An upper bound for the number of solutions of a system of congruences

(1)

LXVI.4 (1994)

An upper bound for the number of solutions of a system of congruences

by

Lyn Dodd (Nottingham)

1. Introduction. Let X denote an indeterminate. For each vector m ∈ Z

^2r

with components satisfying 0 < m

i

≤ h define

f

₁

(X) = Y

r i=1

(X + m

_i

) and f

₂

(X) = Y

2r i=r+1

(X + m

_i

) .

In [1] Burgess showed that, for any prime p > 3 and primitive character χ (mod p

^α

), the estimate

N +H

X

n=N +1

χ(n)

= O(H

^1−1/r

p

^α(r+1)/4r²^+ε

)

holds in the case r = 3. This inequality was obtained by estimating X

m

X

x∈A1

χ

f

₁

f

₂

(x)

where

A

₁

= {x : 0 ≤ x < p

^α

, p - f

₁

(x)f

₂

(x)} .

In order to do this, Burgess found estimates for the inner summation over various subsets of A

₁

and then counted the number of m for which these subsets were non-empty. The counting process was carried out using different methods, one of which concerned the estimation of the cardinality of

S = {m : 0 < m

_i

≤ h, f

₁

(X) ≡ f

₂

(X) (mod p

^µ

)} .

The estimation of such character sums in the case r = 2 is contained in [2].

The case r = 4 has yet to be proved. This paper estimates #S when r = 4, as a step in the direction of a proof. The result obtained is given by the following theorem.

The contents of this paper formed part of the author’s PhD thesis (Nottingham Uni-

versity, 1991) which was supported by Science and Engineering Research Council.

(2)

Theorem 1. Suppose p is a prime greater than 5 and µ is a positive integer. If

H = {m : 0 < m

_i

≤ h for i = 1, . . . , 8 and f

₁

(X) ≡ f

₂

(X) (mod p

^µ

)}

then

#H µ

⁶

h

⁸

p

3µ+[µ/2]−[µ/4]

‘ + h

⁶

p

µ+[µ/2]−[µ/4]

+ h

⁵

p

^µ−[µ/2]

+ h

⁴

. In [4] Hua and Min obtain an asymptotic formula for the number of solutions of the system

x

^h₁

+ . . . + x

^h_s

≡ y

₁^h

+ . . . + y

^h_s

(mod p

^l

) (1 ≤ h ≤ k)

where s, k, h, l are integers such that s ≥ k ≥ 4, l ≥ k

²

and p is a prime greater than k. Assuming that p 6= 5 and letting s

r

(x) = P

₄

i=1

x

^r_i

, in the particular case s = k = 4, l = 16 the number of solutions of the system

(1)

s

1

(x) ≡ s

1

(y) s

₂

(x) ≡ s

₂

(y) s

3

(x) ≡ s

3

(y) s

₄

(x) ≡ s

₄

(y)

 

 



 

 

(mod p

¹⁶

)

is p

⁷⁶

(1 + O(p

^−1/4

)). Writing

σ

₁

(x) = x

₁

+ x

₂

+ x

₃

+ x

₄

,

σ

₂

(x) = x

₁

x

₂

+ x

₁

x

₃

+ x

₁

x

₄

+ x

₂

x

₃

+ x

₂

x

₄

+ x

₃

x

₄

, σ

3

(x) = x

1

x

2

x

3

+ x

1

x

2

x

4

+ x

1

x

3

x

4

+ x

2

x

3

x

4

, σ

₄

(x) = x

₁

x

₂

x

₃

x

₄

,

it follows that

s

₁

(x) = σ

₁

(x) ,

s

₂

(x) = (σ

₁

(x))

²

− 2σ

₂

(x) ,

s

3

(x) = (σ

1

(x))

³

− 3σ

1

(x)σ

2

(x) + 3σ

3

(x) , s

₄

(x) = (σ

₁

(x))

⁴

− 4(σ

₁

(x))

²

σ

₂

(x)

+ 4σ

1

(x)σ

3

(x) + 2(σ

2

(x))

²

− 4σ

4

(x) . Since p > 5 the systems (1) and

(2)

σ

1

(x) ≡ σ

1

(y) σ

₂

(x) ≡ σ

₂

(y) σ

3

(x) ≡ σ

3

(y) σ

₄

(x) ≡ σ

₄

(y)

 

 



 

 

(mod p

¹⁶

)

(3)

are equivalent. But (2) holds if and only if Y

4

i=1

(X + x

_i

) ≡ Y

4 i=1

(X + y

_i

) (mod p

¹⁶

)

for indeterminate X, which, by Theorem 1, has p

⁷⁶

solutions in one complete system of residues. A comparison with the result of Min and Hua shows that, in this case, Theorem 1 is essentially best possible.

