On the number of points with pairwise integral distances on a circle

Discrete Applied Mathematics (

)

–

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On the number of points with pairwise integral distances on a circle Ganbileg Bat-Ochir Olonlog School, Ulaanbaatar, Mongolia

article

info

Article history: Received 16 February 2017 Received in revised form 8 July 2018 Accepted 10 July 2018 Available online xxxx

a b s t r a c t We give a complete solution of the classical problem finding the number of an integral √ points set on the circle with radius R = n 2DD , where n ∈ N and D ∈ {1, 2, 3, 7, 11, 19, 43, 67, 163}. We also find an upper bound for the minimum radius of the circle which contains an n integral points set for n ⩽ 3072. © 2018 Elsevier B.V. All rights reserved.

Keywords: An n integral points set Gaussian prime factorization Minimal radius

1. Introduction A set of n points in the Euclidean plane with pairwise integral distances is called an n integral points set. It is well known that there are infinitely many points with pairwise integral distances on a line, [1]. Moreover, the authors proved three different ways that there exists non-collinear integral points set on the plane. In particular, they showed that there exists a circle such that one can place an n integral points set. If all, but one, points lie on a side of a triangle with integer side lengths (a+b+c)(a+b−c)(a−b+c)(−a+b+c) , see [9]. a, b, c, then the points are in one-to-one correspondence to factors of the number D = gcd(b2 −c 2 +a2 ,2a)2 Another interesting question is for what n do there exist n points in the plane, no three on a line and no four on a circle, with all pairwise distances integral ? Source for these and related problems, √ see [3,5]. To date, the only known examples have n ≤ 7 [8]. The first published heptagon has coordinates of the form (a, b 2002) with a, b ∈ Z. α α α H. Harborth, A. Kemnitz, M. Möller [6] showed that if n = p1 1 p2 2 . . . pk k , where pi is a prime number of the form 3s + 1, √

then there exists 3τ (R) = 3(α1 + 1)(α2 + 1) · . . . · (αk + 1) integral points set on the circle with radius R = n 3 3 . This result becomes a special case of Theorem 3 of the current paper when n is even integer and its prime factorization does not contain the prime number of the form 3s + 2. Theorem 3 also implies that this construction is maximal. A. Antonov and S. Kurz showed [2] that if n is a product of prime numbers of the form 4k + 1, then the number of points with pairwise integral distances is 2τ (n) on the circles with radius n and 2n . Moreover, they conjectured that the number 2τ (n) is maximal. However, in Theorem 1, we recover the construction of A. Antonov and S. Kurz as a special case when n is not divisible by 4 and its prime factorization does not contain the prime numbers of the form 4s + 3 on the circle with radius n/2. Theorem 1 also shows that this construction is maximal. Note that some results in table 2 and table 3 in [2] are the same as Table 1 of the current paper. P. Erdős asked that what is the maximal number f (n) of the odd distances of an n integral points set. This question answered by L. Piepmeyer in [10] so that f (n) = n2 /3 + r(r − 3)/6, where n ≡ r (mod 3), r = 1, 2, 3. Piepmeyer’s construction was as follows: consider an equilateral triangle with side length 7m and rotate m − √1 times with angle θ = arccos 71/98. The maximal construction of an n integral points set on the circle with radius R = n2DD with n ≡ 2 (mod 4) and D ∈ {3, 11, 19, 43, 67, 163} in Corollary 3 imply that the number of odd distances is maximal. E-mail address: [email protected]. https://doi.org/10.1016/j.dam.2018.07.004 0166-218X/© 2018 Elsevier B.V. All rights reserved.

Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.

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Table 1 R(n, 1) = k/2 when n ≤ 3072. R(n, 1)

n

R(n, 1)

1 2

2

1 2

n

· 53 · 13 · 17 · 29 · 37 · 41

193, 256

1 2

·5

3, 4

1 2

· 5 · 13 · 17 · 29 · 37 · 41

257, 288

1 2

· 52

5, 6

1 2

· 54 · 13 · 17 · 29 · 37 · 41

289, 320

1 2

· 5 · 13

7, 8

1 2

· 52 · 13 · 17 · 29 · 37 · 41 · 53

321, 384

1 2

· 52 · 13

9, 12

1 2

· 52 · 132 · 172 · 29 · 37 · 41

385, 432

1 2

· 5 · 13 · 17

13, 16

1 2

· 53 · 13 · 17 · 29 · 37 · 41 · 53

433, 512

3

2

1 2

· 5 · 13

17,18

1 2

· 5 · 13 · 17 · 29 · 37 · 41 · 53

513, 576

1 2

· 52 · 13 · 17

19, 24

1 2

· 54 · 13 · 17 · 29 · 37 · 41 · 53

577, 640

1 2

· 53 · 13 · 17

24, 32

1 2

· 52 · 13 · 17 · 29 · 37 · 41 · 53 · 61

641, 768

1 2

· 52 · 132 · 17

33, 36

1 2

· 52 · 132 · 172 · 29 · 37 · 41 · 53

769, 864

1 2

· 54 · 13 · 17

37, 40

1 2

· 53 · 13 · 17 · 29 · 37 · 41 · 53 · 61

865, 1024

2

2

2

2

1 2

· 5 · 13 · 17 · 29

41, 48

1 2

· 5 · 13 · 17 · 29 · 37 · 41 · 53 · 61

1025, 1152

1 2

· 53 · 13 · 17 · 29

49, 64

1 2

· 54 · 13 · 17 · 29 · 37 · 41 · 53 · 61

1153, 1280

1 2

· 52 · 132 · 17 · 29

65, 72

1 2

· 53 · 132 · 17 · 29 · 37 · 41 · 53 · 61

1281, 1536

1 2

· 54 · 13 · 17 · 29

73, 80

1 2

· 52 · 132 · 172 · 29 · 37 · 41 · 53 · 61

1537, 1728

1 2

· 52 · 13 · 17 · 29 · 37

81, 96

1 2

· 54 · 132 · 17 · 29 · 37 · 41 · 53 · 61

1729, 1920

2

2

2

1 2

· 5 · 13 · 17 · 29 · 37

97, 128

1 2

· 5 · 13 · 17 · 29 · 37 · 41 · 53 · 61 · 73

1921, 2048

1 2

· 52 · 132 · 17 · 29 · 37

129, 144

1 2

· 52 · 132 · 17 · 29 · 37 · 41 · 53 · 61 · 73

2049, 2304

1 2

· 54 · 13 · 17 · 29 · 37

145, 160

1 2

· 54 · 13 · 17 · 29 · 37 · 41 · 53 · 61 · 73

2305, 2560

1 2

· 52 · 13 · 17 · 29 · 37 · 41

161, 192

1 2

· 53 · 132 · 17 · 29 · 37 · 41 · 53 · 61 · 73

2561, 3072

3

3

S. Kurz and A. Wassermann [9], computed the minimal diameter for n ⩽ 192. Table 1 of [9] is the same as Table 2 of the current paper. The main reason for the present article’s existence is to find the maximum number of points with pairwise integral distances on a circle with positive integer diameter for certain cases. Also the author observed using the ring of quadratic integers one can generalize this problem to a positive√non integer diameter. We note that the ring of integers in the field Q( −D), with D negative, is a unique factorization domain when D ∈ {1, 2, 3, 7, 11, 19, 43, 67, 163} and for no other values of D, see [11] for details. 2. Some notation and preliminary results Let N, Z, P and Q be the set of all positive integers, the set of all integers, set of all primes and set of all rational numbers, α α respectively. Let n be a positive integer. Recall that if n = p1 1 . . . pk k with p1 , . . . , pk distinct prime numbers, then the number of all positive integer divisors of n is given by τ (n) := (α1 + 1) . . . (αk + 1). The following rings will be used in the rest of this paper,

