A power series equality involving binomial coefficients

Searching with Approach Zero you can find several references to this identity, for example here. You just need to tweak a little the index and variables.

This is Li Shanlan identity.

Permit me to contribute an algebraic proof. We seek

$${m+k\choose k}^2 = \sum_{q=0}^m {k\choose m-q}^2 {2k+q\choose q}$$

Starting with the RHS we find

$$\sum_{q=0}^m {k\choose q}^2 {2k+m-q\choose m-q} \\ = \sum_{q=0}^m {k\choose q} [z^k] z^q (1+z)^k [w^{m}] w^q (1+w)^{2k+m-q}.$$

Now we may extend $q$ to infinity because the coefficient extractor $[w^m]$ enforces the upper limit. We get

$$[z^k] (1+z)^k [w^m] (1+w)^{2k+m} \sum_{q\ge 0} {k\choose q} z^q w^q (1+w)^{-q} \\ = [z^k] (1+z)^k [w^m] (1+w)^{2k+m} (1 + zw/(1+w))^k \\ = [z^k] (1+z)^k [w^m] (1+w)^{k+m} (1 + w + zw)^k$$

Re-expanding we find

$$[z^k] (1+z)^k [w^m] (1+w)^{k+m} \sum_{q=0}^k {k\choose q} w^q (1+z)^q.$$

We may set the upper limit of the sum to $m.$ (If $k\lt m$ the values $k\lt q\le m$ produce zero from the binomial coefficient and we may raise $q$ to $m.$ If $k\gt m$ the values $m\lt q\le k$ produce zero by the coefficient extractor $[w^m]$ and we may lower $q$ to $m.$) We get

$$[z^k] (1+z)^k [w^m] (1+w)^{k+m} \sum_{q=0}^m {k\choose q} w^q (1+z)^q \\ = \sum_{q=0}^m {k\choose q} {k+q\choose k} {k+m\choose m-q}.$$

Now observe that

$${k+q\choose k} {k+m\choose m-q} = \frac{(k+m)!}{k! \times q! \times (m-q)!} = {m+k\choose k} {m\choose m-q}.$$

This yields for our sum

$${m+k\choose k} \sum_{q=0}^m {k\choose q} {m\choose m-q}.$$

Using Vandermonde we obtain at last

$$\bbox[5px,border:2px solid #00A000]{ {m+k\choose k}^2.}$$