2. Basic estimates. The basic tools used in proving Theorem 1 are the well-known estimate in Lemma 2 and Proposition 3 which is reproduced from [3]. The notation dxe denotes the least integer greater than or equal to x and p

^α

k x means p

^α

| x, p

^α+1

- x.

Lemma 2. Suppose p is an odd prime and ν is a positive integer. If 0 < x ≤ h then the number of solutions of the congruence x

²

+ Ax + B ≡ 0 (mod p

^ν

) is h/p

^[(ν+1)/2]

+ 1.

Proposition 3. Let f be a polynomial of degree n having integer coefficients. Let p be a prime, d be a positive integer , and α, β and γ be non-negative integers satisfying γ = dα/de. If T = {x ∈ a complete set of residues (mod p

^γ

) : p

^α+β

| f (x), p

^β

k f

^(d)

(x)} then #T n.

If g(x) is a polynomial with integer coefficients such that p

^δ

k g

^(d)

(x) then it follows from Proposition 3 that the number of x satisfying 0 < x ≤ h and g(x) ≡ 0 (mod p

^µ

) is h/p

^µ−δ

+ 1 if d = 1 and h/p

[(µ−δ+1)/2]

+ 1 if d = 2. The proof of Theorem 1 will be given by a series of lemmas.

Throughout we shall use the fact that the conditions A

1

+ . . . + A

n

≡ 0 (mod p

^α

) and p

^a^j

k A

_j

for j = 1, . . . , n imply that a

_k

≥ min(min

_j6=k

a

_j

, α) for k = 1, . . . , n.

3. Initial transformations. Making the substitution M

_i

= m

_i

− m

₁

for i = 2, . . . , 8 we see that f

1

(X) ≡ f

2

(X) (mod p

^µ

) if and only if the following congruences hold simultaneously:

(3) M

₂

+ M

₃

+ M

₄

≡ M

₅

+ M

₆

+ M

₇

+ M

₈

(mod p

^µ

) , M

2

M

3

+ M

2

M

4

+ M

3

M

4

≡ M

₅

M

₆

+ M

₅

M

₇

+ M

₅

M

₈

+ M

₆

M

₇

+ M

₆

M

₈

+ M

₇

M

₈

(mod p

^µ

) , M

2

M

3

M

4

≡ M

5

M

6

M

7

+ M

5

M

6

M

8

+ M

5

M

7

M

8

+ M

6

M

7

M

8

(mod p

^µ

) , (4) 0 ≡ M

₅

M

₆

M

₇

M

₈

(mod p

^µ

) .

Eliminating M

₂

from the second and third congruences of the above system produces the pair of congruences

(5) (M

₃

+ M

₄

)(M

₅

+ M

₆

+ M

₇

+ M

₈

− M

₄

) − M

₃²

(4)

≡ M

5

M

6

+ M

5

M

7

+ M

5

M

8

+ M

6

M

7

+ M

6

M

8

+ M

7

M

8

(mod p

^µ

) and

M

₃

M

₄

(M

₅

+ M

₆

+ M

₇

+ M

₈

− M

₃

− M

₄

)

≡ M

5

M

6

M

7

+ M

5

M

6

M

8

+ M

5

M

7

M

8

+ M

6

M

7

M

8

(mod p

^µ

) , which together imply that

(6) (M

₄

− M

₈

)(M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

+ M

₄²

− M

₄

(M

₅

+ M

₆

+ M

₇

))

≡ M

5

M

6

M

7

(mod p

^µ

) . Define γ

5

, γ

6

, γ

7

, γ

8

by

p

^γ⁵

k (M

₅

, p

^µ

), p

^γ⁵^+γ⁶

k (M

₅

M

₆

, p

^µ

) , (7)

p

^γ⁵^+γ⁶^+γ⁷

k (M

₅

M

₆

M

₇

, p

^µ

), γ

₅

+ γ

₆

+ γ

₇

+ γ

₈

= µ .

