Li Shanlan had not given a proof of the identity in
Duoji bilei. The first proof using differential equations and
Legendre polynomials, concepts foreign to Li, was published by
Pál Turán in 1936, and the proof appeared in Chinese in
Yung Chang's paper published in 1939. The proof begins by expressing n \choose q as
Vandermonde's convolution: :{n \choose q } = \sum_{k=0}^q {{n-p} \choose k}{p \choose {q-k}} Pre-multiplying both sides by n\choose p , :{n \choose p}{n \choose q} = \sum_{k=0}^q {n \choose p} {{n-p} \choose k}{p \choose {q-k}}. Using the following relation : {n \choose p} {{n-p} \choose k} = {{p+k} \choose k}{n \choose {p+k}} the above relation can be transformed to : {n \choose p}{n \choose q} = \sum_{k=0}^q {p \choose {q-k}} {{p+k} \choose k}{n \choose {p+k}}. Next the relation :{p \choose {q-k}} {{p+k} \choose {k}} = {q \choose k}{{p+k}\choose {q}} is used to get : {n \choose p}{n \choose q} = \sum_{k=0}^q {q \choose k}{n \choose {p+k}}{ {p+k}\choose q}. Another application of Vandermonde's convolution yields :{ {p+k}\choose q} = \sum_{j=0}^q {p \choose j}{k \choose {q-j}} and hence :{n \choose p}{n \choose q} = \sum_{k=0}^q {q \choose k}{n \choose {p+k}} \sum_{j=0}^q {p \choose j}{k \choose {q-j}} Since p \choose j is independent of
k, this can be put in the form : {n \choose p}{n \choose q} = \sum_{j=0}^q {p \choose j}\sum_{k=0}^q {q \choose k}{n \choose {p+k}}{k \choose {q-j}} Next, the result : {q \choose k}{k \choose {q-j}} = {q\choose j}{j \choose{q-k}} gives : {n \choose p}{n \choose q} = \sum_{j=0}^q {p \choose j}\sum_{k=0}^q {q \choose j}{j \choose {q-k}}{n \choose {p+k}} :::: = \sum_{j=0}^q {p \choose j}{q \choose j}\sum_{k=0}^q {j \choose {q-k}}{n \choose {p+k}} :::: = \sum_{j=0}^q {p \choose j}{q \choose j}{{n+j}\choose {p+q}} Setting
p =
q and replacing
j by
k, :{n \choose p}^2 = \sum_{k=0}^p {p \choose k}^2{{n+k}\choose {2p}} Li's identity follows from this by replacing
n by
n +
p and doing some rearrangement of terms in the resulting expression: :{{n+p} \choose p}^2 = \sum_{k=0}^p {p \choose k}^2{{n+2p-k}\choose {2p}} ==On
Duoji bilei==