{

√

}

O−1 = {a + bi | a, b ∈ Z} , O−2 = a + b 2i | a, b ∈ Z

and

{ O−D =

a 2

+

b√ 2

Di | a, b ∈ Z , a ≡ b

} (mod 2)

for D ∈ {3, 7, 11, 19, 43, 67, 163} . Let denote by Rn the minimal√radius of a circle which contains an n integral points set. We also denote by R(n; D) the minimal radius of the form R = n D D which contains an n integral points set. We need the notion of Legendre symbol defined by

( ) a

p

{

1

= −1 0

if a is a quadratic residue modulo p and a ̸ = 0 (mod p), if a is a quadratic non residue modulo p, if a ≡ 0 (mod p).

√

Let α ∈ O−D . For the complex number α = a + b −D its square of norm is [α] := αα and its trace is tr(α ) := α + α . Next we present some elementary results with their proofs, it will be used several times throughout this paper. Let T be the triangle with integer sides a, b, c; that has area A and inscribed the circle with radius R. By the cosine law it is straightforward Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.

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Table 2 √ R(n, 3) = k 3/3 when n ≤ 3072. R(n, 3)

n

R(n, 3)

n

√1

2, 3

√1

· 72 · 132 · 192 · 31 · 37

289, 324

4, 6

√1

· 73 · 13 · 19 · 31 · 37 · 43

325, 384

7, 9

√1

10, 12

√1

· 7 · 13

13, 18

√1

· 7 · 13 · 19

19, 24

√1

· 7 · 13 · 19 · 31 · 37 · 43 · 61

649, 768

· 72 · 132 · 19 · 31 · 37 · 43 · 61

769, 864

3

√1 3

√1 3

√1 3

√1 3

√1 3

·7 ·7

2

· 7 · 13 2

3 3 3 3 3 3

√1 3

· 7 · 13

25, 27

√1 3

√1

· 72 · 13 · 19

28, 36

√1

· 7 · 13 · 19 · 31

3

2

2

3

· 7 · 13 · 19 · 31 · 37 · 43

385, 432

· 74 · 13 · 19 · 31 · 37 · 43

433, 480

2

2

· 7 · 13 · 19 · 31 · 37 · 43 · 61

481, 576

· 72 · 132 · 192 · 31 · 37 · 43

577, 648

2

3

37, 48

√1 3

2

· 7 · 13 · 19

49, 54

√1

· 72 · 13 · 19 · 31

55, 72

√1

√1 3

· 7 · 13 · 19 · 31 · 37

73, 96

√1 3

· 7 · 13 · 19 · 31 · 37 · 43 · 61 · 73

1297, 1536

√1

· 72 · 132 · 19 · 31

97, 108

√1

· 72 · 132 · 19 · 31 · 37 · 43 · 61 · 73

1537, 1728

· 7 · 13 · 19 · 31 · 37 · 43 · 61 · 73

1729, 1920

· 72 · 13 · 19 · 31 · 37 · 43 · 61 · 73 · 89

1921, 2304

√1 3 √1 3

√1 3

3

2

3 3

3

√1 3

· 7 · 13 · 19 · 31 · 37

109, 144

√1 3

√1

· 7 · 13 · 19 · 31 · 37 · 43

145, 192

√1

193, 216

√1 3

217, 288

√1

3

√1 3 √1 3

2

2

2

· 7 · 13 · 19 · 31 · 37 2

· 7 · 13 · 19 · 31 · 37 · 43

3

3

4

865, 960

2

· 7 · 13 · 19 · 31 · 37 · 43 · 61 · 73

961, 1152

· 72 · 132 · 192 · 31 · 37 · 43 · 61

1153, 1296

· 7 · 13 · 19 · 31 · 37 · 43 · 61

3

4

2

2

2305, 2592

2

· 7 · 13 · 19 · 31 · 37 · 43 · 61 · 73

2593, 3072

3

· 7 · 13 · 19 · 31 · 37 · 43 · 61 · 73 · 89

√

to see that cos γ ∈ Q, where γ is an angle of T . By Heron’s formula A = 14 (a + b + c)(−a + b + c)(a − b + c)(a + b − c). √ Hence A is of the form m D, where m is a positive integer and D is a square free positive integer, which is called the 4 √ characteristic of T . Since R = abc there exists a rational number q such that R = q D. It is well known that the all 4A characteristic numbers of triangles with the vertices from an n integral points set are same [7]. In the next section, Lemma 1 shows that there is deep connection between the set of solutions of the equation√x2 + Dy2 = n √ in the ring Q( −D) and the number of points with pairwise integral distances on the circle with radius R = n2DD . Proposition 1. Let n be a positive integer and D be a square-free positive integer. If the number of points with pairwise integral √ distances is at least 3 on the circle with the radius R = 2n D , then the maximum number of points with pairwise integral distances √ on the circle with radius R =

n 2

√

D and R′ =

n D 2D

are same.

Proof. Let P(R) be the maximum number of points with pairwise integral distances on the circle with radius R. We claim ′ that P(R) √ = P(R ). To prove this we fix P(R) points on the circle centered at the origin O of the complex plane with the radius n D R = 2 . Let ξ1 , ξ2 be the arbitrary points on this circle. Suppose ̸ ξ1 Oξ2 = 2ϕ . Since P(R) ⩾ 3, we see that cos ϕ ∈ Q. On the other hand, we have |ξ1 − ξ2 | ∈ Z and sin ϕ =

|ξ1 − ξ2 | 2R

=

|ξ1 − ξ2 | √ . n D

√ √ n2 D−|ξ1 −ξ2 |2 √ Hence cos ϕ = ± 1 − sin2 ϕ = ± . Since cos ϕ ∈ Q we have n D √ √ ± n2 D − |ξ1 − ξ2 |2 = x D for some x ∈ Z. Hence |ξ1 − ξ2 |2 = n2 D − x2 D. This implies that |ξ1 − ξ2 |2 is divisible by D. Since D is a square free, |ξ1 − ξ2 | is divisible by D. Hence the distance between any two points is divisible by D. Now if we shrink the radius of this circle by D times, then we still have points with pairwise integral distances. Thus P(R) ⩽ P(R′ ). √ On the other hand, if we fix P(R′ ) points with pairwise integral distances on the circle with radius R′ = n2DD , then we enlarge the radius by D times, gives us points with pairwise integral distances. Hence P(R′ ) ⩽ P(R). This completes the proof. □ Proposition 2. Let p ∈ P and q is prime number of the form 4s + 3. Then

(

−q p

)

= 1 if and only if

( ) p q

= 1.

