Intervals between numbers that are sums of two squares - arXiv

arXiv:1706.07380v1 [math.NT] 22 Jun 2017

Alexander Kalmynin

Intervals between numbers that are sums of two squares Abstract. In this paper, we improve the moment estimates for the gaps between numbers that can be represented as a sum of two squares of integers. We consider certain sum of Bessel functions and prove the upper bound for its weighted mean value. This bound provides estimates for the γ-th moments of gaps for all γ ⩽ 2.

§ 1. Introduction Let S = {s1 < s2 < . . . < sn < . . .} = {1, 2, 4, 5, 8, . . .} ⊂ N be the set of all natural numbers that are expressible as the sum of two squares of integers. One of the classical areas in the research of the properties of this set is the study of the distribution of the gaps between consecutive elements, i.e. the quantity sn+1 − sn or, equivalently, the value distribution of the distance function of S R(x) = min |x − n| n∈S

for x → +∞. The following fact is well known (cf.[3],[1]): Theorem 1. The inequality R(x) ≪ x1/4 holds. √ Proof. Let us note that if f (x) = x − [ x]2 , where [y] is an integer part of the √ √ number y, then 0 ⩽ f (x) ≪ x for x ≫ 1 and x − f (x) = [ x]2 is a square of integer. Therefore, x1/4 ≫ f (f (x)) = x − (x − f (x) + f (x) − f (f (x))) ⩾ 0. But x − f (x) and f (x) − f (f (x)) are squares of integers, so for some integers a and b we have |x − a2 − b2 | ≪ x1/4 , which was to be proved. The author is partially supported by Laboratory of Mirror Symmetry NRU HSE, RF Government grant, ag. пїЅ 14.641.31.0001, the Simons Foundation and the Moebius Contest Foundation for Young Scientists

c ⃝

Alexander Kalmynin,

2017

2

ALEXANDER KALMYNIN

This estimate was probably known to L. Euler and, unfortunately, was not improved after that (it is still unknown if the identity R(x) = o(x1/4 ) is true). Conjecturally, the correct order of growth of R(x) is much smaller: Conjecture 1. For any ε > 0 the inequality R(x) ≪ε xε is true. As for the large values of R(x), by the work [4] of I. Richards, we have the following: Theorem 2. For any ε > 0 there exist infinitely many positive integers x such that  R(x) >

 1 − ε ln x. 4

Remark 1. If we assume that S ∩ [1, x] is a typical trajectory of a Possion point process with intensity #S ∩ [1, x] 1 ≍√ , x ln x then we obtain R(x) ≪ ln x. In addition to the the upper and lower bounds, one can also consider the mean values of our function. The best result of this type is due to C. Hooley [2]: Theorem 3. For any 0 < γ < X

5 3

we have

(sn+1 − sn )γ ≪ x(ln x)(γ−1)/2 .

sn+1 ⩽x

Therefore, for almost all n the inequality sn+1 − sn ≪

√

ln sn holds. More precisely:

Corollary 1. Let g(x) be any tends to infinity. Then the number √ function that x of sn ⩽ x with sn+1 − sn ≫ g(x) ln x is o( √ln ). x The main goal of the present work is to improve Hooley’s theorem, i.e. to prove analogous (but somewhat weaker) estimate for the wider range of values of γ. Theorem 4. For any 1 < γ ⩽ 2 the inequality X

(sn+1 − sn )γ ≪ x(ln x)3/2(γ−1) δ(x, γ)

sn+1 ⩽x

is true, where δ(x, γ) =

( 1, ln x,

if γ < 2; if γ = 2.

3

INTERVALS BETWEEN SUMS OF TWO SQUARES

§ 2. Proof of the main theorem In this section, we will prove Theorem 4 by reducing the original problem to the question about the distribution of values of certain sum of Bessel functions. To do this, we need some lemmas. Let us start with the transformation formula for the theta-function. Lemma 1. Let M be a positive real number. For any real x the equality ϑM (x) :=

2 1 X 2πimx −πm2 /M e−πM (x+n) = √ e e M m∈Z n∈Z

X

is true. Proof. 2 Consider the function g(x) = e−πM x on the real line. It is easy to show that g is a Schwartz function and ϑM (x) =

X

g(x + n).

n∈Z

By the Poisson summation formula, for any x ∈ R we get X

g(x + n) =

n∈Z

X

e2πimx gb(m),

m∈Z

−2πiξx

R

where gb(ξ) = R g(x)e dx is a Fourier transform of our function. On the other hand, it is well known that Z 2 2 1 e−πM x −2πiξx dx = √ e−πξ /M . M R Using this relation, we obtain the required result. With the help of Lemma 1 we will prove the following identity, which will be crucial for the subsequent considerations: Lemma 2. Let M and N be some positive real numbers. Then Z π √ √ 1 I(N, M ) := ϑM ( N sin φ)ϑM ( N cos φ)dφ = 2π −π √ 1 X 1 = r2 (n)J0 (2π N n)e−πn/M =: S(N, M ), M M n⩾0

where J0 (x) =

1 2π

Rπ

eix cos φ dφ is the Bessel function of the first kind of order zero

−π

and r2 (n) is the number of pairs (a, b) of integers such that a2 + b2 = n. Proof. Using lemma 1, we find 1 I(N, M ) = 2π

Z

π

−π

X a∈Z

e

√ 2πia N sin φ −πa2 /M

e

! X b∈Z

√ 2πib N cos φ −πb2 /M

e

e

! dφ.