It may be assumed that γ

₅

≥ γ

₆

≥ γ

₇

≥ γ

₈

≥ 0, by re-ordering M

₅

, M

₆

, M

₇

, M

8

if necessary, and that for i = 5, 6, 7 any power of p dividing M

i+1

also divides M

_i

. Let ε, k, m be given by

p

^ε

k (M

5

M

6

+ M

5

M

7

+ M

6

M

7

+ M

₄²

− M

4

(M

5

+ M

6

+ M

7

), p

^µ

) , (8)

p

^k

k (2M

₃

+ M

₄

− M

₅

− M

₆

− M

₇

, p

^µ

) (9)

and

(10) p

^m

k (2M

₄

− M

₅

− M

₆

− M

₇

, p

^µ

) .

Writing M

1

= m

1

it follows that the number of m = (m

1

, . . . , m

8

) satisfying f

₁

(X) ≡ f

₂

(X) (mod p

^µ

) is less than or equal to the number of solutions in M

₁

, . . . , M

₈

of (3)–(6) with |M

_i

| < h for i = 1, . . . , 8. We now present the lemmas which together provide the proof of Theorem 1. In all cases there are h possible values for m

₁

. Given M

₃

, . . . , M

₈

there are h/p

^µ

+1 choices for M

2

from (3). We have h/p

^{max(µ−k,k)}

+ 1 choices for M

3

from (5) and (9), given M

₄

, . . . , M

₈

. The following notation will be used:

A = h

⁸

p

3µ+[µ/2]−[µ/4]

+ h

⁶

p

µ+[µ/2]−[µ/4]

+ h

⁵

p

^µ−[µ/2]

+ h

⁴

and

S = “M

₁

, . . . , M

₈

: |M

_i

| < h for i = 1, . . . , 8 and (3)–(10) hold”.

4. Extending the conditions S. In this section we obtain the required estimate except for the set {S : 0 < m < µ − [µ/4], ε < µ, [µ/4] < k ≤ [µ/2], γ

₈

> 0 and p | M

₄

}.

Lemma 4. If H

1

= {S : m = 0} then #H

1

µ

⁵

A. P r o o f. Given M

4

, M

5

, M

6

, M

7

there are h/p

^µ−ε

+ 1 choices for M

8

from (6) and (8). Since m = 0 there are only non-singular solutions for M

₄

(5)

from (8) and so we have h/p

^ε

+ 1 choices for M

4

given M

5

, M

6

, M

7

. By (7) it follows that

#H

1

h

p

^µ

+ 1 X

k

h

p

^{max(µ−k,k)}

+ 1

× X

γ₅,γ₆,γ₇

h p

^γ⁵

+1

h p

^γ⁶

+1

h

p

^γ⁷

+1 X

ε

h p

^ε

+1

h p

^µ−ε

+1

µ

⁵

A .

Lemma 5. If H

₂

= {S : m > 0 and p - M

₄

} then #H

₂

µ

⁴

A. P r o o f. Given M

₄

, M

₅

, M

₆

, M

₇

there are h/p

^µ−ε

+ 1 choices for M

₈

from (6) and (8). By (8) and (10) there are h/p

^{max(ε−m,m)}

+ 1 choices for M

4

given M

5

, M

6

, M

7

. Since p - M

4

, (7) and (10) imply that p - M

₇

and so γ

₇

= 0 and γ

₅

+ γ

₆

= µ. From (10) it can be seen that 2M

₄

= M

₅

+ M

₆

+ M

₇

+ Rp

^m

for some R ∈ Z. Substituting for M

₄

in (8) produces

4(M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

) ≡ (M

₅

+ M

₆

+ M

₇

)

²

(mod p

^min(2m,ε)

) , from which we have h/p

min(m,[(ε+1)/2])

+ 1 choices for M

₇

, given M

₅

, M

₆

. Therefore

#H

₂

h

p

^µ

+ 1 X

k

h

p

^{max(µ−k,k)}

+ 1

× X

ε,m

h

p

^{max(ε−m,m)}

+ 1

h

p

min(m,[(ε+1)/2])

+ 1

×

h

p

^µ−ε

+ 1 X

γ5

h p

^γ⁵

+ 1

h p

^µ−γ⁵

+ 1

µ

⁴

A .

