By Eric Mendelsohn

ISBN-10: 0080871763

ISBN-13: 9780080871769

ISBN-10: 0444863656

ISBN-13: 9780444863652

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**Example text**

Now, as 2(bz- b1)= 1 (mod m ) it follows that m is odd and that b2- bl = i ( m + 1) (mod m ) . This implies that q 3 3 and that at least one of columns b1 and b2 is an A-column. And, since a2= al + bz - bl (mod m ) , if row al intersects the B-area then a2 intersects the C-area, and if aZintersects the C-area then al intersects the B-area. We must note that if column bl is a mixed column, then in two particular cases the entry of S ( m , p ) in G has second coordinate different from that usual to an entry in that cell of a T @ ) of that type: if a I= i ( m + 1) the T @ ) of ( a l ,b l ) A ( mis ) an A@),but the entry in G comes from the C @ )below, where second coordinates 1 and 2 are interchanged in the entries of S ( m , p ) , compared to A@);and if al = $(m + 1)+ mlq the T @ )is a C@),but the entries in S(m, p ) come from an A@).

N + m } (with diagonal (1,. . , m, 1, . . , m ) ; this follows from Theorem 4 (for example)). Now apply Theorem 16 to embed this square in a half-idempotent symmetric latin square N of side 2n on symbols (1,. . , 2 n } . Corollary 17 was also proved by Hoffman [9]. Lindner wanted this result in connection with a proof of the result of Doyen and Wilson [ 6 ] ,that any Steiner Triple System of order u can be embedded in a Steiner Triple System of order u for any u k 2u + 1, u = 1 or u = 3 (mod 6).

Because of this row, if the cells of L were to be a subsquare of D ( m , p ) also, it would be on the same symbols. If m > 1 then. by Proposition 4@) and (c), L does not contain the cell of S(rn,p) below the one known to be in L, and so the entry, in P ( m , p ) , of that cell is not a symbol of L. In D ( m , p ) , however, it is placed in a cell of L, and so the cells of L do not form a subsquare of D ( m , p). Proposition 7 . D ( m , p ) contains no proper subsquare. Proof. Throughout this proof.

### Algebraic and Geometric Combinatorics by Eric Mendelsohn

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