INTEGERS: Electronic J. of Combinatorial Number Theory 5 (2005), #A12

ON CONSECUTIVE INTEGER PAIRS WITH THE SAME SUM OF DISTINCT PRIME DIVISORS

Douglas E. Iannucci, University of the Virgin Islands, St. Thomas VI 00802, USA

                                diannuc@uvi.edu

Alexia S. Mintos, University of the Virgin Islands, St. Thomas VI 00802, USA

Received: 1/10/05, Accepted: 5/28/05, Published: 6/29/05

Abstract

We define the arithmetic function P by P (1) = 0, and P (n) = p_{1} + p_{2} + ⋯ + p_{k} if n has the unique prime factorization given by n = ∏_{i=1}^{k} p_{i}^{a_{i}}; we also define ω(n) = k and ω(1) = 0. We study pairs (n, n + 1) of consecutive integers such that P (n) = P (n + 1). We prove that (5, 6), (24, 25), and (49, 50) are the only such pairs (n, n + 1) where {ω(n), ω(n + 1)} = {1, 2}. We also show how to generate certain pairs of the form (2^{2n}pq, rs), with p < q, r < s odd primes, and lend support to a conjecture that infinitely many such pairs exist.

Keywords: Ruth–Aaron pairs, cyclotomic polynomials, Pell sequences, primes

Subject Class: 11A25, 11Y55

1. Introduction

For positive integers n, we define the arithmetic function P (n) by P (1) = 0, and, for a positive integer n having as its unique prime factorization n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}},

P (n) = p_{1} + p_{2} + ⋯ + p_{k}.

That is, P (n) gives the sum of prime divisors of n without multiplicity taken into account. The function is additive, in that P (m) + P (n) = P (mn) if (m, n) = 1.

This function compares to the arithmetic function defined for positive integers n by S(1) = 0 and S(n) = ∑_{i=1}^{k} a_{i}p_{i} whenever n = ∏_{i=1}^{k} p_{i}^{a_{i}}; that is, S(n) gives the sum of primes dividing n, taken with multiplicity. Then S(n) is completely additive, in that S(mn) = S(m) + S(n) for any two positive integers m and n. A Ruth–Aaron pair is a pair (n, n + 1) of consecutive integers such that S(n) = S(n + 1). These were first discussed by Pomerance et. al. [4], and have been the subject of several articles (such as by Pomerance [6], Drost [2]) and numerous websites since.

However, in this article we are interested in finding pairs of consecutive positive integers

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(n, n + 1), such that P (n) = P (n + 1). For the sake of easy reference, we may call these Ruth– Aaron pairs of the second kind, or RAP2s for short. Note, however, that a RAP2 is also an ordinary Ruth–Aaron pair if both members n and n + 1 are square-free.

Some observations regarding RAP2s are immediate. For example, the members (n, n + 1) of a RAP2 are of opposite parity, and are relatively prime. Let n be a positive integer. If n has the unique prime factorization n = ∏_{i=1}^{k} p_{i}^{a_{i}}, then the prime powers p_{i}^{a_{i}}, 1 ≤ i ≤ k, are called the components of n, and we define ω(n) = k, ω(1) = 0 (thus ω counts the components of n). For any given RAP2 (n, n + 1), since 2 divides exactly one of the members (all other prime divisors of n and n + 1 being odd), we see that ω(n) and ω(n + 1) are of opposite parity. In this article we shall completely determine all RAP2s (n, n + 1) whose members have one or two components. We will also investigate RAP2s of the form (2^{2n}pq, rs), with p < q, r < s odd primes.

2. Preliminaries

If p is a prime and a, m, are positive integers we write p^{m}∥a if p^{m} | a and p^{m+1} ∤ a. In this case we say p^{m} exactly divides a. For distinct primes p and q we write e_{p}(q) to denote the exponent to which q belongs modulo p.

For positive integers n, we denote the n^{th} cyclotomic polynomial evaluated at x by Φ_{n}(x). The cyclotomic polynomials (as shown by Niven [5], Ch. 3) may be defined inductively by

(1)  x^{n} − 1 = ∏_{d|n} Φ_{n}(x).

By Theorems 94 and 95, Nagell [3], Ch. 5, we have

Lemma 1. Let p be and q be odd primes and let m be a positive integer. Let h = e_{p}(q). Then p | Φ_{m}(q) if and only if m = hp^{j} for some integer j ≥ 0. If j > 0 then p∥Φ_{hp^{j}}(q).