Multiplying (5) by 4 and then rearranging yields

(2M

3

− Σ

1

+ M

4

)

²

≡ 4M

4

(Σ

1

− M

4

) − 4Σ

2

+ (Σ

1

− M

4

)

²

(mod p

^µ

) where Σ

₁

= M

₅

+ M

₆

+ M

₇

+ M

₈

and Σ

₂

= M

₅

M

₆

+ M

₅

M

₇

+ M

₅

M

₈

+ M

6

M

7

+ M

6

M

8

+ M

7

M

8

. Hence it follows from (9) that

p

^min(2k,µ)

| 4M

₄

(Σ

₁

− M

₄

) − 4Σ

₂

+ (Σ

₁

− M

₄

)

²

and so

(11) (M

₄

+ M

₈

− M

₅

− M

₆

− M

₇

)

²

≡ 4(M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

+ M

₄²

− M

₄

(M

₅

+ M

₆

+ M

₇

)) (mod p

^min(2k,µ)

).

(6)

Lemma 6. If H

3

= {S : m > 0, p | M

4

and ε = µ} then #H

3

µ

³

A. P r o o f. From (6) and (8) we know that M

₅

M

₆

M

₇

≡ 0 (mod p

^µ

) and so, by (7), γ

5

+ γ

6

+ γ

7

= µ. Using (8) and (11) we obtain

M

₄

+ M

₈

− M

₅

− M

₆

− M

₇

≡ 0 (mod p

min(k,[(µ+1)/2])

) ,

from which we have h/p

min(k,[(µ+1)/2])

+ 1 choices for M

₈

given M

₄

, M

₅

, M

6

, M

7

. There are h/p

^[(µ+1)/2]

+1 choices for M

4

from (8), given M

5

, M

6

, M

₇

. Thus

#H

₃

h

h p

^µ

+ 1

h

p

^µ−[µ/2]

+ 1

× X

γ5,γ6

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

h

p

^µ−γ⁵^−γ⁶

+ 1

× X

k

h

p

^{max(µ−k,k)}

+ 1

h

p

min(k,[(µ+1)/2])

+ 1

µ

³

A .

Lemma 7. If H

₄

= {S : k > [µ/2], m > 0, ε < µ and p | M

₄

} then

#H

4

µ

⁵

A. P r o o f. As ε < µ it follows from (8) that p

^ε

k RHS and thus p

^ε

k LHS of (11). Hence ε must be even and p

^ε/2

k M

4

+M

8

−M

5

−M

6

−M

7

. Together with (11) this gives h/p

^µ−ε/2

+ 1 choices for M

₈

given M

₄

, M

₅

, M

₆

, M

₇

. There are h/p

^ε/2

+ 1 choices for M

4

from (8), given M

5

, M

6

, M

7

. There- fore

#H

4

h

p

^µ

+ 1 X

k

h

p

^k

+ 1 X

γ5,γ6,γ7

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

×

h

p

^γ⁷

+ 1 X

ε

h p

^ε/2

+ 1

h p

^µ−ε/2

+ 1

µ

⁵

A .

Lemma 8. If H

₅

= {S : 0 ≤ k ≤ [µ/4], m > 0, ε < µ and p | M

₄

} then

#H

₅

µ

⁵

A. P r o o f. There are h/p

^µ−ε

+ 1 choices for M

₈

from (6) and (8), given

M

4

, M

5

, M

6

, M

7

, and h/p

^{max(ε−m,m)}

+1 choices for M

4

from (8) and (10),

given M

₅

, M

₆

, M

₇

. As in Lemma 5 we have h/p

min(m,[(ε+1)/2])

+ 1 choices

(7)

for M

7

, given M

5

, M

6

, and thus

#H

₅

h

p

^µ

+ 1 X

k

h p

^µ−k

+ 1

× X

ε,m

h

p

^{max(ε−m,m)}

+ 1

h

p

min(m,[(ε+1)/2])

+ 1

×

h

p

^µ−ε

+ 1 X

γ5,γ6

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

µ

⁵

A .

