PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 81(95) (2007), 45–52 DOI 102298/PIM0795045M

SUMS OF LIKE POWERS AND SOME DENSE SETS

Žarko Mijajlović, Miloš Milošević and Aleksandar Perović

Abstract. We introduce the notion of the P -sequences and apply their properties in studying representability of real numbers. Another application of P -sequences we find in generating the Prouhet–Tarry–Escott pairs.

1. Introduction

First we define the notion of P -sequence and establish their basic properties which will be used in later sections. In the next section we use the properties of P -sequences in the study of representability of real numbers by sequences of reals. Recall that a real number r is representable by a sequence ⟨a_{n} | n ∈ ℕ⟩ if there is S ⊆ ℕ such that r = ∑_{n∈S} a_{n}.

The following result on representability of real numbers is due to Kakeya [9]: Suppose that A = ⟨a_{n} | n ∈ ℕ⟩ is a decreasing sequence of positive reals which converges to 0 and that s = ∑ a_{n}, 0 < s ⩽ +∞. Then the following assertions are equivalent:

• Each r ∈ (0, s] is representable by means of A;

• a_{n} ⩽ ∑_{k=n+1}^{∞} a_{k}, for each n. Though we do not use Kakeya’s theorem in our proofs, some particular cases of our examples are its consequences. However, the most interesting cases cannot be obtained by it.

Finally, using the P -sequences, we obtain new methods of generating sums of like powers i.e., the Prouhet–Tarry–Escott pairs. For more information about the Prouhet–Tarry–Escott problem see [1, 2, 4, 5, 6, 8].

2. P -sequences

We use symbols ℤ, ℕ, ℕ^{+}, ℝ and ℝ^{+} to represent the sets of integers, nonnegative integers, positive integers, real numbers and positive real numbers, respectively. In addition, we also adopt the convention that 0^{0} = 1.

The notion of a P -sequence is recursively defined as follows:

2000 Mathematics Subject Classification: 11D41, 11B05, 11B75, 11Y50.

• ⟨1, −1⟩ is a P -sequence;

• If ⟨a_{0}, …, a_{k}⟩ is a P -sequence and a_{0} = −a_{k}, then ⟨a_{0}, …, a_{k}, a_{k}, …, a_{0}⟩ is also a P -sequence;

• If ⟨a_{0}, …, a_{k}⟩ is a P -sequence and a_{0} = a_{k}, then ⟨a_{0}, …, a_{k−1}, 0, −a_{k−1}, …, −a_{0}⟩ is also a P -sequence;

• Each P -sequence can be obtained only by finite use of the above clauses. We denote the n-th P -sequence by P_{n} (assuming that they are ordered by their increasing lengths). For instance,

P_{1} = ⟨1, −1⟩, P_{2} = ⟨1, −1, −1, 1⟩, P_{3} = ⟨1, −1, −1, 0, 1, 1, −1⟩ etc.

For an arbitrary positive integer n, the n-th P -sequence P_{n} = ⟨a_{0}, …, a_{k}⟩ and any integer s ⩾ 0 let us define a polynomial function F_{n,s}(x) over ℝ by

F_{n,s}(x) = ∑_{i=0}^{k} a_{i}(i + x)^{s}.

Lemma 1. F_{n,s} ≡ 0 for s = 0, …, n − 1.

Proof. Since

F_{n,s}(x) = ((n − 1) ⋯ (n − 1 − s))^{−1}F_{n,n−1}^{(n−1−s)}(x), 0 ⩽ s < n − 1,

where F_{n,n−1}^{(n−1−s)} is the (n − 1 − s)-th derivative of F_{n,n−1}, it is sufficient to prove that

(1)  F_{n,n−1} ≡ 0.

We prove the lemma by induction on n. Trivially (1) is true for n = 1, so let us assume that for some n ⩾ 1 the equality (1) holds. We have the following two cases:

n = 2m. Assuming that P_{2m} = ⟨a_{0}, …, a_{k}⟩, we have that
P_{2m+1} = ⟨a_{0}, …, a_{k−1}, 0, −a_{k−1}, …, −a_{0}⟩
F_{2m+1,2m}(x) = ∑_{i=0}^{k−1} a_{i}(i + x)^{2m} − ∑_{i=0}^{k−1} a_{i}(2k − i + x)^{2m}

= ∑_{i=0}^{k} a_{i}(i + x)^{2m} − ∑_{i=0}^{k} a_{i}(i − 2k − x)^{2m}

= F_{2m,2m}(x) − F_{2m,2m}(−2k − x).

