Remarks on Balanced Incomplete Block Designs

Proof. Assume the hypothesis and the falsity of either conclusion. Construct matrix A of 2x + 2 rows and 4x+4 columns, with entries + 1 and — 1. The first column contains exclusively +1, and the second column — 1. Set up one-to-one correspondences between rows of A and elements of D ; between columns other than the first two of A and blocks of D. Enter +1 if the element is contained in the block and —1 otherwise. Then each row of A contains exactly 1+0 + (2x + l) = 2x + 2 entries +1, and hence 2x+2 entries —1. Further, each pair of distinct rows contains two +1 's in exactly 1 + 0 +x = x +1 like columns. It follows that each pair of distinct rows of A has in like columns the ordered pairs (1, 1), (1, —1), ( —1, 1), and (—1, —1) each x + 1 times. Select x + 1 rows corresponding to the elements of a block which is either repeated in D or disjoint from another block of D. Let Ao be the x+1 by 4x+4submatrixof A composed of these rows. Then AoAq =(4x+4)7, with the identity matrix of dimension x + 1. A0 has four columns each with all entries equal; these are the first two and those corresponding to the pair of special blocks of D. All pairs of unequal ± 1 entries in like columns of Ao therefore occur in the other 4x columns. Each pair of distinct rows of A 0 contains 2x+2 unlike entries in like columns; thus the total number of pairs of unequal entries within columns is (x + l)x(2x+2)/2 = (x + l)2x. Among the ix columns the average number of unlike pairs is accordingly (x + 1) 2x/4x = (x + l)2/4. A partition of x+1 elements into two classes, with as many as (x + l)2/4 pairs of elements not in the same class, is possible only if x+1 is even. There exist numerous examples of BIBD with parameters as in the theorem and x an odd positive integer. (It is likely in fact that a design exists for each choice of x. This would be a corollary of Paley's