Lemma 2. Let q be an odd prime and let m be a positive integer. Then 2 | Φ_{m}(q) if and only if m = 2^{j} for some integer j ≥ 0. If j > 1 then 2∥Φ_{2^{j}}(q).

Let q be prime and let m > 0 be an integer. Since, by definition,

Φ_{m}(q) = ∏_{\begin{matrix}k=1 \\ (k,m)=1 \end{matrix}}^{m−1}(q − e^{2πik/m}),

and since Φ_{m}(q) > 0, we have

Φ_{m}(q) = ∏_{\begin{matrix}k=1 \\ (k,m)=1 \end{matrix}}^{m−1} |q − e^{2πik/m}| ,

and since |q − e^{2πik/m}| ≥ q − 1 for 1 ≤ k ≤ m − 1, we have Lemma 3. For a prime q and an integer m > 0 we have Φ_{m}(q) ≥ (q − 1)^{ϕ(m)}.

3. RAP2s of the form (2^{a}p^{b}, q^{c})

The smallest numbers of components the members of a RAP2 can have are 1 and 2. In this instance, the members of the RAP2 have the form 2^{a}p^{b} and q^{c} for positive integers a, b, and c, where p and q are necessarily twin (odd) primes (that is, p + 2 = q). We have two cases arising in this instance, those being 2^{a}p^{b} = q^{c} ± 1. In this section we consider the easier case of the two,

(2)  2^{a}p^{b} = q^{c} − 1.

Clearly c > 1, since 2^{a}p^{b} ≥ 2(q − 2) = q + (q − 4) > q − 1. Thus, since q = p + 2, (2) factors as

2^{a}p^{b} = (p + 1)(q^{c−1} + q^{c−2} + ⋯ + q + 1).

Since (p, p + 1) = 1, it follows that p + 1 = 2^{t} for some positive integer t. Hence p = 2^{t} − 1, q = 2^{t} + 1, which is possible only if t = 2; that is, p = 3 and q = 5. Then (2) becomes

2^{a}3^{b} = 5^{c} − 1.

Since 5^{c} ≡ 1 (mod 3), we have 2 | c, so we write c = 2γ for some positive integer γ. Thus

2^{a}3^{b} = (5^{γ} − 1)(5^{γ} + 1).

Since 2∥5^{γ} + 1, we must have 3 | 5^{γ} + 1. Since (5^{γ} + 1, 5^{γ} − 1) = 2, we have 3 ∤ 5^{γ} − 1. Hence

5^{γ} − 1 = 2^{a−1}, 5^{γ} + 1 = 2 · 3^{b}.

Certainly γ is odd (as 3 ∤ 5^{γ} − 1). Suppose γ > 1. Then

5^{γ} − 1 = (5 − 1)(5^{γ−1} + 5^{γ−2} + ⋯ + 5 + 1).

But the second factor is odd, and greater than 1; this contradicts 5^{γ} − 1 = 2^{a−1}. Therefore γ = 1, and so c = 2. Hence (2) becomes 2^{3} · 3 = 5^{2} − 1; that is, a = 3, b = 1, and we have the RAP2 (24, 25). Hence the only RAP2 of the form (2^{a}p^{b}, q^{c}) is (24, 25).

4. RAP2s of the form (q^{c}, 2^{a}p^{b})

Suppose now that

(3)  2^{a}p^{b} = q^{c} + 1

for positive integers a, b, and c, where p and q are primes such that p + 2 = q. This case is more difficult than that in Section 3.

By (1), we see that (3) is equivalent to

(4)  2^{a}p^{b} = ∏_{\begin{matrix}d|2c \\ d∤c \end{matrix}} Φ_{d}(q).

Let h = e_{p}(q); we observe h = e_{p}(2) as well (since q = p + 2). By Lemmas 1 and 2, each divisor d | 2c such that d ∤ c must either have the form hp^{j} for some integer j ≥ 0 or the form 2^{k} for some integer k ≥ 1. Writing c = 2^{m}s for some integer m ≥ 0 and odd integer s, we see that 2^{m+1}∥d for all divisors d | 2c such that d ∤ c. In particular 2^{m+1}∥h.