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Proof. By the law of quadratic residue:

(

−q

)

( =

p

−1

)( ) q

p

p

= (−1)

p−1 q+1 · 2 2

p−1 2

( ) p

= (−1) (−1) q ( ) ( ) p p = . □ q

p−1 q−1 · 2 2

q

3. Main results 3.1. The number of points with pairwise integral distances on a circle In this (section ) we assume that D ∈ {1, 2, 3, 7, 11, 19, 43, 67, 163}. Suppose that n1 is the product of prime ( ) numbers qj s

s

s

̸= 1. Let n2 = p11 p22 . . . pkk , where p1 , p2 , . . . , pk are the distinct prime numbers such that −pD = 1. Recall j the following well known fact that, for 1 ≤ j ≤ k, √ √ pj = (aj + bj Di)(aj − bj Di), √ √ where aj ± bj Di are the prime elements in O−D . Let denote by ωj the argument of aj + bj Di for 1 ≤ j ≤ k. such that

−D qj

Note that we use the above notations throughout the paper. √

Lemma 1. Let ξ0 , ξ , ξ ′ be the points on the circle centered at origin O of complex plane with radius R = with vertices at the points ξ0 , ξ , ξ ′ has integer sides and arg ξ0 = 0, then

∏

ξ = α2 R

n1 n2 D . 2D

If the triangle

e4itj ωj ,

1⩽j⩽k

where α is a unit in O−D and −sj ⩽ tj ⩽ sj whenever 1 ⩽ j ⩽ k. Proof. Let arg(ξ ) = 2ϕ , |ξ − ξ0 | = a ∈ Z. Then it is straightforward to see that sin ϕ =

√ a 2R

=

a D , n1 n2

√ (n1 n2 )2 −a2 D . n1 n2 )2 a2 D Z.

cos ϕ = ±

Since the triangle with vertices at the points ξ0 , ξ , ξ ′ has integral sides, we see that cos ϕ ∈ Q. Thus b =: That is, b2 + a2 D = (n1 n2 )2 . Consequently,

√

(n1 n2

−

∈

n1 |a and n1 |b. Otherwise we have 2

2

So x + Dy =

n22 ,

(

−D

)

qi

= 1 for some qi such that qi |n1 , but this contradicts with our assumption. Let b = n1 x, a = n1 y. √

sin ϕ =

, cos ϕ =

y D n2

x . n2

A direct computation implies that

√

n2 (x + y Di)2 ξ = R(cos 2ϕ + i sin 2ϕ ) = 1 . 4RD √ √ √ Since (x + y Di)(x − y Di) = n22 , x + y Di ∈ O−D , we have √ x + y Di | n22 . Since the prime factorization is unique in the ring O−D we have

√

x + y Di = α

∏

√ (aj − bj

Di)cj (aj + bj

√

Di)2sj −cj ,

1⩽j⩽k

0 ⩽ cj ⩽ 2sj . Let ωj = arccos

√ aj + bj Hence

Di =

√

√

x + y Di = α

aj

√

pj

. By Euler’s formula we have

pj eiωj and aj − bj

∏

√

sj

Di =

√

pj e−iωj .

pj e−icj ωj ei(2sj −cj )ωj = α n2

1⩽j⩽k

∏

e2i(sj −cj )ωj .

1⩽j⩽k

Let sj − cj = tj . Thus

ξ=

n21 4RD

√

(x + y 2i)2 = α 2 R

∏

e4itj ωj , with − sj ⩽ tj ⩽ sj .

1⩽j⩽k

This completes the proof. □ Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.

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Lemma 2. There exists an integer S such that

√

∑

sin(2

(±mj ωj )) = ∏

D

S

mj

1⩽j⩽k

1⩽j⩽k

pj

and (S ,

∏

mj

pj ) = 1 ,

1⩽j⩽k

where mj is a non-negative integer. Proof. Recall

∏

√

(aj ± bj

Di)mj = T + S

√

∏

Di and

1⩽j⩽k

√ (aj ∓ bj

Di)mj = T − S

√

Di.

1⩽j⩽k

Now the first identity in the statement clearly√holds. For the second one, we assume by contradiction that pl |S for some ml > 0. But then if we mod out by mod (al + bl Di) in the following identity

√

2S

∏

Di =

√

∏

Di)mj −

(aj ± bj

1⩽j⩽k

√

Di)mj ,

(aj ∓ bj

1⩽j⩽k

then we see the obvious contradiction. Hence (S , Lemma 3. All points ξl = α 2 R divisor of unity in O−D and

∏

1⩽j⩽k e

4itj ωj

mj 1⩽j⩽k pj )

∏

= 1. □ √

, 0 ⩽ tj ⩽ sj , 1 ⩽ j ⩽ k are distinct on the circle with radius R =

n D , 2D

where α is a

• 1 ⩽ l ⩽ 2τ (n2 ) when D = 1; • 1 ⩽ l ⩽ 3τ (n2 ) when D = 3; • 1 ⩽ l ⩽ τ (n2 ) when D ∈ {2, 7, 11, 19, 43, 67, 163}. Proof. By contradiction we assume there exist two points from the points ξl are not distinct, say ξℓ = ξq . Suppose that

ξℓ = α12 R

∏ 4it ′ ω , ξq = α22 R 1≤j≤k e j j . Let us consider the ratio

4itj ωj 1≤j≤k e

∏

α 2 ∏ 4i(tj −tj′ )ωj ξl = 12 R e , ξq α2 1⩽j⩽k where |tj − tj′ | = mj for 1 ≤ j ≤ k. Thus

( )4mj ∏ ( )4mj α12 ∏ aj bj √ bj √ aj ξl = 2 Di Di √ +√ √ −√ ξq pj pj pj pj α2 tj ≥tj′ tj
j

Consequently,

α12

k ∏

√ (aj + bj

Di)2mj = α22

j=1

k ∏

√ (aj − bj

Di)2mj .

(1)

j=1

But since at least one of mj ’s is positive number and the prime factorization is unique in the ring O−D , we have the obvious contradiction in (1). This completes the proof. □ Theorem 1. Let R = 21 n1 n2 , where n1 is not a product of any prime numbers of the form 4s + 1, n2 is a product of prime numbers of the form 4s + 1. Then there exist at most 2τ (n2 ) points with pairwise integral distances on the circle with radius R. s

s

s

Proof. Let n2 = p11 p22 . . . pkk , where pi are distinct prime numbers of the form 4s + 1. By Fermat–Euler’s theorem there exist positive integers aj , bj such that pj = a2j + b2j . This factorizes as follows pj = (aj + bj i)(aj − bj i). Note that the complex Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.