Then

F′_{2m+1,2m}(x) = 2m \underbrace{F_{2m,2m−1}(x)}_{=0} +2m \underbrace{F_{2m,2m−1}(−2k − x)}_{=0} = 0,

so F_{2m+1,2m} is a constant function. Since F_{2m+1,2m}(−k) = 0, we conclude that F_{2m+1,2m} ≡ 0.

n = 2m+1. Similarly to the previous case one can easily check that F_{2m+2,2m+1} is a constant function. Since F_{2m+2,2m+1}(−(2k + 1)/2) = 0, we conclude that F_{2m+2,2m+1} ≡ 0 as well. □

Theorem 1. For s ⩾ n the degree of F_{n,s} is equal to s − n.

Proof. Clearly, it is sufficient to prove that

F_{n,n} ≡ const. ≠ 0. (2)

Observe that an immediate consequence of Lemma 1 is the fact that each F_{n,n} is a constant function.

The proof goes by induction on n. F_{1,1} ≡ −1, so let us assume that for some n ⩾ 1 the relation (2) holds.

n = 2m. Assuming that P_{2m} = ⟨a_{0}, …, a_{k}⟩, we have

P_{2m+1} = ⟨a_{0}, …, a_{k−1}, 0, −a_{k−1}, …, −a_{0}⟩
F_{2m+1,2m+1}(x) = ∑_{i=0}^{k−1} a_{i}(i + x)^{2m+1} − ∑_{i=0}^{k−1} a_{i}(2k − i + x)^{2m+1}
= F_{2m,2m+1}(x) + F_{2m,2m+1}(−2k − x).

F_{2m+1,2m+1} is a constant function, so

F_{2m+1,2m+1}(x) = F_{2m+1,2m+1}(−k) = 2F_{2m,2m+1}(−k).

Since a_{i} = a_{k−i}, we have F_{2m,2m+1}(−k) = −F_{2m,2m+1}(0). By the induction hypothesis F_{2m,2m+1} is a linear function thus 1–1, hence F_{2m,2m+1}(−k) ≠ 0.

n = 2m + 1. Similarly to the previous case one can easily deduce that

F_{2m+2,2m+2}(x) = 2F_{2m+1,2m+2}(−(2k + 1)/2)
F_{2m+1,2m+2}(−(2k + 1)/2) = −F_{2m+1,2m+2}(1/2),

which combined with the induction hypothesis implies that

F_{2m+1,2m+2}(−(2k + 1)/2) ≠ 0.  □

Corollary 1. Let P_{n} = ⟨a_{0}, …, a_{k}⟩ be a P -sequence. Then:

(1) ∑_{i=1}^{k} a_{i}i^{s} = 0, s = 0, …, n − 1; (2) sgn ∑_{i=1}^{k} a_{i}i^{n} = (−1)^{n}.

Proof. (1) is an immediate consequence of Lemma 1, while (2) can be obtained by induction, using the fact that a_{i} = a_{k−i} for even n, and a_{i} = −a_{k−i} for odd n. □

3. Dense-expandable sequences

Through this section E will denote some denumerable sequence of positive real numbers, E(n) its n-th member, ∑ E =^{def} ∑_{n=0}^{∞} E(n), lim E =^{def} lim_{n→∞} E(n), r + sE =^{def} ⟨r + sE(n) | n ∈ ℕ⟩ and μ_{E} is a measure on ℕ defined by

μ_{E}(S) = ∑_{n∈S} E(n), S ⊆ ℕ.

We say that measure μ_{E} is continuous if for each r ∈ [0, ∞] there is S ⊆ ℕ such that μ_{E}(S) = r.

Sequence E is dense-expandable if the set X(E) =^{def} {∑_{n∈S} ε_{n}E(n) | S ranges over finite subsets of ℕ and ε_{n} ranges over {−1, 1}} is dense in ℝ.

If ∑ E is convergent, then X(E) is bounded in ℝ, so it cannot be dense. On the other hand, if μ_{E} is continuous, then E is obviously dense-expandable.

Theorem 2. Suppose that E is a sequence of positive real numbers such that lim E = 0 and ∑ E = ∞. Then μ_{E} is continuous.