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ALEXANDER KALMYNIN

Both series are absolutely and uniformly convergent, so we can replace the product of their sums by the double sum and interchange summation and integration. Consequently, I(N, M ) =

1 2πM

2

X

e−(a

+b2 )π/M

Z

π

√

e2πi

N (a sin φ+b cos φ)

dφ.

−π

(a,b)∈Z

Let us now compute the inner integral for all integers a and b. If (a, b) = (0, 0), then the integrand is equal to 1 and so the integral equals 2π. If, in contrast, a2 + b2 ̸= 0, then there exists some θ ∈ [−π, π] such that sin θ = √ Therefore, a sin φ + b cos φ = φ = φ1 + θ, we obtain Z

π

e

√

a b , cos θ = √ . a2 + b2 a2 + b2

a2 + b2 cos(φ − θ). Applying the change of variables

√ 2πi N (a sin φ+b cos φ)

−π

π−θ

Z dφ =

√ 2 2 e2πi N (a +b ) cos φ1 dφ1 .

−π−θ

Since the integrand is periodic with the period 2π, the last integral is equal to the integral of the same function over the interval [−π, π]. Thus, we finally find Z

π

√ N (a sin φ+b cos φ)

e2πi

Z

√ 2 2 p e2πi N (a +b ) cos φ1 dφ1 = 2πJ0 (2π N (a2 + b2 )).

π

dφ =

−π

−π

Substituting the obtained result into the formula for I(N, M ), we deduce the identity I(N, M ) =

1 M

X (a,b)∈Z

J0 (2π

p 2 2 1 S(N, M ), N (a2 + b2 ))e−π(a +b )/M = M

which was to be proved. It turns out that if R(N ) is large, then the quantity I(N, M ) is rather small. Lemma 3. If N and M are positive real numbers and N ⩾ 36, then 2

I(N, M ) ⩽ e−πM R(N )

/5N

+ O(e−πM/4 )

Proof. Let us observe that if N ⩾ 36 and for some φ ∈ [−π, π] we have √ √ || N sin φ||2 + || N cos φ||2 = c, where ||x|| is the distance from x to the nearest integer, then c ⩾ Indeed, there exist some integers a and b such that √ √ ( N sin φ − a)2 + ( N cos φ − b)2 = c √ √ and | N sin φ − a| < 1, | N cos φ − b| < 1. Consequently,

R(N )2 5N .

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INTERVALS BETWEEN SUMS OF TWO SQUARES

√ √ √ √ |N − a2 − b2 | = |( N sin φ − a)( N sin φ + a) + ( N cos φ − b)( N cos φ + b)|. On the other hand, √ √ ( N sin φ − a)2 + ( N cos φ − b)2 = c and √ √ √ √ ( N sin φ + a)2 + ( N cos φ + b)2 ⩽ (2 N sin φ + 1)2 + (2 N cos φ + 1)2 = √ √ √ = 4N + 4 N (sin φ + cos φ) + 2 ⩽ 4N + 4 2 N + 2 ⩽ 5N. Next, using the Cauchy-Bunyakovsky-Schwartz inequality, we find R(N ) ⩽ |N − a2 − b2 | ⩽

√ 5cN ,

as needed. Furthermore, it is easy to see that for any x ∈ R the identity 2

ϑM (x) = e−πM ||x|| + O(e−πM/4 ) holds. Therefore, I(N, M ) =

1 2π

Z

π

√

e−πM (||

√ N sin φ||2 +|| N cos φ||2 )

dφ + O(e−πM/4 ).

−π

√ √ But, as we showed before, for any φ we have || N sin φ||2 + || N cos φ||2 ⩾ Thus, 2

I(N, M ) ⩽ e−πM R(N )

/5N

R(N )2 5N .

+ O(e−πM/4 ),

which was to be proved. So, if R(N ) is sufficiently large, then the quantity I(N, M ) is close to 0. Therefore, the same is true for the sum S(N, M ). For arbitrary x, we split the sum S(x, M ) into several parts: X

S(x, M ) = 1 +

Sc (x, M ) + S∞ (x, M ; N ),

0⩽c⩽ln N

where for c ⩾ 0 we have √ r2 (n)J0 (2π xn)e−πn/M

X

Sc (x, M ) =

cM M ln N

The last sum is easily estimated in terms of N and M :

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ALEXANDER KALMYNIN

Lemma 4. If N, M ⩾ 2, then for all x ⩾ 0 the inequality |S∞ (x, M ; N )| ≪

M N3

is true. Proof. Due to the definition of Bessel function, for any real y we have |J0 (y)| ⩽ 1. Consequently, X

|S∞ (x, M ; N )| ⩽

r2 (n)e−πn/M .

n>M ln N

It is clear that the last sum equals ∞ X

Sc∗ (N, M ),

c=0

where X

Sc∗ (N, M ) =

r2 (n)e−πn/M .

M ln N +cM 0 we have X

r2 (n) ≪ y,

0