-$adic valuation (the same exponent of $2$ in their prime factorization) if $x = 2 \sqrt{b}$ for some even $b$, and $y$ is some odd integer. 
I tried to show this by contradiction, but got stuck really quickly. A contradiction isn't really that easy to get... For instance, consider that for some $m>j$ such that both $m$ and $j$ are indices in the summation, 
$$v_2 \left( \dbinom{n}{m} x^{n-1-m} y^m \right) = v_2 \left( \dbinom{n}{j} x^{n-1-j} y^j \right) .$$
(actually we could just disregard the $y$ term altogether...) 
Could someone please suggest some steps or state why proving this is impossible? Thank you. 

A:

It cannot be proven:
Look at the summation for n=8 and b=4. The 5th and 6th terms have the same 2-adic valuation.
That's because, moving from the 5th to the 6th term, the valuation of the binomial coefficient increases by 2 while the valuation of the power of x decreases by the valuation of x which is also 2.

