N. M. AKULINICHEV

ESTIMATES OF RATIONAL TRIGONOMETRIC SUMS OF A SPECIAL FORM

(Presented by Academician I. M. Vinogradov on 30 X 1964)

An estimate of A. Weil ((^2)) for a complete trigonometric sum with a polynomial with integral coefficients, (p) prime, (n < p), (a_n \not\equiv 0 \pmod p),

[
\left|
\sum_{x=0}^{p-1}
\exp \left[
2\pi i\,\frac{a_n x^n+a_{n-1}x^{n-1}+\cdots+a_1x}{p}
\right]
\right|
\leq (n-1)\sqrt p
]

becomes trivial when (n-1>\sqrt p). In the present note we obtain estimates of rational sums with a polynomial of degree (n) of the form

[
S=\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{Ax^n+Bx}{p}\right],
\qquad
S=\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{Ax^n+Bx^k}{p}\right],
]

which are nontrivial for (n>\sqrt p).

Theorem 1. Let (p \geq 3) be prime; (A, B, n) natural numbers; (p>n); (\delta=(n,p-1)); ((A,p)=1); ((B,p)=1). The inequality

[
\left|
\sum_{x=0}^{p-1}
\exp\left[2\pi i\,\frac{Ax^n+Bx}{p}\right]
\right|
\leq \frac{p}{\sqrt{\delta}}
]

holds.

Proof. Denote by (Y) the set of all elements (y) of the reduced residue system modulo (p) for which (y^n\equiv 1 \pmod p). The set (Y) contains (\delta) elements. For each fixed (y\in Y), if (x) runs through a complete residue system modulo (p), then (xy) also runs through a complete residue system modulo (p). Consequently,

[
S=\sum_{x=0}^{p-1}
\exp\left[2\pi i\,\frac{Ax^n+Bx}{p}\right]
=
\sum_{x=0}^{p-1}
\exp\left[2\pi i\,\frac{A(xy)^n+Bxy}{p}\right]
=
]

[

\sum_{x=0}^{p-1}
\exp\left[2\pi i\,\frac{Ax^n+Bxy}{p}\right]
=S_y.
]

Summing over all (y\in Y), we obtain

[
\delta S=
\sum_{y\in Y}\sum_{x=0}^{p-1}
\exp\left[2\pi i\,\frac{Ax^n+Bxy}{p}\right].
]

Squaring the inequality

[
\delta |S|
\leq
\sum_{x=0}^{p-1}
\left|
\sum_{y\in Y}
\exp\left[2\pi i\,\frac{Bxy}{p}\right]
\right|
]

and applying the Cauchy–Bunyakovsky inequality, we obtain

[
\delta^2 |S|^2
\leq
\sum_{x=0}^{p-1}
\left|
\sum_{y\in Y}
\exp\left[2\pi i\,\frac{Bxy}{p}\right]
\right|^2
=
]

[

p\sum_{y_1\in Y}\sum_{y_2\in Y}\sum_{x=0}^{p-1}
\exp\left[2\pi i\,\frac{B(y_1-y_2)x}{p}\right]
=
p^2
\sum_{\substack{y_1\equiv y_2\;(\mathrm{mod}\ p)\ y_1,y_2\in Y}}
1
=
p^2\delta,
]

i.e.

[
|S|\leq p/\sqrt{\delta}.
]

Corollary. Let (p) be a prime number, (n\mid p-1), ((A,p)=1), ((B,p)=1).
Then

[
\left|\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{Ax^n+Bx}{p}\right]\right|\ll p^{5/6}.
]

Proof. Multiplying the squared estimate of Theorem 1 by A. Weil’s estimate

[
\left|\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{Ax^n+Bx}{p}\right]\right|\le (n-1)\sqrt p,
]

we obtain

[
|S|^2|S|\ll \frac{p^2}{n}\,np^{1/2}=p^{5/2}
]

or

[
|S|\ll p^{5/6}.
]

When the two estimates are multiplied, they lose part of their strength; however, since the left-hand side of the resulting upper bound does not depend on (n), the result has an attractive form.

Denote by (B(T)) the number of solutions of the congruence (Ax^n+Bx\equiv y \pmod p) in the numbers (0\le x\le p-1,\;0\le y\le T-1).

