*Ars Nova*, which introduced new rhythmic schemes and musical notation. He had a deep knowledge of philosophy, rhetoric, and mathematics.

In many ways, de Vitry's interests and accomplishments reflected the Pythagorean view that music is a subdivision of arithmetic, as shown, for example, in the simple mathematical relations between pitch and length of a string (see Circles of Dissonance). His work honored the dictum of the Roman philosopher Boethius (480-524) that "music is number made audible."

One of de Vitry's observations concerned numbers that can be written as a power of 2 multiplied by a power of 3. In modern terms, such numbers would be expressed as 2

^{n}x 3^{m}, where*n*and*m*are integers. He called them harmonic numbers.Here's a table of all harmonic numbers less than 1000.

De Vitry focused on pairs of harmonic numbers that differ by 1: 1, 2; 2, 3; 3, 4; and 8, 9. As it happens, those particular pairs correspond to musically significant ratios, representing an octave, fifth, fourth, and whole tone. He wondered whether there are any other such pairs of harmonic numbers.

In 1342, Levi ben Gerson, also known as Gersonides (1288-1344), proved that the original four pairs of harmonic numbers are the only ones that differ by 1. Here's how his proof went (using modern notation).

If two harmonic numbers differ by 1, one must be odd and the other even. The only odd numbers are powers of 3 (top row of chart, above). Hence, one of the two harmonic numbers must be a power of 3 and the other a power of 2 (first row and first column of chart). The basic task then involves solving two equations:

2

^{n}= 3^{m}+ 1 and 2^{n}= 3^{m}− 1.Gersonides had the idea of looking at remainders after division of powers of 3 by 8 and powers of 2 by 8. For example, 27 divided by 8 gives a remainder of 3. For powers of 3, the remainders all turn out to be 1 or 3, depending on whether the power is even or odd. The remainders for powers of 2 are 1, 2, 4, then 0 for all powers higher than 2.

For 2

^{n}= 3^{m}+ 1, when*m*is odd, 3^{m}has remainder 3, and 2^{n}= 3^{m}+ 1 must then have the remainder 4, so*n*= 2 and*m*= 1. That gives the consecutive harmonic numbers 3, 4. When*m*is odd, the equation gives the consecutive harmonic numbers 1, 2.For 2

^{n}= 3^{m}− 1, when*m*is odd, 3^{m}has remainder 3, so 2^{n}= 3^{m}− 1 has remainder 2; as a result,*n*= 1 and*m*= 1, to give the consecutive harmonic numbers 2, 3.The final case, when

*m*is even, is a little trickier and requires substituting 2*k*for*m*, then solving the equation 2^{n}= 3^{2k}− 1 = (3^{k}− 1)(3^{k}+ 1). That gives the consecutive harmonic numbers 8, 9.*QED*.Interestingly, the theorem about harmonic numbers proved by Gersonides has connections with Fermat's last theorem and with the abc conjecture, a topic of some interest among number theorists.

*Originally posted January 25, 1999*

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