Parts and Divisors
[10/100%]Let be any non-negative integers. We have
Let be any prime number and , , , , , , where are not divisible by . The exponent of on the left-hand side is equal to , and the exponent of on the right-hand side is equal to
The least common multiple of the positive integers may be represented as follows:
We have
The general element of the first determinant , if is a (proper or improper) part of , otherwise it (); in the second determinant the general element is equal to the number of common parts of and , i.e. the smaller of the two numbers and ().
Prove that is equal to 1, where denotes the determinant whose general element is equal to the number of common divisors of and , in other words to the number of divisors of the greatest common divisor of and , where .
Let be arbitrary, , . We have
The general element of the determinant is where , . (Generalization of 30.)
Let be arbitrary, , . We have
The general element of the determinant is where , . (Generalization of 31.)
If are arbitrary and , , then we clearly have
If are arbitrary and , , then we have
and in general
where is the Möbius function (see definition in § 5 preceding 38).
Denote by an arbitrary function defined for . Let
where the last sum is extended over the numbers that are and are relatively prime to . We then have
As is well known, we have
Set
where the product is extended over those numbers that are and relatively prime to . ( is termed the th cyclotomic polynomial.) The zeros of are the th roots of unity, the zeros of are the primitive th roots of unity. Derive the formula
If is the Möbius function, then we have