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Fig. 1. 8 integral points set on the circle with radius R = 65/2.

numbers aj + bj i and aj − bj i are Gaussian primes in the ring O−1 . Let us choose three points with pairwise integral distances ξ0 , ξ , ξ ′ on the circle centered at the origin O of complex plane with radius R such that arg(ξ0 ) = 0. By Lemma 1 and since ∏ 4it ′ ω the only units are α ∈ {±1, ±i}, we have that ξ = ±R 1≤j≤k e j j with −sj ⩽ tj ⩽ sj . Choosing p0 = 2, ω0 = π4 gives

∏ ∏ 4it ′ ω ξ = R 0⩽j⩽k e4itj ωj , where 1 ⩽ j ⩽ k, −sj ⩽ tj ⩽ sj and t0 ∈ {0, 1}. Similarly ξ ′ = R 0⩽j⩽k e j j , where −sj ⩽ tj′ ⩽ sj and t0′ ∈ {0, 1}. By Lemma 2 we have ∑ 2RS |ξ − ξ ′ | = 2R · |sin(2 (tj − tj′ )ωj )| = |tj −tj′ | ∏ 0⩽j⩽k

0⩽j⩽k

pj

with |t0 − t0 | ∈ {0, 1} and ′

⎛

⎞

gcd ⎝S ,

∏

|tj −tj′ |

pj

⎠ = 1.

1⩽j⩽k

Consequently in order for |ξ − ξ ′ | to be an integer we must have |tj − tj′ | ⩽ sj for 1 ⩽ j ⩽ k. If we choose a set Sj which consists of distinct tj ’s with the cardinality more than sj + 1, then the difference between the minimum and maximum element of that set is |tj − tj′ | > sj . Hence number of different ∏values of tj is at most sj + 1. It follows that the number of points with pairwise integral distances is less than 2τ (n2 ) = 2 1⩽j⩽k (sj + 1). Finally, we consider the following set of points

ξℓ = R

eitj ωj ; 1 ≤ j ≤ k, 0 ≤ tj ≤ sj , t0 ∈ {0, 1}.

∏ 0≤j≤k

In total this gives us 2τ (n2 ) points with pairwise integral distances. By Lemma 3 these points are distinct. This completes the proof. □ In the next example, we choose n1 = 1 and n2 = 65 for the construction in the end of proof of Theorem 1, gives an explicit construction of an 8 integral points set on the circle with radius 65/2. Example 1. See Fig. 1. √

Theorem 2. Let R = n1 n2 42 , where n1 is not a product of any prime numbers of the form 8s + 1 and 8s + 3, n2 is a product of prime numbers of the form 8s + 1 or 8s + 3. Then there exist at most τ (n2 ) points with pairwise integral distances on the circle with radius R. s

s

s

Proof. Let n2 = p11 p22 . . . pkk , where pi are distinct prime numbers of the form 8s + 1 or 8s + 3. By Fermat–Euler’s theorem for any prime number p there exist positive integers a, b such that p = a2 + 2b2 if and only if p √ ≡ 1 (mod 8)√or p ≡ 3 (mod 8). Hence there exist positive integers aj , bj such that pj = a2j + 2b2j . Thus, pj = (aj + bj 2i)(aj − bj 2i). Note that the

√

complex numbers aj + bj

√

2i and aj − bj

2i are the prime elements in the ring O−2 . Fix ξ0 , ξ , ξ ′ points with pairwise integral

Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.

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√

Fig. 2. 10 integral points set on the circle with radius R = 891 2/4.

distances on the circle centered at the origin O of the complex plane with radius R such that arg(ξ0 ) = 0. By Lemma 1 and ∏ ∏ 4it ′ ω since α = ±1 is the only units in the ring O−2 , we have ξ = R 1⩽j⩽k e4itj ωj , where −sj ⩽ tj ⩽ s. Similarly, ξ ′ = R 1⩽j⩽k e j j , where −sj ⩽ tj′ ⩽ sj . By Lemma 2 we have

√

∑

|ξ − ξ | = 2R · |sin(2 ′

S

2

1⩽j⩽k

pj

(tj − tj )ωj )| = 2R · ′

|tj −tj′ |

∏

1⩽j⩽k

and

⎛

⎞

gcd ⎝S ,

∏

|tj −tj′ |

pj

⎠ = 1.

1⩽j⩽k

Applying the same method as in the end of the proof Theorem ∏1 it is easy to see that the number of points with pairwise integral distances is at most τ (n2 ). Choosing the points ξℓ = R 1⩽j⩽k e4itj ωj ; 1 ⩽ j ⩽ k, 0 ⩽ tj ⩽ sj , t0 ∈ {0, 1} give us in total τ (n2 ) points with pairwise integral distances. By Lemma 3 all these points are distinct. This completes the proof. □ In the next example, we choose n1 = 1 and n2 = 891 for the construction √ in the end of proof of Theorem 2, gives an explicit construction of an 10 integral points set on the circle with radius 891 2/4. Example 2. See Fig. 2. √

Theorem 3. Let R = n1 n2 63 , where n1 is not a product of any prime number of the form 3s + 1 and n2 is a product of prime numbers of the form 3s + 1 on the circle with radius R. (i) If n1 is an odd integer, then there exist at most τ (n2 ) points with pairwise integral distances. (ii) If n1 is an even integer, then there exist at most 3τ (n2 ) points with pairwise integral distances on the circle with radius R. s

s

s

Proof. Let n2 = p11 p22 . . . pkk , where pi are distinct prime numbers of the forms 3s + 1. It is well known that for any prime 2 2 number p there exist positive integers a, b such √ √ that p = a + 3b if only √ and if p = 3s√+ 1 for some s ∈ N. Hence there exist aj , bj ∈ N such that pj = (aj + bj 3i)(aj − bj 3i). We know that aj + bj 3i and aj − bj 3i are the prime elements in O−3 . As before we consider the ξ0 , ξ , ξ ′ points with pairwise integral distances on arg(ξ0 ) = 0. The √ √ the circle with radius R such that∏ units of O−3 are α ∈ {±1, ±ϵ1 , ±ϵ2 }, where ϵ1 = − 12 + 23 i, ϵ2 = − 12 − 23 i. By Lemma 1 we have ξ = α 2 R 1⩽j⩽k e4itj ωj with ∏ t0 ∈ {0, 1, 2} and −sj ⩽ tj ⩽ sj for 1 ⩽ j ⩽ k. Choosing ω0 = π6 gives ξ = R 0⩽j⩽k e4itj ωj with t0 ∈ {0, 1, 2} and −sj ⩽ tj ⩽ sj for 1 ⩽ j ⩽ k. Similarly, ξ ′ = R

∏

|ξ − ξ | = 2R · |sin(2 ′

∑

4itj′ ωj 0⩽j⩽k e

with t0′ ∈ {0, 1, 2} and −sj ⩽ tj′ ⩽ sj for 1 ⩽ j ⩽ k. By Lemma 2 we have

√ S

(tj − tj )ωj )| = 2R ·

0⩽j⩽k

′

2n0

∏

3

1⩽j⩽k

|tj −tj′ |

(2)

pj

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√

Fig. 3. 9 integral points set on the circle with radius R = 49 3/4.

with n0 ∈ {0, 1} and gcd(S ,

∏

mj

pj ) = 1.