Lemma 9. If H

₆

= {S : [µ/4] < k ≤ [µ/2], m > 0, ε < µ, γ

₈

= 0 and p | M

4

} then #H

6

µ

⁴

A. P r o o f. Given M

₄

, M

₅

, M

₆

, M

₇

we have h/p

^{max(µ−ε,k)}

+ 1 choices for M

₈

from (6), (8) and (11). From (8) there are h/p

^[(ε+1)/2]

+ 1 choices for M

4

given M

5

, M

6

, M

7

. Hence, using (7),

#H

₆

h

p

^µ

+ 1 X

γ5,γ6

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

h

p

^µ−γ⁵^−γ⁶

+ 1

× X

ε,k

h

p

^[(ε+1)/2]

+ 1

h p

^µ−k

+ 1

h

p

^{max(µ−ε,k)}

+ 1

µ

⁴

A .

Lemma 10. If H

₇

= {S : [µ/4] < k ≤ [µ/2], m ≥ µ − [µ/4], ε < µ, γ

₈

> 0 and p | M

4

} then #H

7

µ

⁵

A. P r o o f. Given M

₅

, M

₆

, M

₇

, M

₈

there are h/p

^m

+ 1 choices for M

₄

from (10). By (7) it follows that

#H

7

h

p

^µ

+1 X

γ₅,γ₆,γ₇

h p

^γ⁵

+1

h p

^γ⁶

+1

h p

^γ⁷

+1

h

p

^µ−γ⁵^−γ⁶^−γ⁷

+1

× X

k

h

p

^µ−k

+ 1 X

m

h p

^m

+ 1

µ

⁵

A .

The conditions S have now been extended as follows:

S

⁰

= “S : 0 < m < µ−[µ/4], ε < µ, [µ/4] < k ≤ [µ/2], γ

8

> 0 and p | M

4

” .

This notation will be used in the next section.

(8)

5. Extending the conditions S

⁰

. In all the remaining cases we obtain an expression of the form

#H

i

h

p

^µ

+ 1 X h

⁶

p

^D^a

+ h

⁵

p

^D^b

+ h

⁴

p

^D^c

+ h

³

,

where the sum is over a maximum of six variables. It is sufficient to show that D

a

≥ 2µ + [µ/2] − [µ/4], D

b

≥ µ + [µ/2] − [µ/4] and D

c

≥ µ − [µ/2]

since h

h p

^µ

+ 1

h

⁶

p

2µ+[µ/2]−[µ/4]

+ h

⁵

p

µ+[µ/2]−[µ/4]

+ h

⁴

p

^µ−[µ/2]

+ h

³

A . We now continue with further steps in the proof of Theorem 1.

Lemma 11. If H

₈

= {S

⁰

: 2k ≤ ε} then #H

₈

µ

⁶

A. P r o o f. From (8) and (11) we know that p

^k

| M

₄

+ M

₈

− M

₅

− M

₆

− M

₇

. By (10), this implies that p

^min(m,k)

| M

₄

− M

₈

. Hence, by (6) and (7), it follows that

(12) ε + min(m, k) ≤ γ

₅

+ γ

₆

+ γ

₇

< µ .

There are h/p

^µ−ε

+ 1 choices for M

₈

from (6) and (8), given M

₄

, M

₅

, M

₆

, M

7

, and h/p

^{max(ε−m,m)}

+ 1 choices for M

4

from (8) and (10), given M

5

, M

₆

, M

₇

. Therefore

#H

8

h

p

^µ

+ 1 X

k,ε,m

h p

^µ−k

+ 1

h p

^µ−ε

+ 1

h

p

^{max(ε−m,m)}

+ 1

× X

γ5,γ6,γ7

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

h p

^γ⁷

+ 1

h

p

^µ

+ 1 X

k,ε,m γ₅,γ₆,γ₇

h

⁶

p

^D¹

+ h

⁵

p

^D²

+ h

⁴

p

^D³

+ h

³

,

where

D

1

= 2µ − k − ε + max(ε − m, m) + γ

5

+ γ

6

+ γ

7

,

D

₂

= min(2µ − k − ε + max(ε − m, m), 2µ − k − ε + γ

₅

+ γ

₆

+ γ

₇

, µ − ε + max(ε − m, m) + γ

5

+ γ

6

+ γ

7

) , D

₃

= min(µ − ε + max(ε − m, m), µ − ε + γ

₅

+ γ

₆

+ γ

₇

, 2µ − k − ε ,

γ

₅

+ γ

₆

+ γ

₇

+ max(ε − m, m)) .