Proof. For the fixed positive real number c let C be the family of all subsets S of ℕ such that μ_{E}(S) ⩽ c. Note that C ≠ ∅ since lim E = 0. ⟨C, ⊆⟩ is clearly a poset, so it has a maximal chain, say M. It is easy to see that ⋃ M ∈ C. Suppose that μ_{E}(⋃ M) = b < c. Since μ_{E}(ℕ) = ∞, ℕ ∖ ⋃ M is infinite. Now lim E = 0 implies that there is an index n ∈ ℕ∖⋃ M such that b+E(n) < c. But this implies that μ_{E}(⋃ M) < μ_{E}({n} ∪ ⋃ M) < c, which contradicts the maximality of M. □

A converse implication need not be true. Namely, the sequence

E = ⟨1, 2, 2^{−1}, 2^{2}, 2^{−2}, 2^{3}, 2^{−3}, …⟩

clearly generates a continuous measure μ_{E}, but it is not convergent. However, its limes inferior \underline{lim} E is equal to 0. Note also that from \underline{lim} E = 0 and ∑ E = ∞ does not follow necessarily the continuity of μ_{E}.

Theorem 3. Let φ(n) be the Euler function. Then A = {φ(n)/n | n ∈ N} is dense in the real interval [0, 1].

Proof. First, we remind the reader that φ(n)/n = (1 − 1/p_{1}) ⋯ (1 − 1/p_{k}), where p_{1}, …, p_{k} are all prime factors of n. Now, let a_{n} = − log(1 − 1/p_{n}), where ⟨p_{n} | n ∈ N⟩ is the sequence of elements of the set of all primes P. Then it is easy to see that a_{n} satisfies conditions of the previous theorem so for any r ∈ R^{+} there is S ⊆ N such that ∑_{n∈S} − log(1 − 1/p_{n}) = r, i.e., ∏_{n∈S}(1 − 1/p_{n}) = e^{−r}. As r runs over R^{+}, e^{−r} takes all values in [0, 1], so for any t ∈ [0, 1], there is S ⊆ P such that ∏_{p∈S}(1 − 1/p) = t. Thus lim_{n∈S} φ(n)/n = t, hence A is dense in [0,1]. □

The next result is useful in studying of dense-expandability.

Theorem 4. Suppose that E is a sequence of positive real numbers such that lim E = 0 and ∑ E = ∞. Then for any nonnegative real number r the sequence r + E is dense-expandable.

Proof. For fixed r ⩾ 0 we want to prove that \overline{X(r + E)} = ℝ. Since x ∈ X(r + E) iff −x ∈ X(r + E), it is sufficient to prove that for any c ⩾ 0 and an arbitrary small ε > 0 open interval (c − ε, c + ε) and X(r + E) meet each other.

The assumed properties of E provide the existence of positive integers n and m > n such that:

(1) E(i) < ε/2, for all i ⩾ n; (2) ∑_{i=n}^{m} E(i) ⩽ c < ∑_{i=n}^{m+1} E(i) < c + ε.

Let q = ∑_{i=n}^{m+1} E(i) − c, l = m − n + 2 and let δ = q/2l. Since lim E = 0, there is an integer k > m such that E(i) < δ for all i ⩾ k. Then:

c + ε > ∑_{i=n}^{m+1}(r + E(i)) − ∑_{i=k}^{k+l−1}(r + E(i))

= ∑_{i=n}^{m+1} E(i) − ∑_{i=k}^{k+l−1} E(i) ⩾ ∑_{i=n}^{m+1} E(i) − \frac{q}{2} > c.

Finally,

∑_{i=n}^{m+1}(r + E(i)) − ∑_{i=k}^{k+l−1}(r + E(i)) ∈ X(r + E),

so X(r + E) is dense in ℝ. □

An immediate consequence of Theorem 4 is the fact that the property of being dense-expandable is not invariant to asymptotic equivalence. For instance, sequences E_{1} = ⟨1 | n ∈ ℕ⟩ and E_{2} = ⟨1 + \frac{1}{n+1} | n ∈ ℕ⟩ are asymptotically equivalent, but \overline{X(E_{1})} = ℤ and \overline{X(E_{2})} = ℝ.

In general, a cofinite subsequence of a dense-expandable sequence E need not be dense expandable. As we have mentioned earlier, the sequence

E = ⟨1, 2, 2^{−1}, 2^{2}, 2^{−2}, 2^{3}, 2^{−3}, …⟩

is dense-expandable, but its cofinite subsequence E_{1} = ⟨E(n + 2) | n ∈ ℕ⟩ is not since X(E_{1}) ∩ (1, 2) = ∅.

