Prove $\sum_{k=0}^{58}\binom{2017+k}{58-k}\binom{2075-k}{k}=\sum_{k=0}^{29}\binom{4091-2k}{58-2k}$ Show that :

$$\sum_{k=0}^{58}\binom{2017+k}{58-k}\binom{2075-k}{k}=\sum_{k=0}^{29}\binom{4091-2k}{58-2k}$$

It seems interesting, but how to prove?
 A: Imagine having $N+2m$ boxes in a row. We need to pick $N$ boxes to be special, but in such a way that we can choose an $s$ such that of the first $m+s$ boxes, there are exactly $N-s$ boxes special.
This is quite a strange property, so let's see where this can go wrong. What are ways to pick the boxes that do not satisfy the property? Let's just loop through $s$ to see what's going on. We start with $s=0$. The property now says that of the first $m$ boxes, $N$ need to be special. Let's assume this is not true; there are less than $N$ boxes special in the first $m$ boxes. Then we increase $s$ by one; now there need to be $N-1$ boxes special in the first $m+1$ boxes. If there are more boxes special, then this means there were $N-1$ boxes of the first $m$ special, and the $m+1$'th box was special. We know now that there are no other $s$'s for which this can go right, since of the first $m+s$ boxes, there will be $N$ special (not $N-s$, as we should have).

if there are less than $N-s$ boxes special of the first $m+s$ boxes, but more than $N-s-1$ out of the first $m+s+1$, then of the first $m+s$ boxes there are exactly $N-s-1$ special ones, furthermore, the $m+s+1$'st box is special. This results in an invalid way to pick our boxes (note that this only makes sense when $0\leq s<N$).

So the number of ways we can pick boxes in an invalid way, is
$$\sum_{k=0}^{N-1}{m+k\choose N-k-1}{m+N-k-1\choose k}$$
since we pick the $m+k+1$'th box to be special, and then pick $N-k-1$ boxes before that, and $k$ boxes after that.
The number of ways to pick boxes in a valid way, is of course
$$\sum_{k=0}^{N}{m+k\choose N-k}{m+N-k\choose k}$$
Notice that the number of ways to pick boxes in an invalid way, is the same as picking $N-1$ boxes out of $2m+N-1$, but in such a way that we can choose an $s$ such that of the first $m+s$, there are exactly $N-1-s$ boxes special. Can you hear induction screaming to be used?
Let's write $S_{2m}(N)$ to mean the number of ways that we can choose $N$ boxes out of $2m+N$, but in such a way that we can choose an $s$ (between $0$ and $N$) such that of the first $m+s$, there are exactly $N-s$ boxes special. Then, as noted above, we know that the number of ways to not do this, is $S_{2m}(N-1)$. Thus,
$${2m+N\choose N}=S_{2m}(N)+S_{2m}(N-1)$$
We will now use induction to prove

$$S_{2m}(N)=\sum_{k=0}^N(-1)^{N-k}{2m+k\choose k}$$

We know that $S_{2m}(0)=1$, and so $S_{2m}(0)={m+0\choose 0}$ is indeed true. Now, assume the expression is true for $N$. Then:
\begin{align}
S_{2m}(N+1)&={2m+N+1\choose N+1}-S_{2m}(N)\\
&={2m+N+1\choose N+1}-\sum_{k=0}^N(-1)^{N-k}{2m+k\choose k}\\
&=(-1)^{N+1-(N+1)}{2m+N+1\choose N+1}+\sum_{k=0}^N(-1)^{N+1-k}{2m+k\choose k}\\
&=\sum_{k=0}^{N+1}(-1)^{N+1-k}{2m+k\choose k}\\
\end{align}
and so we can conclude that the expression must be true for all $N\geq0$.

So far, we've proven that

$$\sum_{k=0}^{N}{m+k\choose N-k}{m+N-k\choose k}=\sum_{k=0}^N(-1)^{N-k}{2m+k\choose k}$$
Let's now set $2n$, so that we get

$$\sum_{k=0}^{2n}{m+k\choose 2n-k}{m+2n-k\choose k}=\sum_{k=0}^{2n}(-1)^k{2m+k\choose k}$$

And now we just mess around a little bit with the right-hand side (I use ${a\choose-1}=0$ here):
\begin{align}
\sum_{k=0}^{2n}(-1)^k{2m+k\choose k}&=\\
\sum_{k=0}^{n}{2m+2k\choose 2k}-\sum_{k=1}^{n}{2m+2k-1\choose 2k-1}&=\\
\sum_{k=0}^{n}\left({2m+2k-1\choose 2k}+{2m+2k-1\choose 2k-1}\right)-\sum_{k=0}^{n}{2m+2k-1\choose 2k-1}&=\\
\sum_{k=0}^{n}{2m+2k-1\choose 2k}&=\\
\sum_{k=0}^{n}{2m+2(n-k)-1\choose 2(n-k)}&=\\
\sum_{k=0}^{n}{2m+2n-2k-1\choose 2n-2k}
\end{align}
so finally, we can conclude
$$\sum_{k=0}^{2n}{m+k\choose 2n-k}{m+2n-k\choose k}=\sum_{k=0}^{n}{2m+2n-2k-1\choose 2n-2k}$$
(and the requested result is obtained by setting $n=29$ and $m=2017$).
