AN ADDITIVE PROBLEM WITH FRACTIONAL POWERS

S. V. Konyagin

Abstract. For any α ∈ (1, 3/2) a sufficiently large positive ineteger x is representable in a form x = [U^{α}] + [V^{α}] with positive integers U and V.

J.-M. Deshouillers[1] proved that for any α ∈ (1, 4/3) a sufficiently large positive ineteger x is representable in a form x = [U^{α}] + [V^{α}] with nonnegative integers U and V. S. Gritsenko[2] extended the range of α possessing this property to α ∈ (1, 55/41). M. Z. Garaev and Ka-Lam Kueh[3] showed that for any α ∈ (1, 3/2] and N ≥ 2 there are ≫ N numbers x ≤ N representable as x = [U^{α}] + [V^{α}] with positive integers U and V.

In the present paper we prove the following result.

Theorem. For any α ∈ (1, 3/2) a sufficiently large positive ineteger x is representable in a form x = [U^{α}] + [V^{α}] with positive integers U and V.

Proof. Take

β = (3/2 − α)/2.

For positive integers k and n with n/2 ≤ k ≤ n ≤ x^{β} define positive numbers u_{0} = u_{0}(n, k) and v_{0} = v_{0}(n, k) in such a way that

(1)  u_{0}^{α} + v_{0}α = x and nu_{0}^{α−1} = kv_{0}^{α−1}.

So,

(2) u_{0} = \frac{k^{1/(α−1)}}{L} x^{1/α}, v_{0} = \frac{n^{1/(α−1)}}{L} x^{1/α}, L = (k^{α/(α−1)} + n^{α/(α−1)})^{1/α} .

Note that

(3)  u_{0} ≍ v_{0} ≍ x^{1/α}.

Denote u = [u_{0}], v = [v_{0}]. By (3), we have

0 ≤ u_{0}^{α} − u^{α} ≪ u_{0}^{α−1} ≪ x^{(α−1)/α},
0 ≤ v_{0}^{α} − v^{α} ≪ x^{(α−1)/α}.

Therefore,

(4)  0 ≤ x − u^{α} − v^{α} ≪ x^{(α−1)/α}.

Typeset by AMS-TEX

Our aim is to show that for suitable n, k, positive integers w, U = u−wn, V = v+wk we have

(5)  x = [U^{α}] + [V^{α}].

We consider the function

f(w) = (u − wn)^{α} + (v + wk)^{α}, 0 ≤ w ≤ u/n.

Taking into account (3), we have

(u − wn)^{α} = u^{α} − αu^{α−1}wn + \frac{α(α − 1)}{2}u^{α−2}(wn)^{2} + O(u^{α−3}(wn^{3}))

= u^{α} − αu_{0}^{α−1}wn + O(u_{0}^{α−2}wn) + \frac{α(α − 1)}{2}u^{α−2}(wn)^{2} + O(u^{α−3}(wn^{3}))

= u^{α} − αu_{0}^{α−1}wn + \frac{α(α − 1)}{2}u^{α−2}(wn)^{2} + O(x^{(α−2)/α}wn)

+O(x^{(α−3)/α}(wn)^{3}). (6)

By the similar way,

(v + wk)^{α} = v^{α} + αv_{0}^{α−1}wk + \frac{α(α − 1)}{2}v^{α−2}(wk)^{2} + O(x^{(α−2)/α}wn)

+O(x^{(α−3)/α}(wn)^{3}). (7)

Here we have used that k ≍ n. Taking the sum of (6) and (7) and using (1) we get

f(w) = u^{α} + v^{α} + \frac{α(α − 1)}{2}w^{2}(u^{α−2}n^{2} + v^{α−2}k^{2}) + O(x^{(α−2)/α}wn)

+O(x^{(α−3)/α}(wn)^{3}). (8)

Now, take wn = Ax^{1/(2α)} with big A. We find from (8)

f(w) ≥ u^{α} + v^{α} + \frac{α(α − 1)}{2}A^{2}u^{α−2}x^{1/α} + O(x^{(α−3/2)/α}A^{3}).

By (3), there is a constant c = c(α) > 0 such that \frac{α(α−1)}{2}u^{α−2} ≥ cx^{(α−2)/α}. Hence,

f(w) ≥ u^{α} +v^{α} +cA^{2}x^{(α−1)/α} +O(x^{(α−3/2)/α}A^{3}) = u^{α} +v^{α} +(cA^{2} +o(1))x^{(α−1)/α}.

Now we can see from (4) that if we fix a big A then for sufficiently large n the inequality f(w) > x + 1 holds and, also, f(0) < x + 1. Thus, there is w_{0} > 0 such that

(9)  (u − w_{0}n)^{α} + (v + w_{0}k)^{α} = f(w_{0}) = x + 1 and w_{0}n ≪ x^{1/(2α)}.

To proceed further, we need the following estimate.

AN ADDITIVE PROBLEM WITH FRACTIONAL POWERS 3

Lemma 1. Let z_{0} > 0, z_{1} ≥ z_{0}/2, z_{2} ≥ z_{0}/2. Then

|z_{1}^{α} − z_{2}^{α} − α(z_{1} − z_{2})z_{0}^{α−1}| ≪ |z_{1} − z_{2}| max_{z∈[z_{1},z_{2}]} |z − z_{0}|z_{0}^{α−2}.

Proof. We have

z_{1}^{α} − z_{2}^{α} = ∫_{z_{2}}^{z_{1}} αz^{α−1}dz.