Theorem 2. Let ((A,p)=1,\;(B,p)=1,\;n\mid p-1).
Then

[
B(T)=T+\theta p^{5/6}\ln p,\qquad |\theta|\le 1.
]

Proof. Representing the number (B(T)) in the form

[
B(T)=\sum_{y=0}^{T-1}\sum_{x=0}^{p-1}\frac1p\sum_{a=0}^{p-1}
\exp\left[2\pi ia\,\frac{Ax^n+Bx-y}{p}\right]
]

and separating the terms with (a=0), we obtain

[
B(T)=T+\frac1p\sum_{a=1}^{p-1}
\left(\sum_{y=0}^{T-1}\exp\left[-2\pi i\,\frac{ay}{p}\right]\right)
\left(\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{Ax^n+Bx}{p}\right]\right).
]

From the known estimate

[
\left|\sum_{a=1}^{p-1}\left(\sum_{y=0}^{T-1}\exp\left[-2\pi i\,\frac{ay}{p}\right]\right)\right|<p\ln p
]

and the estimate of the corollary to Theorem 1, the assertion of Theorem 2 follows.

Theorem 3. Let (p) be a prime number; (a,\ b,\ k,\ n) natural numbers; (n\mid p-1); ((n,k)=1); (\delta=(k,p-1)); ((a,p)=1); ((b,p)=1).
Then

[
\left|\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{ax^n+bx^k}{p}\right]\right|
\ll \frac{p}{\sqrt n}+\sqrt{\delta-1}\,p^{3/4}.
]

Proof. Choose the set (Y) to be the same as in Theorem 1. Similarly to the proof of Theorem 1, we have

[
\left|\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{a x^n+b x^k}{p}\right]\right|^2
\leq
\frac{p}{n^2}\sum_{y_1\in Y}\sum_{y_2\in Y}\sum_{x=0}^{p-1}
\exp\left[2\pi i\,\frac{b x^k\left(y_1^k-y_2^k\right)}{p}\right]
=
]

[

\frac{p^2}{n}+\frac{p}{n^2}
\sum_{\substack{y_1,y_2\in Y\ y_1\ne y_2}}
\sum_{x=0}^{p-1}
\exp\left[2\pi i\,\frac{b\left(y_1^k-y_2^k\right)}{p}x^k\right].
]

Since, by the conditions of the theorem, (y_1^k\not\equiv y_2^k\pmod p) for (y_1\ne y_2), the estimate of I. M. Vinogradov is applicable to the sum

[
\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{b\left(y_1^k-y_2^k\right)x^k}{p}\right]
]

[
\left|\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{b\left(y_1^k-y_2^k\right)}{p}x^k\right]\right|
\leq
(\delta-1)\sqrt p.
]

Consequently,

[
\left|\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{a x^n+b x^k}{p}\right]\right|^2
\leq
\frac{p^2}{n}+\frac{p}{n^2}\,n(n-1)(\delta-1)\sqrt p.
]

From the inequality (\sqrt{a'+b'}\leq \sqrt{a'}+\sqrt{b'}) (for (a'\geq 0,\ b'\geq 0)) Theorem 3 follows.

In the case ((k,p-1)=1) we obtain the estimate

[
\left|\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{a x^n+b x^k}{p}\right]\right|
\leq
\frac{p}{\sqrt n}.
]

This estimate is stronger than A. Weil’s estimate for (n>p^{1/3}). The estimate of Theorem 3 is stronger than A. Weil’s estimate for

[
n>\max\left((1/2p)^{1/3},\ (1/2p)^{1/4}\sqrt{\delta-1}\right).
]

The method used in proving the estimates can be applied to deriving a variety of other inequalities. For example, the following assertion is true.

Theorem 4. Let ((a,p)=(b,p)=(c,p)=1;\ n_1\mid p-1;\ n_2\mid p-1;\ (n_1,n_2)=1;\ (k,p-1)=1;\ (k,n_1)=1).

Then

[
\left|\sum_{x=0}^{p-1}\exp\left[2\pi i\,\frac{a x^{n_1}+b x^{n_2}+c x^k}{p}\right]\right|
\leq
\sqrt 2\,\frac{p}{\sqrt[4]{n_2}}.
]

I express my gratitude to Academician I. M. Vinogradov for his help in the writing of this paper.

Received
28 IX 1964

REFERENCES

1. I. M. Vinogradov, Selected Works, Moscow, 1952.
2. A. Weil, Proc. Nat. Acad. Sci. U.S.A., 34, 204 (1948).