1≤j≤k

Finally, the number of distinct values of tj is at most sj + 1, where t0 ∈ {0, 1, 2}. Note that in (2) if t0 = t0′ , then n0 = 0, and if t0 ̸ = t0′ then n0 = 1. Hence if n1 is an even integer, then the number of points with pairwise integral distances is at most 3τ (n2 ), proving (i). Similarly, if n1 is an odd integer, then the number of points with pairwise integral distances is at most τ (n2 ), proving (ii). □ In the next example, we choose n1 = 2 and n2 = 49 for the construction √ in the end of proof of Theorem 3, gives an explicit construction of an 9 integral points set on the circle with radius 49 3/3. Example 3. See Fig. 3. √

Theorem 4. Let R = 2s0 n1 n2 147 , where s0 is a non-negative integer, n1 is not multiplicity of any prime numbers of the form 7s + 1, 7s + 2, 7s + 4 and n2 is only divisible by odd prime numbers of the form 7s + 1, 7s + 2, 7s + 4. Then there exist either at most τ (n2 ) points with pairwise integral distances when s0 = 0 or at most s0 τ (n2 ) points with pairwise integral distances when s0 ⩾ 1 on the circle with radius R. Before we prove Theorem 4, let us consider the following lemma. Lemma 4. Let {xn }n≥1 , {yn }n≥1 be a sequence of integer numbers such that

(

1+

√ )n 7i

2

√ =

xn + yn 2

7i

.

Then xn , yn are odd integers for all n. Proof. The sequences {xn }, {yn } satisfy the following recurrence formula: x1 = 1, x2 = −3 and xn+2 = xn+1 − 2xn for n ≥ 1; y1 = 1, y2 = 1 and yn+2 = yn+1 − 2yn for n ≥ 1. Since x1 , x2 and y1 , y2 are integers, xn , yn are integers for all n ≥ 3. □ Proof (Proof of Theorem 4). It is well known that for any prime number there exist positive integers a, b such that p = a2 + 7b2 sk s1 s2 if and only if p ≡ 1, 9, 11 √, 15, 23, 25√(mod 28), see [4]. √ Assume that√n2 = p1 p2 . . . pk with pi distinct. Observe that 2 2 pj = aj + 7bj = (aj + bj 7i)(aj − bj 7i) and aj + bj 7i and aj − bj 7i are prime elements in the ring O−7 . Also, note √

that 2 =

1+ 7i 2

√

·

1− 7i . 2

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9

√

Fig. 4. 6 integral points set on the circle with radius R = 44 7/7.

As before we consider the ξ0 , ξ , ξ ′ points with pairwise integral distances on the circle with radius ∏ R such that arg(ξ0 ) = 0. 1 Recall that α = ±1 are the units of O−7 . Choosing p0 = 2 and ω0 = arccos √ gives ξ = R 0⩽j⩽k e4itj ωj , −sj ⩽ tj ⩽ sj . Similarly, ξ ′ = R

∏

0⩽j⩽k e

x

cos nω0 =

2n+1 By Lemma 2 we have

4itj′ ωj

2 2

, −sj ⩽ tj′ ⩽ sj . By Lemma 4 there exist odd integers x, y such that

√

and sin nω0 =

y 7 2n+1

. √

|ξ − ξ | = 2R|sin(2 ′

∑

(tj − tj )ωj )| = 2R ·

S

′

|t0 −t0′ |

2

0≤j≤k

∏

7 |tj −tj′ |

1≤j≤k

pj

and ′

∏

gcd(S , 2|t0 −t0 |

|tj −tj′ |

pj

) = 1.

1≤j≤k

Finally, rest of the proof is the same with the proof of Theorem 1. □ Example 4. See Fig. 4. √

Let R = n 2DD , where n ∈ N, and D ∈ {11, 19, 43, 67, 163}. We ( p )write n as the product of three positive integers n1 , n2 , n3 such that n is the product of the prime numbers p such that ̸= 1; n2 is the product of the prime numbers p such that 1 D (p) 2 2 = 1 and p is not a prime number of the form a + b D, where a, b ∈ Z; n3 is the product of the prime numbers p such D ( ) p that D = 1 and p is a prime number of the form a2 + b2 D, where a, b ∈ Z. √

Theorem 5. The number of points with pairwise integral distances on the circle with radius R = n 2DD is at most (a) τ (n2 n3 ) if n1 is even, (b) ⌊ 13 (τ (n2 ) + 2)⌋τ (n3 ) if n1 is odd, where ⌊x⌋ is the greatest integer less than or equal to x. Before we prove Theorem 5, let us define the following sets: Z0 =

{

{ Z1 =

1 2

{ Z2 =

√

}

a + b Di : 2 ∤ a + b , a, b ∈ Z ,

1 2

√

}

√

}

(a + b Di) : 2 ∤ ab , 4 | a − b , a, b ∈ Z , (a + b Di) : 2 ∤ ab, 4 | a + b , a, b ∈ Z .

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Lemma 5. Let z1 ∈ Zj and z2 ∈ Zℓ with j, ℓ ∈ {0, 1, 2}. If |z1 − z2 | is an integer, then |z1 − z2 | is an even integer when j = ℓ and is an odd integer when j ̸ = ℓ.

√

Proof. If z = a + b Di √ ∈ Z0 , then one of a, b is an even integer and other is an odd integer. Hence [z ] = a2 + Db2 is an odd 1 integer. Let z = 2 (a + b Di) ∈ Zj with j ∈ {1, 2}. Since a, b are an odd integer, we have a2 ≡ b2 ≡ 1 (mod 8). Also, since D ≡ 3 (mod 8), we have a2 + Db2 ≡ 1 + 3 · 1 ≡ 4 (mod 8). Hence [z ] = 41 (a2 + Db2 ) is an odd integer. Thus, for j ∈ {1, 2}, [zj ] is an odd integer. Next we have to check several cases. √ √ First, if z1 = a1 + b1 Di ∈ Z0 and z2 = a2 + b2 Di ∈ Z0 , then tr(z1 z2 ) = 2(a1 a2 + b1 b2 D) is an even integer. Using the identity

|z1 − z2 |2 = [z1 ] + [z2 ] − tr(z1 z2 ),

√

√

we see that |z1 − z2 | is an even integer. Second, if z1 = a1 + b1 Di ∈ Z0 and z2 = 12 (a2 + b2 Di) ∈ Zj for j ∈ {1, 2}, then D ≡ −1 (mod 4). Also, since among the integers a1 , a2 , b1 , b2 only one is an even √ integer. Hence, tr(z1 z2 ) =√a1 a2 + b1 b2 D is an odd integer. Hence |z1 − z2 | is an odd integer. Finally, if z1 = 21 (a1 + b1 Di) ∈ Zj , z2 = 12 (a2 + b2 Di) ∈ Zℓ for j, ℓ ∈ {1, 2}, then tr(z1 z2 ) = 21 (a1 a2 + b1 b2 D) is an integer. If j = ℓ, then a1 ≡ ±b1 (mod 4) and a2 ≡ ±b2 (mod 4). So a1 a2 + b1 b2 D ≡ a1 a2 + a1 a2 (−1) ≡ 0 (mod 4). Hence |z1 − z2 | is an even integer. Otherwise, we have a1 ≡ ±b1 (mod 4) and a2 ≡ ∓b2 (mod 4). Thus a1 a2 + b1 b2 D ≡ a1 a2 + (±a1 )(∓a2 )(−1) ≡ 2a1 a2 ≡ 2 (mod 4). It follows that |z1 − z2 | is an odd integer. This completes the proof. □ Lemma 6. With the above notation, (i) If π1 ∈ Zj j ∈ {0, 1, 2} , π2 ∈ Z0 then π1 π2 ∈ Zj ; (ii) If π1 , π2 ∈ Zj , then π1 π2 ∈ Zl , where {j, l} = {1, 2}; (iii) If π1 ∈ Z1 and π2 ∈ Z2 , then π1 π2 ∈ Z0 . Proof.

