Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm not quite sure how to even start this problem. I'm really just looking for direction on how to begin.

The $t$ mutually orthogonal Latin squares $A_1, A_2, ... , A_t$ of side $n$ have mutually orthogonal subsquares $ S_1, S_2, ... S_t$ occupying their upper left $s$x$s$ corners. Prove that $n$ is greater than or equal to $(t+1)s$.

I know that $t$ must be less than $n$, but I can't find any other information to help me.

share|cite|improve this question
up vote 3 down vote accepted

Consider the cells in row $s+1$ to the right of column $s$. These $n-s$ cells must contain each of the $s$ symbols of the subsquares, since none of those symbols appear in the $s$ cells in the left part of the row. Furthermore, this must be true for the cells in corresponding positions in each of the $t$ MOLS, and no single position in two distinct orthogonal squares can contain multiple symbols from the subsquares, since this would create a pairing that already exists within the orthogonal subsquares. Thus $t$ copies of $s$ symbols must be distributed among only $n-s$ positions, yielding the desired inequality.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.