It can be seen from (12) that D

1

≥ 2µ − k + max(ε − m, m) + min(m, k) ≥

2µ+[µ/2]−[µ/4]. By (7) we know that γ

₅

+γ

₆

+γ

₇

≥ µ−[µ/4] > µ−k. Also,

if max(ε−m, m) > γ

5

+γ

6

+γ

7

then ε−m > γ

5

+γ

6

+γ

7

. This is possible only

if m < [µ/4], in which case (12) implies that ε + m ≤ γ

₅

+ γ

₆

+ γ

₇

< ε − m,

(9)

a contradiction. We conclude that max(ε − m, m) ≤ γ

5

+ γ

6

+ γ

7

. Hence D

₃

= min(µ − ε + max(ε − m, m), 2µ − k − ε) ≥ µ − [ε/2] ≥ µ − [µ/2] and D

₂

= 2µ − k − ε + max(ε − m, m). If ε > [µ/2] + [µ/4] then (12) implies that m ≤ [µ/4] and so D

2

= 2µ − k − m ≥ 2µ − [µ/2] − [µ/4]. If ε ≤ [µ/2] + [µ/4]

then, as k ≤ max(ε − m, m), we have D

₂

≥ 2µ − ε ≥ 2µ − [µ/2] − [µ/4].

It may now be assumed that µ ≥ 2k > ε. Consequently, it follows from (8) and (11) that ε is even and

(13) p

^ε/2

k M

₄

+ M

₈

− M

₅

− M

₆

− M

₇

. Denote by Q the expression

(14) Q = M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

+ M

₄²

− M

₄

(M

₅

+ M

₆

+ M

₇

) . Lemma 12. If H

₉

= {S

⁰

: 2k > ε, 2m > ε} then #H

₉

µ

⁵

A. P r o o f. Given M

4

, M

5

, M

6

, M

7

there are h/p

max(µ−ε,2k−ε/2)

+ 1 choices for M

₈

from (6), (8), (11) and (13). By (10), p

^2m

k (2M

₄

− M

₅

− M

₆

− M

₇

)

²

and so

4Q ≡ 2(M

5

M

6

+ M

5

M

7

+ M

6

M

7

) − M

₅²

− M

₆²

− M

₇²

(mod p

^2m

) . As p

^ε

k Q we deduce that

(15) p

^ε

k M

₅²

+ M

₆²

+ M

₇²

− 2(M

5

M

6

+ M

5

M

7

+ M

6

M

7

) .

It follows from (10) and (13) that p

^ε/2

| M

₄

− M

₈

, which, together with (6) and (7), implies that

(16) 3ε/2 ≤ γ

₅

+ γ

₆

+ γ

₇

< µ . Thus from (6) we obtain

Q

p

^ε

· M

4

− M

8

p

^ε/2

≡ M

5

M

6

M

7

p

^3ε/2

(mod p

^µ−3ε/2

) and from (11) and (13) we have

Q

p

^ε

· M

₄

+ M

₈

− M

₅

− M

₆

− M

₇

p

^ε/2

₂

≡ 4Q

³

p

^3ε

(mod p

^2k−ε

) .

Since M

4

+ M

8

− M

5

− M

6

− M

7

= 2M

4

− M

5

− M

6

− M

7

− (M

4

− M

8

), combining the above two congruences produces

Q(2M

₄

− M

₅

− M

₆

− M

₇

) − M

₅

M

₆

M

₇

p

^3ε/2

₂

≡ 4Q

³

p

^3ε

(mod p

min(µ−3ε/2, 2k−ε)

), which simplifies to

(17) Q

²

(M

₅²

+ M

₆²

+ M

₇²

− 2(M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

))

≡ M

₅

M

₆

M

₇

(2Q(2M

₄

−M

₅

−M

₆

−M

₇

)−M

₅

M

₆

M

₇

) (mod p

min(µ+3ε/2, 2k+2ε)

).

(10)

From (15) it can be seen that p

^3ε

k LHS of (17). By (16) we must have min(µ + 3ε/2, 2k + 2ε) > 3ε and thus p

^3ε

k RHS of (17), or

p

^3ε−γ⁵^−γ⁶^−γ⁷

k 2Q(2M

₄

− M

₅

− M

₆

− M

₇

) − M

₅

M

₆

M

₇

. This together with (10) implies that

3ε − γ

₅

− γ

₆

− γ

₇

≥ min(ε + m, γ

₅

+ γ

₆

+ γ

₇

) .