√

√

√

(i) Let π1 ∈ Z0 , π2 ∈ Z0 . Assume that π1 = a1 + b1 Di, π2 = a2 + b2 Di , π1 π2 = a + b Di. On the other hand, the Gaussian norm is multiplicative, so N(π1 π2 ) = N(π1 )N(π2 ). That is, a2 + b2 D = (a21 + b21 D)(a22 + b22 D). By assumption 2 2 a21 + b21 D and a22 + b22 D are odd numbers, so a + b. This means π1 π2 ∈ Z0 . √the same is true for √a + b D. Hence 2 ∤ √ Let π1 ∈ Zj , j ∈ {1, 2} and π1 = 21 (a1 + b1 Di) ; π2 = a2 + b2 Di , π1 π2 = 12 (a + b Di) . Observe a = a1 a2 − b1 b2 D and b = a1 b2 + a2 b1 are odd numbers. Since a + b = (a1 + b1 )(a2 + b2 ) − b1 b2 (D + 1), 4 | D + 1 and a2 + b2 is odd, if 4 | a1 + b1 then 4 |√ a + b and if 4 ∤ a1 + b√ 1 then 4 ∤ a + b. Hence π1 π2 ∈ Zj . (ii) Let π1 = 12 (a1 + b1 Di), π2 = 12 (a2 + b2 Di), √ 2x = a1 a2 − b1 b2 D , and 2y = a1 b2 + a2 b1 . Then π1 π2 = 21 (x + y Di). Assume π1 ∈ Zj , π2 ∈ Zj for j ∈ {1, 2}. a1 ≡ ±b1 (mod 4), a2 ≡ ±b2 (mod 4) with a1 , a2 , b1 , b2 are odd numbers. Since D ≡ −1 (mod 4), we have 2x = a1 a2 − b1 b2 D ≡ 2a1 a2 ≡ 2 (mod 4); 2y = a1 b2 + a2 b1 ≡ 2a1 b2 ≡ 2 (mod 4). Hence x, y are odd numbers. So 2x ∓ 2y = a1 a2 − b1 b2 D ∓ (a1 b2 + a2 b1 ) = (a1 ∓ b1 )(a2 ∓ b2 ) − b1 b2 (D + 1). Since 16 | (a1 ∓ b1 )(a2 ∓ b2 ) and 8 ∤ b1 b2 (D + 1) we have 4 ∤ x ∓ y i.e. 4 | x ± y. If π1 , π2 ∈ Z1 then π1 π2 ∈ Z2 and if π1 , π2 ∈ Z2 then π1 π2 ∈ Z1 . ( ( √ ) √ ) (iii) Let π1 ∈ Z1 , π2 ∈ Z2 . Hence we have π1 = 21 a1 + b1 Di , π2 = 12 a2 + b2 Di , where a1 , a2 , b1 , b2 are odd numbers such that a1 ≡ b1

(mod 4) , a2 ≡ −b2

(mod 4).

Thus a1 a2 − b1 b2 D ≡ a1 a2 − a1 (−a2 )(−1) ≡ 0

(mod 4).

Similarly, a1 b 2 + a2 b 1 ≡ 0

(mod 4).

Let 4x = a1 a2 − b1 b2 D and 4y = a1 b2 + a2 b1 , where x, y are integers. Hence,

√ π1 π2 = x + y Di.

On the other hand, observe that 4x + 4y = (a1 + b1 )(a2 + b2 ) − b1 b2 (D + 1).

(3)

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11

On the right hand of (3), we have 8 | (a1 + b1 )(a2 + b2 ). Also, since D ∈ {11, 19, 43, 67, 163} we see that 8 ∤ b1 b2 (D + 1). We conclude that 2 ∤ x + y. Consequently, π1 π2 ∈ Z0 . □ Lemma 7. Let π01 , . . . , π0k ∈ Z0 , π11 , . . . , π1s ∈ Z1 and π21 , . . . , π2q ∈ Z2 . Then the product where s − q = 3m + r, 0 ⩽ r ⩽ 2, m ∈ Z.

∏k

∏k

∏k

i=1

π0i

∏k

j=1

π1j

∏k

ℓ=1 π2ℓ

is in Zr

∏k

Proof. U3 = {1, ε1 , ε2 } be the set of 3rd roots of unity. By Lemma 6, in the product i=1 π0i j=1 π1j ℓ=1 π2ℓ if we replace π0i by 1, replace π1j by ε1 and replace π2ℓ by ε2 , then as a result if we get 1 then the product belongs to Z0 , if we get ε1 , then the q −q s−q product belongs to Z1 and if we get ε2 , then the product belongs to Z2 . After this assume that we get 1k ε1s ε2 = ε1s ε1 = ε1 ; s − q = 3m + r , 0 ⩽ r ⩽ 2 , m ∈ Z. If r = 0 then the product belongs to Z0 , if r = 1 then the product belongs to Z1 , if r = 2 then the product belongs to Z2 . □ Lemma 8. Let πj ∈ Z1 for 1 ⩽ j ⩽ k and πj ∈ Z2 for k + 1 ⩽ j ⩽ k + q. For given positive integers s1 , s2 , . . . , sk+q , let S be the ∏ tj set of all complex numbers π such that π = 1⩽j⩽k+q πj , 0 ⩽ tj ⩽ sj . If S = S0 ∪ S1 ∪ S2 and Sj ⊂ Zj for 0 ⩽ j ⩽ 2, then 1 ∏ max {|S0 |, |S1 |, |S2 |} = ⌊ ( (sj + 1) + 2)⌋. 3 1⩽j⩽k+q

(Here |A| denotes the number of elements in a set A) Proof. In the product

t

∏

1⩽j⩽k+q

πj j if πj ∈ Z1 , then we replace πj by ε1 , if πj ∈ Z2 , then we replace πj by ε2 . Let

(t1 + · · · + tk ) − (tk+1 + · · · + tk+q ) = 3m + r , m ∈ Z , 0 ⩽ r ⩽ 2. By Lemma 7 if r = 0, then π ∈ Z0 , if r = 1, then π ∈ Z1 , if r = 2, then π ∈ Z2 . Define the following sets Aj = 0, 1, 2, . . . , sj

{

}

for 1 ⩽ j ⩽ k

and Aj = 0, −1, −2, . . . , −sj

{

}

for k + 1 ⩽ j ⩽ k + q.

