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I. The Idea of Analytic Number Theory
6
sums then are the terms of the product
2
4
3
6
2
5
3
4
5
6
(z + z + z + z + z + z )(z + z + z + z + z + z )
5
2
4
3
6
z + 2z + 3z + 4z + 5z + 6z 7
8
9
10
11
+ 5z + 4z + 3z + 2z + z 12
The correspondence, for example, says that there are 3 ways for the
10
10 to show up, the coefficients of z being 3, etc. The question is: Is
there any other way to number these two cubes with positive integers
so as to achieve the very same alternatives?
Analytically, then, the question amounts to the existence of
positive integers, a 1 ,...,a 6 ; b 1 ,...,b 6 , so that
(z + ··· + z )(z + ··· + z )
a 6
a 1
b 6
b 1
10
3
12
4
11
2
z + 2z + 3z + ··· + 3z + 2z + z .
These would be the “Crazy Dice” referred to in the title of this sec-
tion. They look totally different from ordinary dice but they produce
exactly the same results!
So, repeating the question, can
b 1
a 6
a 1
b 6
(z + ··· + z )(z + ··· + z )
5
4
6
2
3
(z + z + z + z + z + z ) (6)
5
6
4
2
3
× (z + z + z + z + z + z )?
To analyze this possibility, let us factor completely (over the ratio-
5
6
4
3
2
nals) this right-hand side. Thus z+z +z +z +z +z z 1−z 6
1−z
2
2
3
2
zà 1+z+z )(1+z ) zà 1+z+z )(1+zð( 1−z+z ). We conclude
from (6) that the “a-polynomial” and “b-polynomial” must consist of
these factors. Also there are certain side restrictions. The a’s and b’s
are to be positive and so a z-factor must appear in both polynomials.
2
The a-polynomial must be 6 at z 1 and so the (1 + z + z )(1 + zð
factor must appear in it, and similarly in the b-polynomial. All that
2
is left to distribute are the two factors of 1 − z + z . If one apiece are
given to the a- and b-polynomials, then we get ordinary dice. The
only thing left to try is putting both into the a-polynomial.