I've watched the video as far as Timothy Gowers sets out the problem. Now here's my attempt at a mental-arithmetic solution plan: 1. For any n, attempt to do

I've watched the video as far as Timothy Gowers sets out the problem. Now here's my attempt at a mental-arithmetic solution plan: 1. For any n, attempt to do an LU decomposition of the Pascal matrix P into the product of lower-triangular matrix L and upper-triangular matrix U. 2. By symmetry, these should be mirror-images of one another in the leading diagonal. 3. det(L) det(U) = det(P), which we know is 1. Also, by symmetry det(L) = det(U), which must then be either 1 or -1. The simplest approach would be to guess that is 1 and that the leading diagonal is all 1s. 4. Induct the first 3 cases, by mental arithmetic to reach: 1 0 0 1 1 1 1 1 1 1 1 0 * 0 1 2 = 1 2 3 1 2 1 0 0 1 1 3 6 5. Guess that L and U are also Pascal's triangle arranged as rows and columns respectively. 6. Compute the bottom right corner for a couple more cases as a check: 1^2 + 3^2 + 3^2 + 1^2 = 20 1^2 + 4^2 + 6^2 + 4^2 + 1^2 = 70 ... seems OK. So I guess, for the bottom right corner there's a binomial identity something like: $\sum_{k=0}^n \binom{n}{k}\binom{n}{k} = \binom{2n}{n}$ and a more general version of that identity for the cell (i,j) with 0 <= i, j <= n $\sum_{k=0}^n \binom{i}{k}\binom{j}{k} = \binom{i+j}{n}$ on an (n+1) x (n+1) matrix.