Then the number of sum α ∈A αj which has the property that j j Consider the following generating function

∑

f (x) =

∏

=

∑

(1 + x + x2 + · · · + xsj )

1⩽j⩽k

∏

αj ∈Aj αj

∑

≡ r (mod 3) is |Sr | for r = 0, 1, 2.

(1 + x−1 + x−2 + · · · + x−sj )

k+1⩽j⩽k+q

ak x . k

It is immediate that

|S0 | =

∑

ak , |S1 | =

3|k

Set ε =

1 ( 2

∑

ak+1 and |S2 | =

3|k

∑

ak+2 .

3|k

√ −1 +

|S0 | = |S1 | = |S2 | =

1 3 1 3 1 3

3i). Then

(f (1) + f (ε ) + f (ε 2 )) ; (f (1) + ε 2 f (ε ) + ε f (ε 2 )) ; (f (1) + ε f (ε ) + ε 2 f (ε 2 )).

Let f1 (x) = 1 + x + x2 + · · · + xn . Thus,

ε n+1 − 1 ε 2(n+1) − 1 and f1 (ε 2 ) = . ε−1 ε2 − 1 Suppose n + 1 = 3k + r , 0 ⩽ r ⩽ 2 , k ∈ Z. Assume first r = 0. Then f1 (ε ) = f1 (ε 2 ) = 0. Next if r = 1 then f1 (ε ) = f1 (ε 2 ) = 1. Finally if r = 2 then f1 (ε ) = −ε 2 and f1 (ε 2 ) = −ε . Since 1 + ε −1 + ε −2 + · · · + ε −n = f1 (ε −1 ) = f1 (ε 2 ) and 1+ε −2 +ε −4 +· · ·+ε −2n = f1 (ε −2 ) = f1 (ε ) we have |f (ε )| ⩽ 1 , |f (ε 2 )| ⩽ 1. Let m0 = f (ε )+f (ε 2 ) , m1 = ε 2 f (ε )+ε f (ε 2 ) , m2 = ε f (ε ) + ε 2 f (ε 2 ). Since mj = 3|Sj | − f (1) for 0 ⩽ j ⩽ 2 we can see that mj are integers and m0 ≡ m1 ≡ m2 (mod 3). Also, since |S0 | + |S1 | + |S2 | = f (1) we have m0 + m1 + m2 = 0. Hence |m0 | ⩽ |f (ε )| + |f (ε 2 )| ⩽ 2, |m1 | ⩽ 2 , |m2 | ⩽ 2. Consequently, {m0 , m1 , m2 } ∈ {{0, 0, 0} , {1, 1, −2} , {−1, −1, 2}}. Thus, we conclude that f (1) + 2 1 ∏ max {|S0 |, |S1 |, |S2 |} = ⌊ ⌋=⌊ ( (sj + 1) + 2)⌋. □ f 1 (ε ) =

3

3

1⩽i⩽k+q

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Proof (Proof of Theorem 5). By Proposition 2 we know that

(

−D

)

p

)

–

= 1 if and only if

(p) D

= 1. Since the prime factorization is

unique in the ring O−D , there exist integers a, b such that 4p = a2 + b2 D. s s s s Let n2 = p11 . . . pmm , n3 = pmm++11 . . . pkk , where pj are distinct prime numbers. Then the primes pj in the ring O−D can be factorized as follows

√

pj =

aj + b j

√

Di

aj − b j

·

Di

2 2 with aj , bj are odd positive integers, 1 ⩽ j ⩽ m and

√

pj = (aj + bj

√

Di)(aj − bj

Di)

with aj , bj are positive integers, m + 1 ⩽ j ⩽ k. As before we consider the ξ0 , ξ , ξ ′ points with pairwise integral distances on the circle with radius R such that arg(ξ0 ) = 0. The units of O−D are α = ±1. By Lemma 1 we have e4itj ωj , −sj ⩽ tj ⩽ sj

∏

ξ =R

1⩽j⩽k

and

ξ′ = R

∏

e

4itj′ ωj

, −sj ⩽ tj ⩽ sj .

1⩽j⩽k

By Lemma 2 we have

√

|ξ − ξ | = 2R|sin( ′

∑

(tj − tj )ωj )| = 2R · 2s0

1≤j≤k

D

S

′

|tj −tj′ |

∏

1≤j≤k

pj

and gcd(S , 2s0

∏

|tj −tj′ |

pj

) = 1.

1≤j≤k

In order for |ξ − ξ ′ | to be an integer we must have |tj − tj′ | ⩽ sj for 1 ⩽ j ⩽ k. Arguing as in the proof of Theorem 1, if we choose a set Sj which consists of distinct tj ’s with the cardinality more than sj + 1, then the difference between the minimum and maximum element of that set is |tj − tj′ | > sj . Hence number of different values of tj is at most sj + 1. By Lemma 1 we get 4RD n21

ξ = n22

∏

e4itj ωj

1≤j≤k

( ∏

=

aj − b j

1≤j≤m

∏ (

·

√ ) cj ( Di

aj + bj

2

√ )2sj −cj Di

2

aj − b j

√ )cj ( Di

aj + b j

√ )2sj −cj Di

.

m+1≤j≤k

Consequently, by Lemma 6 we have 4RD n21

ξ ∈ Zj for j ∈ {0, 1, 2}.

Suppose that

4RD n21

ξ ∈ Zj and 4RD ξ ′ ∈ Zℓ with j, ℓ ∈ {0, 1, 2}. If j = ℓ, then by Lemma 5 and since 4RD is an even integer we have n2 n2 1

s0 = 0. Similarly, if j ̸ = ℓ, then since

4RD n21

1

is an odd integer we have s0 = 1. Hence we can see that if n1 is an even integer, then

similar way in proof √ of Theorem √ 1 we can construct at most τ (n2 n3 ) points with pairwise integral distances on the circle. If ′ n1 is odd integer, 2n D B ∈ Zj , 2n D B′ ∈ Zl and j ̸ = l, then |ξ − ξ ′ | cannot be an integer. Hence we only have the choice having 1

1

the points with pairwise integral distances is j = l. Finally, by Lemma 5 we can construct at most ⌊ 13 (τ (n2 ) + 2)⌋τ (n3 ) points with pairwise integral distances on the circle with radius R. This completes the proof. □ Example 5. See Fig. 5.

Corollary 1. The constructions of points with pairwise integral distances in Theorems 1–4 and 5(a) are symmetric and unique up to rotations. √

Corollary 2. If the number of points with pairwise integral distances is more than 3 on the circle radius R = n D D , where D ∈ {1, 2, 3, 7} with n is an integer or D ∈ {1, 2, 3, 7, 11, 19, 43, 67, 163} with n is even integer, then we can construct the maximum number of points with pairwise integral distances. Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.

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)

–

√

13

√

Fig. 5. 10 integral points set and 4 integral points set on the circle with radius R = 405 11/11 and R = 405 11/22, respectively.

