Number theory in Wu formula I want to prove the Wu formula, and I have managed to understand that it is equivalent to the following fact:
For any integer $m>k>0$ the number $\sum\limits_{j=0}^{k}{{m-j-1}\choose{k-j}}{{k+m}\choose{j}}$ is even.
I've wasted a lot of time trying to prove it. Can anyone help?
 A: First, if $m=k$ the expression is zero, so we can prove for $m\geq k>0$.
Let us prove it by induction on $m-k$.
We verify the base for $m=k$: $\sum\limits_{j=-\infty}^{+\infty}{{m-j-1}\choose{k-j}}{{k+m}\choose{j}}=0$ and for $m=k+1$: $\sum\limits_{j=-\infty}^{+\infty}{{k-j}\choose{k-j}}{{2k+1}\choose{j}}=\frac{1}{2}\sum\limits_{j=-\infty}^{+\infty}{{2k+1}\choose{j}}=2^{2k}$.
Now let us prove the induction step. All further equalities are written mod 2.
We will constantly use that ${{x}\choose{y}}={{x-2}\choose{y}}+2{{x-2}\choose{y-1}}+{{x-2}\choose{y-2}}={{x-2}\choose{y}}+{{x-2}\choose{y-2}}$
It is clear that $\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j}}{{k+m-2}\choose{j}}=\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j}}{{k+m-2}\choose{j}}+\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j-2}}{{k+m-2}\choose{j}}+\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j-2}}{{k+m-2}\choose{j}}$.
Therefore, $\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j}}{{k+m-2}\choose{j}}=\sum\limits_{j=-\infty}^{+\infty}{{m-j-1}\choose{k-j}}{{k+m-2}\choose{j}}+\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j-2}}{{k+m-2}\choose{j}}$
We change $j$ to $i$ in the second summand and obtain
$\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j}}{{k+m-2}\choose{j}}=\sum\limits_{j=-\infty}^{+\infty}{{m-j-1}\choose{k-j}}{{k+m-2}\choose{j}}+\sum\limits_{i=-\infty}^{+\infty}{{m-i-3}\choose{k-i-2}}{{k+m-2}\choose{i}}$
Now we set $j=i+2$ in the second summand and obtain
$\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j}}{{k+m-2}\choose{j}}=\sum\limits_{j=-\infty}^{+\infty}{{m-j-1}\choose{k-j}}{{k+m-2}\choose{j}}+\sum\limits_{i=-\infty}^{+\infty}{{m-j-1}\choose{k-j}}{{k+m-2}\choose{j-2}}$
Hence,
$\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j}}{{k+m-2}\choose{j}}=\sum\limits_{j=-\infty}^{+\infty}{{m-j-1}\choose{k-j}}{{k+m}\choose{j}}$
By the induction hypothesis, $\sum\limits_{j=-\infty}^{+\infty}{{m-j-3}\choose{k-j}}{{k+m-2}\choose{j}}=0$, therefore, $\sum\limits_{j=-\infty}^{+\infty}{{m-j-1}\choose{k-j}}{{k+m}\choose{j}}=0$.
