Problem 8.58

Problems and Theorems in Analysis · 6

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Problem 8.58

Consider all possible factorizations of an even integer of the form n=αβn=\alpha \beta, such that α\alpha is odd (possibly 1) and β\beta is even. Then

∑β−∑α \sum \beta - \sum \alpha

is equal to the sum of the divisors of n/2n/2.

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Problem 8.58.1

Let P(n,k)P(n,k) denote the number of decompositions of nn into a product of kk factors; only positive integers greater than 1 are admissible as factors, and two decompositions are regarded as equal if, and only if, they involve the same factors in the same order. For instance, P(12,2)=4P(12,2)=4 and P(12,3)=3P(12,3)=3 since

12=2⋅6=6⋅2=3⋅4=4⋅3=3⋅2⋅2=2⋅3⋅2=2⋅2⋅3. 12 = 2\cdot 6 = 6\cdot 2 = 3\cdot 4 = 4\cdot 3 = 3\cdot 2\cdot 2 = 2\cdot 3\cdot 2 = 2\cdot 2\cdot 3 .

Show that, for n≥2n\geq 2,

μ(n)=−P(n,1)+P(n,2)−P(n,3)+⋯ , \mu (n) = -P(n,1)+P(n,2)-P(n,3)+\cdots ,

the difference between the number of even'' and odd'' decompositions, that is, decompositions into an even and an odd number of factors, respectively. For instance

μ(12)=−1+4−3=0. \mu (12) = -1+4-3 = 0 .
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Problem 8.58.2

(continued). Show that

P(n,1)−12P(n,2)+13P(n,3)−⋯ P(n,1) - \tfrac 12 P(n,2) + \tfrac 13 P(n,3) - \cdots

equals 1/m1/m if n=pmn=p^m, the mmth power of a prime number pp, but equals 0 if nn is divisible by more than one prime. In the case n=12n=12 we have

1−42+33=0. 1 - \tfrac 42 + \tfrac 33 = 0 .
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Problem 8.58.3

(continued). If nn is the product of mm different prime factors, then

P(n,k)=k!Skm P(n,k) = k!S_k^m

(cf. the introduction to Problem 186 of Part I).

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Problem 8.58.4

If pp is a prime and mm a positive integer, then

P(pm,k)=(m−1k−1). P(p^m,k) = \binom {m-1}{k-1} .

(For this particular case, verify 58.1 by using the binomial theorem; also cf. 58.2 with Problem 38 of Part I.)

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Problem 8.58.5

For n>1n>1 let Q(n)Q(n) denote the number of different decompositions of the integer nn into a product of integers greater than 1; two decompositions are not regarded as different if they involve the same factors, and so the order of factors does not matter (in contrast to 58.1); define Q(1)=1Q(1)=1. For instance Q(12)=4Q(12)=4. If nn is the product of mm different prime factors, then

Q(n)=Tm Q(n) = T_m

(cf. the introduction to Problem 186 of Part I).

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Problem 8.58.6

(continued). Show that

∑n=1∞Q(n)n−s=exp⁡(∑n=1∞(ζ(ns)−1)/n). \sum _{n=1}^\infty Q(n)n^{-s} = \exp \left(\sum _{n=1}^\infty (\zeta (ns)-1)/n\right) .
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Problem 8.58.7

Let nn be a positive integer,

n=2jp1k1p2k2⋯pμkμq1l1q2l2⋯qνlν, n = 2^j p_1^{k_1}p_2^{k_2}\cdots p_\mu ^{k_\mu }q_1^{l_1}q_2^{l_2}\cdots q_\nu ^{l_\nu } ,

where 2,p1,…,pμ,q1,…,qν2, p_1,\ldots ,p_\mu ,q_1,\ldots ,q_\nu are distinct primes, p1≡⋯≡pμ≡1p_1\equiv \cdots \equiv p_\mu \equiv 1, q1≡⋯≡qν≡3(mod4)q_1\equiv \cdots \equiv q_\nu \equiv 3 \pmod4. Set

δ(n)=(k1+1)(k2+1)⋯(kμ+1) \delta (n) = (k_1+1)(k_2+1)\cdots (k_\mu +1)

if l1,l2,…,lνl_1,l_2,\ldots ,l_\nu are all even, and δ(n)=0\delta (n)=0 otherwise. Show that δ(n)\delta (n) is, in fact, the difference between the numbers of two different kinds of divisors of nn:

δ(n)=∑t′∣n(−1)(t′−1)/2, \delta (n) = \sum _{t'\mid n} (-1)^{(t'-1)/2} ,

where the summation is extended over the odd divisors t′t' of nn.

(4δ(n)4\delta (n) is the number of lattice points on the circle x2+y2=nx^2+y^2=n. This fact was discovered by Gauss.)

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Problem 8.58.8

Which of the functions P(n,2),P(n,3),…,Q(n)P(n,2), P(n,3), \ldots , Q(n) and δ(n)\delta (n) are multiplicative?

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