Extremely cool square. (1 Viewer)

[Fus Ro Dah]

The square given below has the unique property, that the products of the rows, columns, and the two diagonals, yield the same value k.

A B C
D E F
G H I

So ABC=DEF=GHI=ADG=BEH=CFI=AEI=CEG=k.

Prove that if A,B,..,I are all integers, then k must be a perfect cube.

 

[johnpap]

This seems untrue, e.g. what about k=8 with
2 2 2
2 2 2
2 2 2

Also just in general, if any grid worked with k a square, you could multiply every entry by a non-square, preserve the property you were talking about and make the new k non-square...

 

[Fus Ro Dah]

My apologies John. I meant for k to be a perfect cube. I had my mind set on the word square due to the nature of the configuration.

 

[RealiseNothing]

Would be amazing if largarithmic posted.

 

[johnpap]

So if you can prove that it's true for any particular prime factor you're done, so we'll show that if a prime p is a factor of k, then the exponent it has in the prime factorisation of k is congruent to 0 mod 3. So note that each column has the same power of p dividing it and then let a,b,c,d...h,i be the exponents of p in the prime factorisations of A,B,C...H,I. Now, since ABC = DEF for example, a+b+c = d+e+f. So essentially we now need to know if it's possible to fill in the box with numbers such that the sum of every row, column and diagonal is the same, but with that sum not divisible by 3. But this breaks quickly by noting the following:
b+e = g+i (Pivot at H) -> b = g+i-e
b+c = e+i (Pivot at A)
a+b = e+g (Pivot at C), so sum the last two and a+b+c + b = 2e + i+g, but then sub b = g+i-e and a+b+c = 3e, so 3 divides the sum on the top row, hence all the rows, hence any prime that divides k has exponent divisible by 3 and so k is a cube.