The basic strategy in proving that a certain sequence E is dense-expandable is in choosing countably many pairwise disjoint finite subsets S_{n} of ℕ and appropriate ε_{n,i}s such that the sequence ⟨ ∑_{i∈S_{n}} ε_{n,i}E(i) | n ∈ ℕ⟩ satisfies conditions of theorem 4. As an illustration we will prove that sequence E = ⟨ln n | n > 0⟩ is dense-expandable. First, note that the sequence ⟨ln(1 + \frac{1}{2n}) | n > 0⟩ satisfies the conditions of Theorem 4, so it is dense-expandable. The sets S_{n} = {2n, 2n + 1}, n > 0 are pairwise disjoint and ln(1+1/2n) = ln(2n+1)−ln 2n, so X(⟨ln(1+1/2n) | n > 0⟩) ⊆ X(E). Hence E is dense-expandable.

Theorem 5. The sequence ⟨n^{δ} | n ∈ ℕ^{+}⟩ is dense-expandable if and only if δ = −1 or δ > −1 and δ ∉ ℤ.

Proof. If δ < −1, then ∑_{n=1}^{∞} n^{δ} converges, so X(⟨n^{δ} | n ∈ ℕ^{+}⟩) is bounded in ℝ. By Theorem 4 sequence ⟨n^{δ} | n ∈ ℕ^{+}⟩ is dense-expandable for each δ ∈ [−1, 0). If δ is a positive integer, then X(⟨n^{δ} | n ∈ ℕ^{+}⟩) ⊆ ℤ. It remains to prove that ⟨n^{δ} | n ∈ ℕ^{+}⟩ is dense-expandable for any δ ∈ ℝ^{+} ∖ ℤ.

Fix δ ∈ ℝ^{+} ∖ ℤ. Then there is a unique positive integer m such that m − 1 <

δ < m. Let E(n) =^{def} ∑_{i=0}^{k} a_{i}(n − i)^{δ}, n > k, where P_{m} = ⟨a_{0}, …, a_{k}⟩ is the m-th

P -sequence. Then:

E(n) = ∑_{i=0}^{k} a_{i}(n − i)^{δ} = n^{δ} ∑_{i=0}^{k} a_{i}(1 − \frac{i}{n})^{δ}

= n^{δ} ∑_{i=0}^{k} a_{i}(∑_{j=0}^{∞}(−1)^{j}\binom{δ}{j}i^{j}n^{−j}) = n^{δ} ∑_{j=0}^{∞}(−1)^{j}n^{−j}\binom{δ}{j}(∑_{i=0}^{k} a_{i}i^{j})

= n^{δ} ∑_{j=m}^{∞}(−1)^{j}n^{−j}\binom{δ}{j}(∑_{i=0}^{k} a_{i}i^{j}) (Corollary 1)

= (−1)^{m}n^{δ−m}\binom{δ}{m} ∑_{i=0}^{k} a_{i}i^{m} + o(n^{δ−m}).

Since \binom{δ}{m} > 0 and sgn ∑_{i=0}^{k} a_{i}i^{m} = (−1)^{m}, sequence ⟨E(n) | n > k⟩ is ultimately positive. Now −1 < δ − m < 0 implies that ⟨E(n) | n > k⟩ is dense-expandable (Theorem 4). The same is obviously true for the sequence E = ⟨E(kn) | n > k⟩.

Finally, sets S_{n} = {kn − i | i ∈ {0, …, k}}, n > k are pairwise disjoint and each E(kn) is equal to ∑_{j∈S_{n}} ε_{n,j}j^{δ}, where ε_{n,j} are the corresponding coordinates of P_{m}. Thus the sequence ⟨n^{δ} | n > 0⟩ is dense-expandable. □

Theorem 6. Let ⟨p_{n} | n ∈ ℕ⟩ be the sequence of all prime numbers and let E = ⟨p_{n}^{δ} | n ∈ ℕ⟩. Then:

(1) The Riemann hypothesis implies that E is dense-expandable for any 0 < δ < 1/2;

(2) Hypothesis lim_{n→∞}(\sqrt{p_{n+1}} − \sqrt{p_{n}}) = 0 implies that E is dense-expandable for any 0 < δ ⩽ 1/2.

