Section 4

Parts and Divisors

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Problem 8.28

Let a,b,c,…,k,la,b,c,\ldots ,k,l be any non-negative integers. We have

max⁡(a,b,c,…,k,l)=a+b+c+⋯+k+l \max (a,b,c,\ldots ,k,l) = a+b+c+\cdots +k+l −min⁡(a,b)−min⁡(a,c)−⋯−min⁡(k,l) {}-\min (a,b)-\min (a,c)-\cdots -\min (k,l) +min⁡(a,b,c)+⋯ {}+\min (a,b,c)+\cdots ⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots ±min⁡(a,b,c,…,k,l). \pm \min (a,b,c,\ldots ,k,l) .
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Problem 8.29

Let pp be any prime number and a=pαa′a=p^\alpha a', b=pβb′b=p^\beta b', c=pγc′c=p^\gamma c', , k=pκk′k=p^\kappa k', l=pλl′l=p^\lambda l', where a′,b′,c′,…,k′,l′a',b',c',\ldots ,k',l' are not divisible by pp. The exponent of pp on the left-hand side is equal to max⁡(α,β,γ,…,κ,λ)\max (\alpha ,\beta ,\gamma ,\ldots ,\kappa ,\lambda ), and the exponent of pp on the right-hand side is equal to

α+β+γ+⋯+κ+λ−min⁡(α,β)−min⁡(α,γ)−⋯−min⁡(κ,λ) \alpha +\beta +\gamma +\cdots +\kappa +\lambda -\min (\alpha ,\beta )-\min (\alpha ,\gamma )-\cdots -\min (\kappa ,\lambda ) +min⁡(α,β,γ)+⋯ {}+\min (\alpha ,\beta ,\gamma )+\cdots ⋯⋯⋯⋯⋯⋯⋯⋯ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots ±min⁡(α,β,γ,…,κ,λ). \pm \min (\alpha ,\beta ,\gamma ,\ldots ,\kappa ,\lambda ) .
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Remark.
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The least common multiple MM of the positive integers a,b,c,…,k,la,b,c,\ldots ,k,l may be represented as follows:

M=abc⋯kl(a,b)−1(a,c)−1⋯(k,l)−1(a,b,c)⋯(a,b,c,…,k,l)±1. M = abc\cdots kl(a,b)^{-1}(a,c)^{-1}\cdots (k,l)^{-1}(a,b,c)\cdots (a,b,c,\ldots ,k,l)^{\pm 1} .
Problem 8.30

We have

1=∣1000…01100…01110…01111…0\multicolumn6c\dotfill1111…1∣2=∣1111…11222…21233…31234…4\multicolumn6c\dotfill1234…n+1∣. 1 = \begin{vmatrix} 1& 0& 0& 0& \ldots & 0\\ 1& 1& 0& 0& \ldots & 0\\ 1& 1& 1& 0& \ldots & 0\\ 1& 1& 1& 1& \ldots & 0\\ \multicolumn {6}{c}{\dotfill }\\ 1& 1& 1& 1& \ldots & 1 \end{vmatrix}^2 = \begin{vmatrix} 1& 1& 1& 1& \ldots & 1\\ 1& 2& 2& 2& \ldots & 2\\ 1& 2& 3& 3& \ldots & 3\\ 1& 2& 3& 4& \ldots & 4\\ \multicolumn {6}{c}{\dotfill }\\ 1& 2& 3& 4& \ldots & n+1 \end{vmatrix} .
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Remark.
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The general element of the first determinant ηλμ=1\eta_{\lambda \mu }=1, if μ\mu is a (proper or improper) part of λ\lambda, otherwise it =0=0 (λ,μ=0,1,…,n\lambda ,\mu =0,1,\ldots ,n); in the second determinant the general element cλμc_{\lambda \mu } is equal to the number of common parts of λ\lambda and μ\mu, i.e. the smaller of the two numbers λ+1\lambda +1 and μ+1\mu +1 (λ,μ=0,1,…,n\lambda ,\mu =0,1,\ldots ,n).

Problem 8.31

Prove that ∣cλμ∣\left|c_{\lambda \mu }\right| is equal to 1, where ∣cλμ∣\left|c_{\lambda \mu }\right| denotes the determinant whose general element cλμc_{\lambda \mu } is equal to the number of common divisors of λ\lambda and μ\mu, in other words to the number of divisors of the greatest common divisor of λ\lambda and μ\mu, where λ,μ=1,2,…,n\lambda ,\mu =1,2,\ldots ,n.

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Problem 8.32

Let a0,a1,…,ana_0,a_1,\ldots ,a_n be arbitrary, Aν=∑t≤νatA_\nu = \sum_{t\leq \nu } a_t, ν=0,1,…,n\nu =0,1,\ldots ,n. We have

∣A0A0A0A0…A0A0A1A1A1…A1A0A1A2A2…A2A0A1A2A3…A3\multicolumn6c\dotfillA0A1A2A3…An∣=a0a1a2⋯an. \begin{vmatrix} A_0& A_0& A_0& A_0& \ldots & A_0\\ A_0& A_1& A_1& A_1& \ldots & A_1\\ A_0& A_1& A_2& A_2& \ldots & A_2\\ A_0& A_1& A_2& A_3& \ldots & A_3\\ \multicolumn {6}{c}{\dotfill }\\ A_0& A_1& A_2& A_3& \ldots & A_n \end{vmatrix} = a_0a_1a_2\cdots a_n .
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Remark.
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The general element cλμc_{\lambda \mu } of the determinant is =Ar=A_r where r=min⁡(λ,μ)r=\min (\lambda ,\mu ), λ,μ=0,1,…,n\lambda ,\mu =0,1,\ldots ,n. (Generalization of 30.)