Therefore,

|z_{1}^{α} − z_{2}^{α} − α(z_{1} − z_{2})z_{0}^{α−1}| ≤ ∫_{z_{2}}^{z_{1}} |αz^{α−1} − αz_{0}^{α−1}||dz|

≤ |z_{1} − z_{2}| max_{z∈[z_{1},z_{2}]} α|z^{α−1} − z_{0}^{α−1}|.

Note that, since 0 < α − 1 < 1, we have for z ∈ [z_{1}, z_{2}]

|z^{α−1} − z_{0}^{α−1}| ≤ (α − 1)|z − z_{0}|(min(z, z_{0}))^{α−2} ≪ |z − z_{0}|z_{0}^{α−2},

and the lemma follows.

Denote w_{1} = [w_{0}], w_{2} = [w_{0}] + 1. We can write several useful inequalities based on Lemma 1. Lemma 2. There exists ε = ε(α, x) such that

ε ≪ nx^{(α−3/2)/α}

and the following inequalities hold:

(10)  |(u − w_{j}n)^{α} − (u − w_{0}n)^{α} − α(w_{0} − w_{j})nu_{0}^{α−1}| ≤ ε (j = 1, 2),
(11)  |(v + w_{j}k)^{α} − (v + w_{0}k)^{α} − α(w_{j} − w_{0})kv_{0}^{α−1}| ≤ ε (j = 1, 2).

The proof is straightforward and follows immediately from Lemma 1 and (9). Note that, by the choice of n,

(12)  ε ≪ x^{γ}, γ = (α − 3/2)(2 − α)/(2α) < 0.

The following lemma is the crucial part of the proof of the theorem. Lemma 3. If ∥αnu_{0}^{α−1}∥ > 6ε then either [(u − w_{1}n)^{α}] + [(v + w_{1}k)^{α}] = x or [(u − w_{2}n)^{α}] + [(v + w_{2}k)^{α}] = x. Proof. Taking the sum of (10) and (11) we have, by (1) and (9),

(13)  |(u − w_{j}n)^{α} + (v + w_{j}k)^{α} − x − 1| ≤ 2ε (j = 1, 2).

Consider the case if for some j = 1, 2 we have ∥(u − w_{j}n)^{α}∥ > 2ε. Then 2ε < {(u − w_{j}n)^{α}} + {(v + w_{j}k)^{α}} < 2 − 2ε.

Hence, by (13),

x − 1 < [(u − w_{j}n)^{α}] + [(v + w_{j}k)^{α}] < x + 1,

or [(u − w_{j}n)^{α}] + [(v + w_{j}k)^{α}] = x. Thus, the conclusion of the lemma holds unless

(14)  ∥(u − w_{1}n)^{α}∥ ≤ 2ε, ∥(u − w_{2}n)^{α}∥ ≤ 2ε.

It suffices to study the case if (14) holds and, therefore,

(15)  ∥(u − w_{1}n)^{α} − (u − w_{2}n)^{α}∥ ≤ 4ε.

On the other hand, by (10),

|(u − w_{1}n)^{α} − (u − w_{2}n)^{α} − αnu_{0}^{α−1}| ≤ 2ε.

Thus, ∥αnu_{0}^{α−1}∥ ≤ 6ε. This completes the proof of the lemma.

Take n = [x^{β}]. To prove the theorem, it suffices to show the existence of an integer k ∈ [n/2, n] such that

(16)  ∥αnu_{0}^{α−1}∥ > 6ε,

where ε satisfies (12). Let

g(u) = \frac{αu}{(1 + u^{α/(α−1)})^{(α−1)/α}} .

By (2), the inequality (16) can be rewritten as

(17)  ∥nx^{(α−1)/α}g(k/n)∥ > 6ε.

Fix a positive integer l so that (l−1)β > (α−1)/α and an interval [u_{1}, u_{2}] ⊂ (1/2, 1) so that u_{2} > u_{1} and g^{(j)}(u) ≠ 0 for u ∈ [u_{1}, u_{2}], j = 1, …, l. Then for the exponential sum

S = ∑_{u_{1}n≤k≤u_{2}n} e (nx^{(α−1)/α}g(k/n))

we have from [1, Theorem 2.9]

|S| ≪ n(n^{−l}F)^{1/(2^{l}−2)} + nF^{−1},

where F = nx^{(α−1)/α}. Since F → ∞ and, by the choice of l, n^{−l}F → 0 as n → ∞, we get S = o(n) as n → ∞. Therefore, for a large n there exists k ∈ [u_{1}n, u_{2}n] such that

∥nx^{(α−1)/α}g(k/n)∥ > 0.2 > 6ε.

Indeed, if we assume the contrary, then

ℜS ≥ ∑_{u_{1}n≤k≤u_{2}n} cos(0.4π) ≫ n,

but this inequality does not agree with S = o(n).

So, (17) holds. The proof of the theorem is complete.

The author is much thankful to Professor D. Tolev for useful discussions.

References

[1] J.-M. Deshouillers. Un problme binaire en thorie additive. Acta Arith. 25 (1973–1974), 393–403.

[2] S. A. Gritsenko. Three additive problems. Izv. Ross. Akad. Nauk Ser. Mat. 56 (1992), no. 6, 1198–1216; translation in Russian Acad. Sci. Izv. Math. 41 (1993), no. 3, 447–464

[3] M. Z. Garaev and Ka-Lam Kueh. L_{1}-norms of exponential sums and the corresponding additive problem. J. Anal and its Appl. 20 (2001), 999–1006.

[4] S. W. Graham and G. Kolesnik. Van der Corput’s method of exponential sums. London Mathematical Society, London Note Series. 126. Cambridge University Press, 1991.