Necessary conditions for a Sudoku puzzle to have no repetitions Is it true that if a Sudoku puzzle has the following features there will be no repetitions in rows, columns and $3 \times 3$ subsquares?


*

*The sum of each row must be $45$

*The sum of each column must be $45$

*The sum of each $3 \times 3$ subsquare must be $45$


If so, why? Is there a mathematical proof? If not, why? Is there a case where these conditions are satisfied, but is there at least one repetition?
Thanks!
 A: No. For instance, this "sudoku" fulfills your conditions, but has some repetitions:
$$
\begin{array}{|ccc|ccc|ccc|}
\hline 5&5&5&5&5&5&5&5&5\\
 5&5&5&5&5&5&5&5&5\\
 5&5&5&5&5&5&5&5&5\\
\hline 5&5&5&5&5&5&5&5&5\\
 5&5&5&5&5&5&5&5&5\\
 5&5&5&5&5&5&5&5&5\\
\hline 5&5&5&5&5&5&5&5&5\\
 5&5&5&5&5&5&5&5&5\\
 5&5&5&5&5&5&5&5&5\\
\hline
\end{array}
$$
A: If all the cells are distinct 1 through 9 then the sum is 1+2+....+9 =45.  But there is utterly no reason earth to assume the converse, that is $a+b+.... +i = 45$ then they are all distinct.
For any $b,...,h =N$ we can have $a$ be any $1 \le a \le 45-N $ and $i = 45-N-a$.  And we can determine values for the other rows and columns.  Yes, it takes a bit of thought to actually work this out but there is no reason that  that keeping them distinct will be a requirement.
Let's suppose for instance we have a grid labeled A1....A9..... I1.... I9 where every row, column and quadrant add up to 45.  Then lets say we replace mk (where $A \le m \le I$ and $1\le k \le 9$) with mk + 1.  Then we replace mj in the same column and quadrant with mk - 1$, replace nk in the same column and quadrant with nk-1 and nj with nj + 1.  Then all the quadrants, columns and rows still add to 45 but one or the other or both grids are no longer distinct.
e.g  suppose we have:
$\begin{array}{|ccc|ccc|ccc|}
\hline 1&2&3&4&5&6&7&8&9\\
 4&5&6&7&8&9&1&2&3\\
 7&8&9&1&2&3&4&5&6\\
\hline 2&3&4&5&6&7&8&9&1\\
 5&6&7&8&9&1&2&3&4\\
 8&9&1&2&3&4&5&6&7\\
\hline {\color{red}3}&4&{\color{red}5}&6&7 &8&9&1&2\\
 6&7&8&9&1&2&3&4&5\\
 {\color{red}9}&1&{\color{red}2}&3&4&5&6&7&8\\
\hline
\end{array}$
and we replace it with 
$\begin{array}{|ccc|ccc|ccc|}
\hline 1&2&3&4&5&6&7&8&9\\
 4&5&6&7&8&9&1&2&3\\
 7&8&9&1&2&3&4&5&6\\
\hline 2&3&4&5&6&7&8&9&1\\
 5&6&7&8&9&1&2&3&4\\
 8&9&1&2&3&4&5&6&7\\
\hline {\color{blue}4}&4&{\color{blue}4}&6&7 &8&9&1&2\\
 6&7&8&9&1&2&3&4&5\\
 {\color{blue}8}&1&{\color{blue}3}&3&4&5&6&7&8\\
\hline
\end{array}$
Note, the sums must be the same but values need not be distinct.
A: imagine each digit is a 5, then all summation to 45 are met and we clearly have repition, all that's necessary is a pattern with an average of 5 to pull this off.