If ε + m ≤ γ

5

+ γ

6

+ γ

7

then 2ε − m ≥ γ

5

+ γ

6

+ γ

7

≥ ε + m, contradicting 2m > ε. Hence, from (7), (16) and the above we conclude that

(18) µ −

µ 4

≤ 3ε

2 = γ

5

+ γ

6

+ γ

7

< µ . It follows from (8), (10) and (18) that

p

^5ε/2

k 2Q(2M

₄

−M

₅

−M

₆

−M

₇

)(M

₅²

+M

₆²

+M

₇²

−2(M

₅

M

₆

+M

₅

M

₇

+M

₆

M

₇

))

−2M

₅

M

₆

M

₇

(2Q + (2M

₄

− M

₅

− M

₆

− M

₇

)

²

) . This is the derivative of (17) with respect to M

₄

and so there are h/p

min(µ−ε, 2k−ε/2)

+ 1 choices for M

₄

given M

₅

, M

₆

, M

₇

. By (7) and (18) it follows that

#H

₉

h

p

^µ

+ 1 X

k,ε

h

p

min(µ−ε,2k−ε/2)

+ 1

h

p

+ 1

×

h

p

^µ−k

+ 1 X

γ5,γ6,γ7

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

h p

^γ⁷

+ 1

µ

³

h

p

^µ

+ 1 X

k,ε

h

⁶

p

^D⁴

+ h

⁵

p

^D⁵

+ h

⁴

p

^D⁶

+ h

³

, where

D

₄

= 2µ + k ≥ 2µ +

µ 2

−

µ 4

, D

₅

= min

2µ + k − 3ε

2 , 2µ − k + ε

2 , µ + k + ε

> min(2µ − ε, µ + k) ≥ µ +

µ 2

−

µ 4

, D

6

= min

µ + 2k − 3ε

2 , 2µ − k − ε, µ + k − ε

2 , µ − k + 3ε 2

> µ − k ≥ µ −

µ 2

.

Lemma 13. If H

₁₀

= {S

⁰

: 2k > ε > 2m} then #H

₁₀

µ

⁶

A.

(11)

P r o o f. Given M

4

, M

5

, M

6

, M

7

there are h/p

+ 1 choices for M

₈

from (6), (8), (11) and (13). Since p 6= 2, from (8) and (14) we know that p

^ε

k 4Q. This can be rewritten as

(2M

₄

− M

₅

− M

₆

− M

₇

)

²

≡ M

₅²

+ M

₆²

+ M

₇²

− 2(M

5

M

6

+ M

5

M

7

+ M

6

M

7

) (mod p

^ε

) , which, together with (10), implies that

(19) p

^2m

k M

₅²

+ M

₆²

+ M

₇²

− 2(M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

) .

From (10) and (13) we see that p

^m

| M

4

− M

8

and so from (6) and (7) we have

(20) ε + m ≤ γ

5

+ γ

6

+ γ

7

< µ . Using (6), (11), (13) and (20) we deduce that

Q

p

^ε

· M

4

− M

8

p

^m

≡ M

5

M

6

M

7

p

^ε+m

(mod p

^µ−ε−m

) and

Q

p

^ε

· M

₄

+ M

₈

− M

₅

− M

₆

− M

₇

p

^m

₂

≡ 4Q

³

p

^2ε+2m

(mod p

^2k−2m

) . Combining these two congruences as in the previous lemma, we obtain (21) Q

²

(M

₅²

+ M

₆²

+ M

₇²

− 2(M

5

M

6

+ M

5

M

7

+ M

6

M

7

))

≡ M

₅

M

₆

M

₇

(2Q(2M

₄

−M

₅

−M

₆

−M

₇

)−M

₅

M

₆

M

₇

) (mod p

min(µ+ε+m,2k+2ε)

).

By (8) and (19), p

^2ε+2m

k LHS of (21). But from (20) we know that min(µ+

ε + m, 2k + 2ε) > 2ε + 2m and so p

^2ε+2m

k RHS of (21), or

p

^2ε+2m−γ⁵^−γ⁶^−γ⁷

k 2Q(2M

₄

− M

₅

− M

₆

− M

₇

) − M

₅

M

₆

M

₇

.

