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A 3x3 square is filled out with 9 positive integers such that the product of each row, column, and diagonals are equal. The sum of all 4 corners is less that 10. Find all possible configurations
In this case, a square cannot be rotated to make another one. You can repeat numbers too.
I thought of this as a systematic question, carefully writing down all possibilities, but that took way too long. Is there a faster way?

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  • $\begingroup$ So you're allowed repetitions (which would make a square of only $1$'s allowed), but you're not allowed to make a square that is rotationally symmetric (which makes a square of only $1$'s not allowed after all). Is this right? $\endgroup$ – Arthur Mar 5 '17 at 1:59
  • $\begingroup$ @Arthur No. Say i had the square (which is not a legitimate one that answers my question) (1st row): 123 (2nd row): 456 (3rd row): 789. Then I had the square (1st row): 369 (2nd row): 258 (3rd row) 147. These two would not be different squares because I can rotate one of them to become the other.\ $\endgroup$ – Gerard L. Mar 5 '17 at 2:07
  • $\begingroup$ @Arthur: I think OP means that rotations are not counted as distinct. I would also ask—are reflections not counted as distinct as well? $\endgroup$ – Brian Tung Mar 5 '17 at 2:15
  • $\begingroup$ There can't have been that many possibilities. I can only come up with (a) all $1$'s, (b) all $2$'s, (c) $1, 4, 2; 4, 2, 1; 2, 1, 4$. I used this short white paper as a starting point, but instead of using the numbers directly, allowed zero elements, then used those elements as exponents to prime factors. Does any prime besides $2$ work? $\endgroup$ – Brian Tung Mar 5 '17 at 2:23
  • $\begingroup$ Reflections also don't count. $\endgroup$ – Gerard L. Mar 5 '17 at 2:23
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Every magic multiplication 3x3 square is of the form: $$\begin{bmatrix} \frac{a}{b} & abc & \frac{a}{c} \\ \frac{ab}{c} & a & \frac{ac}{b} \\ ac & \frac{a}{bc} & ab \end{bmatrix}$$ with the magic product $a^3$

If we assume all entries are integers, the square is rewritten as $$\begin{bmatrix} ac & ab^2c^2 & ab \\ ab^2 & abc & ac^2 \\ abc^2 & a & ab^2c \end{bmatrix}$$ with the magic product $a^3b^3c^3$

The sum of the corners is then $ac+ab+abc^2+ab^2c=a(1+bc)(b+c)<10$. Since $a,b,c$ must be positive integers, it follows that $1\leq a,b,c \leq 2$.

The only valid solutions are $(a,b,c)=(1,1,1),(2,1,1),(1,2,1),(1,1,2)$ corresponding to the squares: $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}, \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}, \begin{bmatrix} 1 & 4 & 2 \\ 4 & 2 & 1 \\ 2 & 1 & 4 \end{bmatrix}, \begin{bmatrix} 2 & 1 & 4 \\ 4 & 2 & 1 \\ 1 & 4 & 2 \end{bmatrix}$$ The last two squares actually just the same square rotated. So there are 3 possible such squares. Definitely not too many to count. $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}, \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}, \begin{bmatrix} 1 & 4 & 2 \\ 4 & 2 & 1 \\ 2 & 1 & 4 \end{bmatrix}$$

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