Evans conjecture for symmetric latin squares The Evans conjecture ( which was proved later by Smetaniuk) states that for any $n$, if at most $n-1$ entries of a partial $n\times n$ latin square are filled,  it can be completed to the full latin square.
My question pertains to whether this is applicable to symmetric (or commutative) latin square? That is, given $n\times n$ latin square, with at most $n-1$ prefilled entries ( which are symmetric with respect to diagonal), can it be completed to a symmetric latin square? Note that this corresponds to pre- total coloring of a complete graph of order $n$ with at most $n-1$ previous filled colors.  Any hints? Thanks beforehand.
 A: No, when it comes to symmetric latin squares it is no longer true that as many as $n-1$ cells can be prescribed unconditionally. This is explained in the Ph.D. thesis of Matthew Henderson.

The key point here is that in a symmetric latin square, precisely
because of the symmetry, every symbol $\sigma$ occurs an even number
of times in cells outside of the main diagonal. Therefore, as every
symbol $\sigma$ occurs $n$  times in total it follows that the number
of cells of the main diagonal containing symbol $\sigma$ is congruent
to $n$ modulo 2. A partial latin square can be incompletable because
there are more symbols which occur on the main diagonal a number of
times (zero included) incongruent to $n$  modulo 2 than there are
empty cells on the main diagonal.
If this obstruction does not occur the diagonal is called "admissible". For $n$ odd a diagonal is admissible if and only if no symbol occurs more than once on it.

The generalization of Evans theorem to symmetric latin squares by Andersen and Hilton is that the symmetric latin square is completable if the diagonal is admissible and at most $n-1$ cells in total are prescribed.
In response to the comment by Richard Stanley: it is not sufficient to only count cells on or above the main diagonal, as this counter example shows. The diagonal is admissible ($n=5$ is odd and no symbol appears more than once on the diagonal), and the number of prescribed cells on or above the main diagonal is $n-1=4$, and yet the latin square is not completable.