Proof. In order to prove (1), assume the Riemann hypothesis. Then, the following relation holds for the consecutive primes:

p_{n+1} − p_{n} ≪ \sqrt{p_{n}} log p_{n} (3)

Suppose that 0 < δ < 1/2. Then

p_{n+1}^{δ} − p_{n}^{δ} = p_{n+1}^{δ} [1 − (1 − \frac{p_{n+1} − p_{n}}{p_{n+1}} )^{δ}]

= p_{n+1}^{δ} [1 − (1 − \binom{δ}{1}\frac{p_{n+1} − p_{n}}{p_{n+1}} + O(\frac{p_{n+1} − p_{n}}{p_{n+1}} ))]

= δ \frac{p_{n+1} − p_{n}}{p_{n+1}^{1−δ}} + O(\frac{p_{n+1} − p_{n}}{p_{n+1}^{1−δ}} ) ≪ \frac{\sqrt{p_{n}} log p_{n}}{p_{n+1}^{1−δ}} + o(1) → 0

as n → ∞ (see [3]). Taking u_{n} = p_{n+1}^{δ} − p_{n}^{δ}, n ∈ N^{+}, we see that for 0 < δ < 1/2,

∑_{k⩽n} u_{k} = p_{n+1}^{δ} − 2^{δ} → ∞, as n → ∞,

u_{n} > 0 and lim_{n} u_{n} = 0. Thus, by Theorem 4 E is dense-expandable.

In order to prove (2), let us assume that lim_{n→∞} (\sqrt{p_{n+1}} − \sqrt{p_{n}}) = 0. If f(δ) = x^{δ} − y^{δ}, 0 < δ and y < x, then f′(δ) = x^{δ} ln(x) − y^{δ} ln(y) > 0, so f(δ) is increasing for δ > 0. Hence, if 0 < δ ⩽ 1/2, then 0 ⩽ p_{n+1}^{δ}−p_{n}^{δ} ⩽ \sqrt{p_{n+1}}−\sqrt{p_{n}} → 0, as n → ∞, so by an argument as in (1), the assertion follows. □

4. Sums of like powers

Finite disjoint subsets U and V of ℤ will be called a Prouhet-Tarry-Escott pair for the given integer n > 1 if they have the same cardinality and

(4)  ∑_{u∈U} u^{s} = ∑_{v∈V} v^{s}, s = 0, …, n − 1, and ∑_{u∈U} u^{n} ≠ ∑_{v∈V} v^{n}.

The sums satisfying the left-hand conjunct of (4) are also known as sums of like powers.

If ⟨U_{1}, V_{1}⟩, …, ⟨U_{m}, V_{m}⟩ are Prouhet-Tarry-Escott pairs for the given integer n and if sets U_{1}, …, U_{m}, V_{1}, …, V_{m} are pairwise disjoint, then clearly sets U = ⋃_{i=1}^{m} U_{i} and V = ⋃_{i=1}^{m} V_{i} form another Prouhet-Tary-Escott pair for n.

Now let us describe how one can use the P -sequences in order to generate the Prouhet-Tarry-Escott pairs:

1. Let n ⩾ 2 be an arbitrary integer and let P_{n} = ⟨a_{0}, …, a_{k}⟩ be the n-th P -sequence. By lemma 1 we have that

∑_{i=0}^{k} a_{i}(pi + l)^{s} = 0, s = 0, …, n − 1, l ∈ ℤ, p ∈ ℤ ∖ {0}

(observe that ∑ i = 0^{k}a_{i}(pi + l)^{s} = p^{s}F_{n,s}(l/p)). Since each P -sequence has the same number of 1s and −1s, we have that sets U_{p,l} and V_{p,l} defined by

U_{p,l} = {pi + l | 0 ⩽ i ⩽ k ∧ a_{i} = −1} and V_{p,l} = {pi + l | 0 ⩽ i ⩽ k ∧ a_{i} = 1}

form a Prouhet-Tarry-Escott pair for the given integer n ⩾ 2. Note that

∑_{i=0}^{k} a_{i}(pi + l)^{n} ≠ 0,

since ∑_{i=0}^{k} a_{i}(pi + l)^{n} = p^{n}F_{n,n}(1/p) (see Theorem 1).