Suppose s is composite. Then t | s for some odd integer t such that 1 < t < s. Then by (4), Φ_{s}(q)Φ_{t}(q) | 2^{a}p^{b}. This is impossible as 2 ∤ Φ_{s}(q)Φ_{t}(q) by Lemma 2, and, as 2 | h, we have h ∤ s, and so p ∤ Φ_{s}(q)Φ_{t}(q) by Lemma 1. Hence either s is prime or s = 1.

Suppose s is prime. Then h = 2^{m+1}s. For, if this were not the case then we would have h = 2^{m+1}; since 2 ∤ Φ_{2c}(q) by Lemma 2, it follows that either p ∤ Φ_{2c}(q) (if s ≠ p) or p∥Φ_{2c}(q) (if s = p) by Lemma 1. The former possibility clearly contradicts (4); the latter implies Φ_{2c}(q) = p, which is impossible as Φ_{2c}(q) > p by Lemma 3.

Therefore, since h = 2^{m+1}s, we have

(5)  2^{a}p^{b} = Φ_{2^{m+1}s}(q)Φ_{2^{m+1}}(q)

by (4). This implies m = 0 because otherwise (5) is impossible since we have p ∤ Φ_{2^{m+1}}(q), 2∥Φ_{2^{m+1}}(q), and Φ_{2^{m+1}}(q) > 2 by Lemmas 1, 2, and 3 respectively. Therefore h = 2s and

2^{a}p^{b} = Φ_{2}(q)Φ_{2s}(q),

with 2^{a} = Φ_{2}(q) = q+1 and p^{b} = Φ_{2s}(q). But then q = 2^{a}−1, so that p = 2^{a}−3. It is clear that a > 2, hence p ≡ 5 (mod 8). Thus 2 is a quadratic nonresidue of p, and hence 2^{(p−1)/2} ≡ −1 (mod p) by Euler’s criterion. Since 2^{2} | p − 1 = 2^{a} − 4, it follows that 2^{2} | e_{p}(2). But e_{p}(2) = h, and since h = 2s, we have 2∥h, a contradiction.

Therefore s = 1 and hence c = 2^{m} for some integer m ≥ 0. Thus (3) becomes

(6)  2^{a}p^{b} = q^{2^{m}} + 1.

First let us suppose that m > 1. Then q^{2^{m}} ≡ 1 (mod 4) so that a = 1. Hence

(7)  2p^{b} = q^{2^{m}} + 1.

Since p | q^{2^{m}} + 1 = Φ_{2^{m+1}}(q), it follows from Lemma 1 that h = 2^{m+1}; recalling as well h = e_{p}(2), it follows that p ≡ 1 (mod 2^{m+1}). Since e_{p}(2) = 2^{m+1} and Φ_{2^{m+1}}(2) = 2^{2^{m}} + 1, it follows from Lemma 1 that

(8)  p | 2^{2^{m}} + 1.

Suppose p = 2^{m+1}t + 1 for some odd integer t. Then, as 2^{2^{m}} ≡ −1 (mod p) by (8),

2^{(p−1)/2} = 2^{2^{m}t} = (2^{2^{m}})^{t} ≡ (−1)^{t} ≡ −1 (mod p),

INTEGERS: ELECTRONIC J. OF COMBINATORIAL NUMBER THEORY X (2005), #A12 5

and hence (\frac{2}{p}) = −1 by Euler’s criterion, where (\frac{·}{·}) denotes the Legendre symbol. But, p ≡ 1 (mod 8), which implies (\frac{2}{p}) = +1, a contradiction. Therefore

(9)  p ≡ 1 (mod 2^{m+2}).

Suppose b > 2^{m}. Then from (7) we have

2p^{b−2^{m}} = (1 + \frac{2}{p})^{2^{m}} + \frac{1}{p^{2^{m}}} .

By (9), p > 2^{m+2}, and so

2p^{b−2^{m}} < (1 + \frac{1}{2^{m}} )^{2^{m}} + 1 < e + 1 < 4,

which implies 2p < 4, a contradiction. On the other hand, suppose b < 2^{m}. Then by (6),

2 = (1 + \frac{2}{p})^{b} (p + 2)^{2^{m}−b} + \frac{1}{p^{b}} > (p + 2)^{2^{m}−b} ≥ p + 2,

a contradiction. Therefore we must have b = 2^{m}, so that (7) becomes

(10)  2p^{2^{m}} = q^{2^{m}} + 1.

