Section 1

Lattice Points in Two and Three Dimensions

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Problem 8.238

All three vertices of an equilateral triangle cannot be lattice points.

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Problem 8.238.1

A polygon is equiangular, has nn sides, and all its vertices are lattice points. Then nn is either 4 or 8.

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Problem 8.238.2

A regular polygon has nn sides and all its vertices are lattice points. Then n=4n=4.

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Problem 8.239

How thick must the trunks of the trees in a regularly spaced circular forest grow if they are to block completely the view from the center?

Let ss be a given positive integer. Let every lattice point p,qp,q that satisfies the inequality 1≤p2+q2≤s21\leq p^2+q^2\leq s^2 be the center of a circle of radius rr. If rr is sufficiently small, there exist rays running from the point 0, 0 to infinity without meeting the circles described above (the forest is transparent); such rays no longer exist if rr is sufficiently large (for r=12r=\tfrac 12 the circles touch). Let r=ρr=\rho be the value of rr which divides the two cases (limit of transparency). Then we have

1s2+1≤ρ<1s. \frac{1}{\sqrt{s^2+1}} \leq \rho < \frac1s .
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Problem 8.240

In the plane draw the widest possible infinite straight ``path'', contained between two parallels and avoiding all the lattice points, that makes an angle equal to arctan⁡x\arctan x with the yy-axis. Denoting the width of this path by φ(x)\varphi (x), determine f(x)=φ(x)1+x2f(x)=\varphi (x)\sqrt{1+x^2}.

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Problem 8.241

Two lattice points x,yx,y and x′,y′x',y' are termed congruent mod nn, if

x≡x′(modn),y≡y′(modn). x\equiv x' \pmod n , \qquad y\equiv y' \pmod n .

Among any kn2+1kn^2+1 distinct lattice points there are always k+1k+1 that are congruent mod nn.

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Problem 8.242

In the plane consider a domain of area (Jordan content) FF. It may contain no lattice points, but can always be moved parallel to itself so that [F]+1[F]+1 lattice points lie in it.

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Problem 8.243

The lattice points in the closed first quadrant are enumerable. I.e. there exist functions f(x,y)f(x,y) defined for ordered pairs of non-negative integers x,yx,y that combine the following two properties:

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(1)

The values assumed by f(x,y)f(x,y) form the set of numbers 1,2,3,4,5,…1,2,3,4,5,\ldots.

(2)

The function f(x,y)f(x,y) takes different values at distinct points; i.e. if x,y,x′,y′x,y,x',y' are integers, x≥0x\geq 0, y≥0y\geq 0, x′≥0x'\geq 0, y′≥0y'\geq 0, (x−x′)2+(y−y′)2>0(x-x')^2+(y-y')^2>0, then f(x,y)≠f(x′,y′)f(x,y)\neq f(x',y').

There even exist rational entire functions f(x,y)f(x,y) that enumerate the lattice points, i.e. that have the properties (1) (2). An example is the function

f(x,y)=12(x2+2xy+y2+3x+y+2)=(x+y+12)+x+1 f(x,y) = \tfrac 12(x^2+2xy+y^2+3x+y+2) = \binom {x+y+1}{2}+x+1

and the function arising from it by interchanging xx and yy [the lattice points are counted successively on the line segments

x+y=0,x+y=1,x+y=2,…,x≥0,y≥0. x+y=0, \quad x+y=1, \quad x+y=2, \quad \ldots , \quad x\geq 0, \quad y\geq 0 .

Let f(x,y)f(x,y) be a rational entire function of degree mm

f(x,y)=φ0(x,y)+φ1(x,y)+⋯+φm(x,y), f(x,y) = \varphi _0(x,y)+\varphi _1(x,y)+\cdots +\varphi _m(x,y) ,

where φμ(x,y)\varphi_\mu (x,y) denotes a rational entire homogeneous function of degree μ\mu. If f(x,y)f(x,y) enumerates the lattice points, then clearly φm(x,y)≥0\varphi_m(x,y)\geq 0 for x≥0x\geq 0, y≥0y\geq 0. If φm(x,y)>0\varphi_m(x,y)>0 for x≥0x\geq 0, y≥0y\geq 0, x+y>0x+y>0 and f(x,y)f(x,y) enumerates the lattice points, then the degree of f(x,y)f(x,y) is

m=2. m=2 .

[The number of lattice points contained inside the contour line f(x,y)=const.f(x,y)=\text{const.} is related to the enclosed area.]]

Problem 8.244

Consider the square domain 12≤x≤n+12\tfrac 12\leq x\leq n+\tfrac 12, 12≤y≤n+12\tfrac 12\leq y\leq n+\tfrac 12 as a chess board with $n^2$ squares'', i.e. by means of straight lines parallel to the axis divide it into $n^2$ squares of unit area. The problem of the nn queens'' requires us to choose from the n2n^2 lattice points that constitute the centers of the n2n^2 squares nn lattice points (x1,y1),(x2,y2),…,(xn,yn)(x_1,y_1),(x_2,y_2),\ldots ,(x_n,y_n) that satisfy the 2n(n−1)2n(n-1) inequalities

xμ≠xν,yμ≠yν,xμ−xν≠yμ−yν,xμ−xν≠−(yμ−yν), x_\mu \neq x_\nu , \qquad y_\mu \neq y_\nu , \qquad x_\mu -x_\nu \neq y_\mu -y_\nu , \qquad x_\mu -x_\nu \neq -(y_\mu -y_\nu ) ,

where μ≠ν\mu \neq \nu; μ,ν=1,2,…,n\mu ,\nu =1,2,\ldots ,n. Replace the inequalities by the more restrictive ``incongruence relations''

xμ≢xν,yμ≢yν,xμ−xν≢yμ−yν,xμ−xν≢−(yμ−yν) x_\mu \not\equiv x_\nu , \qquad y_\mu \not\equiv y_\nu , \qquad x_\mu -x_\nu \not\equiv y_\mu -y_\nu , \qquad x_\mu -x_\nu \not\equiv -(y_\mu -y_\nu )

(modn)\pmod n and show that these have a solution if and only if nn is relatively prime to 6.

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Problem 8.244.1

A regular polygon has nn sides and all its vertices are lattice points. Then nn is equal to 3, 4 or 6.

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Problem 8.244.2

(continued). Give examples showing that the three cases n=3,4n=3,4 and 6 can actually occur.

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Problem 8.244.3

All the vertices of a regular polyhedron are lattice points. Then the polyhedron is either a cube or a tetrahedron or an octahedron.

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Problem 8.244.4

(continued). Give examples showing that the three cases mentioned above can actually occur.

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