2. Let P_{n} = ⟨a_{0}, …, a_{k}⟩ be the n-th P -sequence (n ⩾ 2). We define the sequence Q_{n} = ⟨b_{0}, …, b_{k+2}⟩ as follows:

b_{i} = \begin{cases}a_{i}, & i ∈ \{0, 1\} \\ a_{i} + a_{i−2}, & 1 < i < k − 1 \\ a_{i−2}, & i ∈ \{k + 1, k + 2\}. \end{cases}

For example, we obtain Q_{3} from P_{3} = ⟨1, −1, −1, 0, 1, 1, −1⟩ in the following manner:

P_{3}  1 −1 −1 0 1 1 −1
1 −1 −1 0 1 1 −1
Q_{3}  1 −1 0 −1 0 1 0 1 −1

By induction one can show that each b_{i} ∈ {−1, 0, 1} and that each Q_{n} has the same number of 1s and −1s. Now for any non-negative integer s < n we have that

∑_{i=0}^{k} b_{i}(i + 1)^{s} = ∑_{i=0}^{k} a_{i}(i + 1)^{s} + ∑_{i=0}^{k} a_{i}(i + 3)^{s} = 0

∑_{i=0}^{k} b_{i}(i + 1)^{n} = ∑_{i=0}^{k} a_{i}(i + 1)^{n} + ∑_{i=0}^{k} a_{i}(i + 3)^{n}
= F_{n,n}(1) + F_{n,n}(3) = 2F_{n,n}(1) ≠ 0,

so U = {i+1 | b_{i} = 1} and V = {i+1 | b_{i} = −1} represents a Prouhet–Tarry–Escott pair.

3. For the n-th P -sequence P_{n} = ⟨a_{0}, …, a_{k}⟩ let
X_{n} = {i ∈ ℕ^{+} | i ⩽ k ∧ a_{i} = −1} and Y_{n} = {i ∈ ℕ^{+} | i ⩽ k ∧ a_{i} = 1}.

Clearly, X_{n} and Y_{n} are disjoint and |X_{n}| = |Y_{n}| + 1. Furthermore, using the definition of the notion of a P -sequence one can easily check that X_{2n+1} ⊂ X_{2n+2} and Y_{2n+1} ⊂ Y_{2n+2}, and the sets U = X_{2n+2} ∖ X_{2n+1} and V = Y_{2n+2} ∖ Y_{2n+1} are disjoint and have the same cardinality. Bearing in mind the corollary 1, we see that for each nonnegative integer s ⩽ 2n holds

∑_{i∈U} i^{s} = ∑_{i∈X_{2n+2}} i^{s} − ∑_{i∈X_{2n+1}} i^{s} = ∑_{i∈Y_{2n+2}} i^{s} − ∑_{i∈Y_{2n+1}} i^{s} = ∑_{i∈V} i^{s}.

For instance, if n = 4, then X_{4} = {1, 2, 6, 7, 11, 12}, Y_{4} = {4, 5, 8, 9, 13}, U = {7, 11, 12}, V = {8, 9, 13} and

7^{s} + 11^{s} + 12^{s} = 8^{s} + 9^{s} + 13^{s}, s = 1, 2.

References

[1] P. Borwein, C. Ingalls, The Prouhet–Tarry–Escott problem revisited, Enseign. Math. 40 (1994), 3–27

[2] P. Borwein, P. Lisonek, C. Percival, Computational investigations of the Prouhet–Tarry–Escott problem, Math. Comp. 72, 244 (2002), 2063–2070

[3] H. Cramer, Some theorems concerning prime numbers, Arch. Math. Astr. Fysik 15:5 (1920), 1–33

[4] L. E. Dickson, History of the Theory of Numbers, Carnegie Institute, Washington, 1920.

[5] A. Gloden, Mehrgradige Glaichungen, Noordhoff, Groningen, 1944, 103pp.

[6] R. K. Guy, Unsolved Problems in Number Theory, Sections A8 (Gaps between primes. Twin primes) and D1 (Sums of like powers. Euler’s conjecture), second edition, Springer-Verlag, New York, 1994

[7] A. Ivić, An Introduction to Analytic Number Theory, Zoran Stojanović, Novi Sad, 1996, 389pp.

[8] W. Narkiewicz, Classical Problems in Number Theory, PWN, Warszawa, 1986, 363pp.

[9] P. Ribenboim, My Numbers, My Friends, Springer-Verlag, New York, 2000, 307pp.

Matematički fakultet

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    zarkom@eunet.yu, mionamil@eunet.yu, pera@sf.bg.ac.yu