Since q = p + 2, (10) becomes

p^{2^{m}} = −p^{2^{m}} + (p + 2)^{2^{m}} + 1

= ∑_{k=1}^{2^{m}} \binom{2^{m}}{k}p^{2^{m}−k}2^{k} + 1. (11)

Since by (9) p > 2^{m+2}, we have for each k such that 1 ≤ k ≤ 2^{m},

\binom{2^{m}}{k}p^{2^{m}−k}2^{k} = \frac{2^{m}(2^{m} − 1)(2^{m} − 2) ⋯ (2^{m} − k + 1)}{k!} · p^{2^{m}−k}2^{k}

< \frac{2^{mk}}{k!} · p^{2^{m}−k}2^{k}

= \frac{1}{2^{k}} · \frac{1}{k!} · p^{2^{m}}.

Hence by (11),

p^{2^{m}} < ∑_{k=1}^{2^{m}} \frac{1}{2^{k}} · \frac{1}{k!} · p^{2^{m}} + 1 < p^{2^{m}} (\sqrt{e} − 1 + \frac{1}{p^{2^{m}}} ) < 0.8p^{2^{m}},

a contradiction.

Hence we have (6) with either m = 0 or m = 1. If m = 0 then (6) becomes 2^{a}p^{b} = q + 1 = p + 3, implying p | 3, and hence p = 3, q = 5. Therefore 2^{a}3^{b} = 6, and so a = b = 1, and we have the RAP2 (5, 6).

If m = 1 then (6) becomes

2^{a}p^{b} = q^{2} + 1,

which implies a = 1 since q^{2} ≡ 1 (mod 4). Therefore

2p^{b} = q^{2} + 1 = (p + 2)^{2} + 1 = p^{2} + 4p + 5,

implying p | 5, and hence p = 5, q = 7. Therefore 2 · 5^{b} = 50, and so b = 2, and we have the RAP2 (49, 50). Therefore the only RAP2s of the form (q^{c}, 2^{a}p^{b}) are (5, 6) and (49, 50).

We summarize our results from this and the previous section:

Theorem 1. The only RAP2s (n, n+1) such that {ω(n), ω(n+1)} = {1, 2} are (5, 6), (24, 25), and (49, 50).

5. RAP2s of the form (2^{2n}pq, rs)

We now turn our attention to RAP2s (n, n + 1) where {ω(n), ω(n + 1)} = {2, 3}. There are 88 such pairs less than 10^{9}. Of these, 41 have the form (4pq, rs), six have the form (16pq, rs), and three have the form (64pq, rs), with p < q, r < s odd primes. Among the remaining 38 pairs, no discernable patterns emerged. These data led us to narrow our investigation to those pairs of the form (2^{2n}pq, rs), n ≥ 1.

Given such a pair, we have

(12)  2 + p + q = r + s,
(13)  2^{2n}pq + 1 = rs.

By (12) we have integers x, y, and z such that

(14)  r = x − y,  s = x + y,
(15)  p = x − 1 − z,  q = x − 1 + z.

Substituting (14) and (15) into (13), and simplifying, gives us

((2^{2n} − 1)x − (2^{2n} + 1))(x − 1) = (2^{n}z − y)(2^{n}z + y),

which may be expressed as

(16)  \frac{(2^{2n} − 1)x − (2^{2n} + 1)}{2^{n}z − y} = \frac{2^{n}z + y}{x − 1} = \frac{a}{b},

where a/b represents the fractions in (16) in their lowest terms; thus (a, b) = 1. Separating the variables x, y, and z in (16) gives us

(2^{2^{n}} − 1)bx + ay − 2^{n}az = (2^{2^{n}} + 1)b,
ax − by − 2^{n}bz = a,

which we solve for x, y, in terms of z:

(17)  (a^{2} + (2^{2^{n}} − 1)b^{2})x = 2^{n+1}abz + a^{2} + (2^{2^{n}} + 1)b^{2},
(18)  (a^{2} + (2^{2^{n}} − 1)b^{2})y = 2^{n}(a^{2} − (2^{2^{n}} − 1)b^{2})z + 2ab.

