Proof. Let . By Lemma 1, we can suppose that A and are a pair of orthogonal DLSs over , B and are a pair of orthogonal DLSs over , and C and are a pair of orthogonal DLSs over . Write , , , , , , where , , and . Let F be an normal row-square magic rectangle over and write . Let H be a normal row-square magic rectangle over and write . Let , and denote the row magic sum of F, column magic sum of F and row magic sum of , respectively, and let , and denote the row magic sum of H, column magic sum of H and row magic sum of , respectively. Let , , , and denote the magic sums of A, , , and , respectively. Let
where
Next, we prove that is a TMP.
(i) We prove that E and G are DLSs. Noting that A, B, and C are all diagonal Latin squares, for , we have
Similarly, we obtain
Therefore, E is a DLS. Similarly, one can prove that G is a DLS.
(ii) We prove that E and G are orthogonal. Let and , where , , , . Suppose that and . We have
Noting that
we obtain
Since F and H are both normal, we get
Therefore, we have
Since A and are orthogonal, B and are orthogonal, and C and are orthogonal, we obtain
which indicates that and . It follows that E and G are orthogonal. By (i), E and G are orthogonal DLSs over .
(iii) We prove that is an MS. For , we can write , where , , and . We get
Similarly, we have
where . Thus, is an MS.
(iv) We prove that is an MS. For , we can write , where , , and . We get
Similarly, we have
where . Thus, is an MS.
(v) We prove that is an MS. For , we can write , where , , and . We get
Similarly, we have
where . It follows that is an MS.
In summary, we know that is a TMP. Based on Lemma 5, there exists an NMS, . □