Problem 8.33

Let a1,a2,…,ana_1,a_2,\ldots ,a_n be arbitrary, Aν=∑t∣νatA_\nu = \sum_{t\mid \nu } a_t, ν=1,2,…,n\nu =1,2,\ldots ,n. We have

∣A1A1A1A1…A1A1A2A1A2…A(2,n)A1A1A3A1…A(3,n)A1A2A1A4…A(4,n)\multicolumn6c\dotfillA1A(n,2)A(n,3)A(n,4)…An∣=a1a2a3⋯an. \begin{vmatrix} A_1& A_1& A_1& A_1& \ldots & A_1\\ A_1& A_2& A_1& A_2& \ldots & A_{(2,n)}\\ A_1& A_1& A_3& A_1& \ldots & A_{(3,n)}\\ A_1& A_2& A_1& A_4& \ldots & A_{(4,n)}\\ \multicolumn {6}{c}{\dotfill }\\ A_1& A_{(n,2)}& A_{(n,3)}& A_{(n,4)}& \ldots & A_n \end{vmatrix} = a_1a_2a_3\cdots a_n .
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Remark.
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The general element cλμc_{\lambda \mu } of the determinant is =Ar=A_r where r=(λ,μ)r=(\lambda ,\mu ), λ,μ=1,2,…,n\lambda ,\mu =1,2,\ldots ,n. (Generalization of 31.)

Problem 8.34

If a0,a1,a2,…a_0,a_1,a_2,\ldots are arbitrary and An=∑t≤natA_n = \sum_{t\leq n} a_t, n=0,1,2,…n=0,1,2,\ldots, then we clearly have

a0=A0,a1=A1−A0,a2=A2−A1,…,an=An−An−1,…. a_0=A_0, \quad a_1=A_1-A_0, \quad a_2=A_2-A_1, \quad \ldots , \quad a_n=A_n-A_{n-1}, \quad \ldots .

If a1,a2,a3,…a_1,a_2,a_3,\ldots are arbitrary and An=∑t∣natA_n = \sum_{t\mid n} a_t, n=1,2,3,…n=1,2,3,\ldots, then we have

a1=A1,a2=A2−A1,a3=A3−A1,a4=A4−A2, a_1=A_1, \quad a_2=A_2-A_1, \quad a_3=A_3-A_1, \quad a_4=A_4-A_2, a5=A5−A1,a6=A6−A3−A2+A1,… a_5=A_5-A_1, \quad a_6=A_6-A_3-A_2+A_1, \quad \ldots

and in general

an=∑t∣nμ(t)An/t,n=1,2,3,…, a_n = \sum _{t\mid n} \mu (t) A_{n/t} , \qquad n=1,2,3,\ldots ,

where μ(n)\mu (n) is the Möbius function (see definition in § 5 preceding 38).

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Problem 8.35

Denote by ψ(y)\psi (y) an arbitrary function defined for 0≤y≤10 \leq y \leq 1. Let

g(n)=∑ν=1nψ(νn),f(n)=∑(r,n)=1ψ(rn), g(n) = \sum _{\nu =1}^n \psi \left(\frac\nu n\right) , \qquad f(n) = \sum _{(r,n)=1} \psi \left(\frac rn\right) ,

where the last sum is extended over the numbers rr that are ≤n\leq n and are relatively prime to nn. We then have

f(n)=∑t∣nμ(t)g(nt)=∑t∣nμ(nt)g(t). f(n) = \sum _{t\mid n} \mu (t) g\left(\frac nt\right) = \sum _{t\mid n} \mu \left(\frac nt\right) g(t) .
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Problem 8.36

As is well known, we have

∏ν=1n(x−e2πiνn)=xn−1. \prod _{\nu =1}^n \left(x - e^{\frac{2\pi i\nu }{n}}\right) = x^n - 1 .

Set

∏(r,n)=1(x−e2πirn)=Kn(x), \prod _{(r,n)=1} \left(x - e^{\frac{2\pi ir}{n}}\right) = K_n(x) ,

where the product is extended over those numbers rr that are ≤n\leq n and relatively prime to nn. (Kn(x)K_n(x) is termed the nnth cyclotomic polynomial.) The zeros of xn−1x^n-1 are the nnth roots of unity, the zeros of Kn(x)K_n(x) are the primitive nnth roots of unity. Derive the formula

Kn(x)=∏t∣n(xn/t−1)μ(t). K_n(x) = \prod _{t\mid n} \left(x^{n/t}-1\right)^{\mu (t)} .
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Problem 8.37

If μ(n)\mu (n) is the Möbius function, then we have

∑(r,n)=1e2πirn=μ(n). \sum _{(r,n)=1} e^{\frac{2\pi ir}{n}} = \mu (n) .
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