Our data of RAP2s less than 10^{9} revealed to us many different rational numbers for the quotient a/b in (16), but some persisted more than others, especially 2/1 and 7/4 in the cases where n = 1. Recognizing these values as solutions to the Pell equation a^{2} − 3b^{2} = 1, we decided to assume that a, b, solved the Pell equation

(19)  a^{2} − (2^{2^{n}} − 1)b^{2} = 1

in the general case for n ≥ 1. Under this hypothesis, (17) and (18) simplify to

(2a^{2} − 1)x = 2^{n+1}abz + 2a^{2} + 2b^{2} − 1,
(20)  (2a^{2} − 1)y = 2^{n}z + 2ab.

It is well known (e.g., as shown by Shockley [7], Ch. 12) that all positive solutions to (19) are given by

a_{1} = 2^{n},  b_{1} = 1,
a_{j+1} = 2^{n}a_{j} + (2^{2^{n}} − 1)b_{j}  (j ≥ 1),
(21)  b_{j+1} = a_{j} + 2^{n}b_{j}  (j ≥ 1).

One shows by induction that 2^{n} | a_{j}b_{j} for all j ≥ 1. Hence we may parametrize z from (20): since y is an integer it follows that 2a^{2} − 1 divides 2^{n}z + 2ab, and since 2a^{2} − 1 is odd we have

z ≡ −\frac{2ab}{2^{n}} (mod 2a^{2} − 1).

Thus z has the form given by

(22)  z = (2a^{2} − 1)k + 2a^{2} − 1 − \frac{2ab}{2^{n}}

for integers k ≥ 0. Substituting (22) into (17) and (18) gives us

(23)  x = 2^{n+1}abk + 2^{n+1}ab − 2b^{2} + 1,
(24)  y = 2^{n}k + 2^{n}.

Substituting (22), (23), and (24) into (14) and (15) gives us

Theorem 2. Let integral n ≥ 1 be given and let a, b, be solutions to the Pell equation (19). Then (2^{2n}pq, rs) is a RAP2 if, for an integer k ≥ 0, the following four quantities are all prime:

p = 2(2^{n+1}ab − 2a^{2} + 1)k + (2^{n+1} − 2b^{2} − 2a^{2} + 1 + \frac{2ab}{2^{n}}) ,
q = 2(2^{n+1}ab + 2a^{2} − 1)k + (2^{n+1} − 2b^{2} + 2a^{2} − 1 − \frac{2ab}{2^{n}}) ,
r = 2^{n+1}(2ab − 1)k + 2^{n}(2ab − 1) − 2b^{2} + 1,
s = 2^{n+1}(2ab + 1)k + 2^{n}(2ab + 1) − 2b^{2} + 1.

Note that we substituted 2k instead of k to ensure p and q as given in Theorem 2 are odd. We also kept 2ab in the numerators above (rather than reduce to ab/2^{n−1}) since the Pell sequences (21) have the property b_{2j} = 2a_{j}b_{j} (as well as a_{2j} = 2a_{j}^{2} −1). Moreover, one shows by induction that for all n and k, if a_{3j}, b_{3j} in (21) are the solutions used in applying Theorem 2, then at least one of p, q, r, and s is divisible by 3 (hence no RAP2 is produced).

There are 149 RAP2s of the form (2^{2n}pq, rs) whose elements are less than 2^{34}. Of these, 116 correspond to n = 1, and 16 of these involve the solutions a_{1} = 2, b_{1} = 1 of the Pell equation a^{2} − 3b^{2} = 1, while an additional 3 involve a_{2} = 7, b_{2} = 4. Also, 16 such RAP2s correspond to n = 2, 3 of which involve the solutions a_{1} = 4, b_{1} = 1 of the equation a^{2} − 15b^{2} = 1, and 9 of the RAP2s correspond to n = 3, 3 of which involve a_{1} = 8, b_{1} = 1 (a^{2} − 63b^{2} = 1). Finally, 3 of the RAP2s involve n = 4.

We had found the RAP2s less than 2^{34} by a straightforward computer search. Later on, we applied Theorem 2 to search for the RAP2s of the special form described in that theorem. We found literally thousands of them. We computed them on a PC, using the UBASIC software package. Primality of p, q, r, s, were verified by the APR primality test due to Adleman, Pomerance, and Rumely [1].