Corollary 3. 3 | τ (n2 ) and n1 ≡ 2 (mod 4) or 3 ∤ τ (n2 ) and n1 = 2; n3 = 1 in Theorem 5, also s0 = 1 in Theorem 3, the number 2 r(r −3) of edges with odd length connecting the points with pairwise integral distances is exactly N3 + 6 , where N ≡ r (mod 3) with r = 1, 2, 3. Proof. Let V be the N integral points set. We divide the set V into three disjoint sets as follows: Vi = {ξ | ξ ∈ V and 4RD 2 ξ ∈ Zi } with the cardinality mi for i = 1, 2, 3; D ∈ {3, 11, 17, 19, 43, 67, 163} (see Fig. 6). n1

ξ ′ ∈ Vj , then |ξ − ξ ′ | is odd integer when i ̸= j and even integer when i = j. Hence the number of edges with odd length is f := m0 m1 + m1 m2 + m2 m0 . From the proof of From the proofs of Theorems 3 and 5 we can see that if

4RD n21

ξ ∈ Vi and

4RD n21

Lemma 8, we can see that

{m0 , m1 , m2 } = {k, k, k} when N = 3k, {m0 , m1 , m2 } = {k, k, k + 1} when N = 3k + 1, {m0 , m1 , m2 } = {k, k + 1, k + 1} when N = 3k + 2. Consequently,

⎧ N2 ⎪ ⎪ , ⎪ ⎪ ⎪ ⎨ 23 N −1 f = , ⎪ ⎪ 23 ⎪ ⎪ N − 1 ⎪ ⎩ , 3

when N = 3k when N = 3k + 1 when N = 3k + 2.

This simplifies to f =

N2 3

+

r(r − 3) 6

.

□

Corollary 4. There exists an n integral points set on a circle such that exactly one distance between two points is an odd integer but all other distances are even integers. Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.

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Fig. 6. The associated graph to the proof of Corollary 3.

n

√

1 . Then Proof. Let us consider the points ξj = Re4itj ω , 0 ⩽ j ⩽ n − 1 on the circle with radius R = 2 14 7 , where ω = arccos √ 2 2 this forms an n points integral set. Theorem 4 implies that |ξ0 − ξn−1 | is an odd integer , but the others |ξj − ξk | are even integers. □

3.2. The minimum value of the radius of a circle, which contains n integral points α

α

√

By Theorems 1–5 we have R(n, D) = α · p1 1 · · · pk k DD , where n ≥ 3 , D ∈ {1, 2, 3, 7, 11, 19, 43, 67, 163} and p1 < p2 < · · · < pk . Here we take α = 1 when D ∈ {1, 2} and α = 2 when D ∈ {3, 7, 11, 19, 43, 67, 163}. Also, if n = α ′ (α1 + 1) · · · (αk + 1), then we take 2, α = 3, 1,

when D = 1 when D = 3 when D ̸ = 1, 3

{

′

Clearly, α1 ≥ α2 ≥ · · · ≥ αk . Theorem 6. With the above notation, (a) (b) (c) (d)

mαj ≤ αi when pm i < pj ; +1 αi ≤ 2m when pk+1 < pm ; i αj = 1 when pi pj > pk+1 and αi ≥ 3; α1 ≥ 8 when αi = 2, αj = 2 and pi pj > p1 pk+1 .

Proof. α

αj

α +m αj −1

(a) Since pi i pj > pi i

pj

αi αj

, we have (αi + 1)(αj + 1) > (αi + m + 1)αj . Assume by the contradiction that mαj > αi . α +m αj −1

α

α

√

Now if we replace pi pj by pi i pj in the product R(n, D) = α · p1 1 · · · pk k 2DD then we have at least n points with pairwise integral distances on the resulting circle with smaller radius. But this contradicts the fact that R(n, D) is the minimal. Hence mαj ≤ αi . α α −m−1 α α −m−1 (b) Since pi i > pi i √ pk+1 , we have αi + 1 > (αi − m) · 2. If not, we replace pi i by pi i in the product R(n, D) = α

α

α · p1 1 · . . . · pk k

D 2D

, then we have n points with pairwise integral distances on a circle with smaller radius. This contradicts the fact that R(n, D) is the minimal. Hence αi ≤ 2 m. α αj α −1 αj −1 (c) Assume by contradiction that αj ≥ 2. Then pi i pj > pi i pj pk+1 and R(n, D) is minimal, we have (αi + 1)(αj + 1) > α

αj

α −1 αj −1pk+1

α

α

√

2αi αj . If not, we replace pi i pj by pi i pj in the product R(n, D) = α · p1 1 · · · pk k 2DD then we get more than n points with pairwise integral distances on a circle with smaller radius than R(n, D), which is a contradiction. Hence (αi − 1)(αj − 1) < 2, but this is again contradicts as α ⩾ 2. Therefore αj = 1. α α +1 (d) Since p1 1 p2i p2j > p1 1 p1 p2 · pk+1 , we have (α1 + 1) · 3 · 3 > (α1 + 2) · 2 · 2 · 2. Hence α1 > 7. □ Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.

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Fig. 7. n integral points set on the circle with radius Rn for 2 ≤ n ≤ 6.

√

Fig. 8. 7 integral points set on the circle with radius R = 64 15/15.

Example 6. In this example using Theorem 6 we find the value of R(n, 1) for smaller value of n. By Theorem 1 we have pi = 4s + 1, so p1 = 5, p2 = 13, p3 = 17, . . . . Clearly, R(2, 1) = 1/2. Note that we have α = 1. If R(n; 1) = 21 5α1 13α2 , then by Theorem 6(a) and since 5 < 13 we must have α2 ≤ α1 . By Theorem 6(b) and since 17 < 52 , we have α1 ≤ 2. Hence R(4, 1) = 5/2, R(6, 1) = 12 52 = 25/2, R(8, 1) = 21 5 · 13 = 65/2, R(12, 1) = 12 52 · 13, R(18, 1) = 21 52 · 132 . Exactly the same way we find √R(n, 1) and R(n, 3) for n ≤ 3072, see Tables 1 and 2. Clearly, R2 = 1/2, R3 = √ find that R4 =

8 15 15

, R5 = R6 =

7 3 3

. See Fig. 7 and 8.

In Fig. 8 we constructed 7 integral points set on the circle with radius R = √

√

√ 64 15 . 15

3 . 3

We also √

We conjecture that R7 =

64 15 15

and

R8 = R9 = 493 3 . From Tables 1 and 2, we can see that R(n, 3) < R(n, 1) for 8 ≤ n ≤ 3072. Hence we obtain an upper bound for Rn for 8 ≤ n ≤ 3072, namely Rn ≤ R(n, 3).

Acknowledgments

The author would like to owe a special thanks to the anonymous referee for having a careful reading and corrections. Also, I am very thankful to Dr B. Bayasgalan who brought to my attention this nice problem. I am also thankful to B. Bayarjargal for several helpful suggestions. Finally, I am thankful to Dr Batzorig Undrakh for translated this article from Mongolian to English, and several comments on the proofs and to improve the current presentation of this paper. Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.

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Please cite this article in press as: G. Bat-Ochir, On the number of points with pairwise integral distances on a circle, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.07.004.