6. Concluding Remarks

It is unknown if there are infinitely many RAP2s. The question of infinitude also remains open for ordinary Ruth–Aaron pairs—see Pomerance [6] for a detailed history. In light of Theorem 2, fixing n at say n = 1, if one could show that for each solution a_{j}, b_{j}, 3 ∤ j, to (19), there exists at least one k for which p, q, r, s, are all prime, then a proof of infinitely many RAP2s of the form (4pq, rs) would be obtained. We have not been able to produce such a proof, but we conjecture the existence of infinitely many RAP2s nonetheless.

We have also considered RAP2s (n, n + 1) for which {ω(n), ω(n + 1)} = {1, 4}. These would be obtained by finding distinct odd primes p_{1}, p_{2}, p_{3}, q, and positive integers a, b_{1}, b_{2}, b_{3}, c, such that

(25)  2 + p_{1} + p_{2} + p_{3} = q,

and such that

(26)  2^{a}p_{1}^{b_{1}}p_{2}^{b_{2}}p_{3}^{b_{3}} = q^{c} ± 1.

Let h = [e_{p_{1}}(q), e_{p_{2}}(q), e_{p_{3}}(q)]. Then p_{1}, p_{2}, p_{3}, all divide q^{c} −1 only if h | c, in which case q^{h} −1 divides 2^{a}p_{1}^{b_{1}}p_{2}^{b_{2}}p_{3}^{b_{3}}. Thus if q^{h} − 1 is found to contain any prime factors other than 2, p_{1}, p_{2}, p_{3}, then a contradiction is obtained. Using modular arithmetic, we can find α, β_{1}, β_{2}, β_{3}, such that 2^{α}∥q^{h} − 1 and p_{i}^{β_{i}}∥q^{h} − 1 (1 ≤ i ≤ 3). A contradiction is obtained if 2^{α}p_{1}^{β_{1}}p_{2}^{β_{2}}p_{3}^{β_{3}} < q^{h} − 1.

In the case of q^{c} + 1, (26) becomes

2^{a}p_{1}^{b_{1}}p_{2}^{b_{2}}p_{3}^{b_{3}} = ∏_{\begin{matrix}d|2c \\ d∤c \end{matrix}} Φ_{d}(q)

INTEGERS: ELECTRONIC J. OF COMBINATORIAL NUMBER THEORY X (2005), #A12 9

by (1). By Lemma 1, the primes p_{1}, p_{2}, p_{3}, all divide q^{c} + 1 only if e_{p_{1}}(q), e_{p_{2}}(q), and e_{p_{3}}(q) are all even such that each quantity is exactly divisible by the same power of 2. In this case we have q^{h/2} + 1 | 2^{a}p_{1}^{b_{1}}p_{2}^{b_{2}}p_{3}^{b_{3}}. Thus a contradiction is obtained if q^{h/2} + 1 contains any prime factors other than 2, p_{1}, p_{2}, or p_{3}.

For all odd primes q < 20000, we found all triples of odd primes p_{1} < p_{2} < p_{3} satisfying (25), and then we disproved the possibility of (25) and (26) by computation. We conjecture the nonexistence of RAP2s (n, n + 1) for which {ω(n), ω(n + 1)} = {1, 4}, although we have not yet obtained a proof.

References

[1] L. Adleman, C. Pomerance, and R. Rumely, On distinguishing prime numbers from composite numbers, Ann. of Math. 117 (1973), no. 2, 173–206.

[2] J. Drost, Ruth/Aaron Pairs, J. Recreational Math. 28 (1996), 120–122.

[3] T. Nagell, Introduction to Number Theory, AMS Chelsea Publishing, Providence, RI, 1981.

[4] C. Nelson, D. Penney, and C. Pomerance, 714 and 715, J. Recreational Math. 7 (1974), 87–89.

[5] I. Niven, Irrational Numbers (Carus Mathematical Monograph no. 11), Mathematical Association of America (Wiley and Sons, dist.), New York, 1967.

[6] C. Pomerance, Ruth-Aaron numbers revisited, Paul Erdős and His Mathematics, I (Series: Bolyai Soc. Math. Stud., Vol. 11) (G. Halasz, L. Lovasz, M. Simonovits, and V. Sos, eds.), Springer-Verlag, New York, 2002, pp. 567–579.

[7] J. Shockley, Introduction to Number Theory, Holt, Rinehart, and Winston, Inc., New York, 1967.