Hence, by (10), we see that 2ε+2m− γ

5

−γ

6

−γ

7

≥ min(ε+m, γ

5

+ γ

6

+γ

7

), which, together with (7) and (20), implies that

(22) µ −

µ 4

≤ ε + m = γ

5

+ γ

6

+ γ

7

< µ . Also, from (10), (19) and (22) it can be seen that

p

^4m

k (2(2M

₄

− M

₅

− M

₆

− M

₇

)

²

+ 4Q)(M

₅²

+ M

₆²

+ M

₇²

−2(M

5

M

6

+ M

5

M

7

+ M

6

M

7

)) − 12M

5

M

6

M

7

(2M

4

− M

5

− M

6

− M

7

) .

This expression is the second derivative of (21) with respect to M

4

and

so there are h/p

[(min(µ+ε−3m,2k+2ε−4m)+1)/2]

+ 1 choices for M

₄

, given

(12)

M

5

, M

6

, M

7

. Hence, by (22),

#H

₁₀

h

h p

^µ

+ 1

× X

k,ε,m

h

p

+1

h

p

min([(µ+ε−3m+1)/2],k+ε−2m)

+1

×

h

p

^µ−k

+ 1 X

γ₅,γ₆,γ₇

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

h p

^γ⁷

+ 1

µ

³

h

p

^µ

+ 1 X

k,ε,m

h

⁶

p

^D⁷

+ h

⁵

p

^D⁸

+ h

⁴

p

^D⁹

+ h

³

, where

D

₇

= max

2µ − k, µ + k + ε 2

+ min

µ + ε − m + 1 2

, k + ε − m

≥ 2µ + ε

2 ≥ 2µ +

µ 2

−

µ 4

, D

8

= µ−k +min

max

µ− ε

2 , 2k

+min

µ − 3m + 1 2

, k + ε

2 −2m

, max

µ + m, 2k + ε 2 + m

, ε + min

µ + ε − m + 1 2

, k + ε − m

> µ + k ≥ µ +

µ 2

−

µ 4

, D

₉

= min

µ − k + min

ε + m, max

µ − ε, 2k − ε 2

, min

µ + ε − 3m + 1 2

, k + ε − 2m

, min

µ − 3m + 1 2

, k + ε

2 − 2m

+ max

µ − ε

2 , 2k

> µ − k ≥ µ −

µ 2

.

It may now be assumed that ε = 2m. There are h/p

max(µ−2m,2k−m)

+1 choices for M

8

from (6), (8), (11) and (13) given M

4

, M

5

, M

6

, M

7

.

Lemma 14. If H

11

= {S

⁰

: 2k > ε = 2m and γ

5

+ γ

6

+ γ

7

≥ 2k + m}

then #H

₁₁

µ

⁵

A.

(13)

P r o o f. From (10) there are h/p

^m

+1 choices for M

4

given M

5

, M

6

, M

7

and it follows that

#H

11

h

p

^µ

+ 1 X

k,m

h

p

max(µ−2m,2k−m)

+ 1

h p

^µ−k

+ 1

h p

^m

+ 1

× X

γ5,γ6,γ7

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

h p

^γ⁷

+ 1

µ

³

h

p

^µ

+ 1 X

k,m

h

⁶

p

^D¹⁰

+ h

⁵

p

^D¹¹

+ h

⁴

p

^D¹²

+ h

³

, where

D

₁₀

= µ + k + 2m + max(µ − 2m, 2k − m) ≥ 2µ + k ≥ 2µ +

µ 2

−

µ 4

, D

₁₁

= min(µ + k + 2m, min(µ − k + m, 2k + 2m) + max(µ − 2m, 2k − m))

≥ µ + k ≥ µ +

µ 2

−

µ 4

,

D

₁₂

= min(max(µ − m, 2k), µ − k + m, 2k + 2m)

≥ min(µ − k, 2k) ≥ µ −

µ 2

.

By (10) and (13) we know that p

^m

| M

4

− M

8

and so, by (6), (7) and (8), (23) 3m ≤ γ

₅

+ γ

₆

+ γ

₇

< µ .

Also, with Q given by (14), from (6), (11) and (13) we obtain Q

p

^2m

· M

4

− M

8

p

^m

≡ M

5

M

6

M

7

p

^3m

(mod p

^µ−3m

) and

Q

p

^2m

· M

4

+ M

8

− M

5

− M

6

− M

7

p

^m

₂

≡ 4Q

³

p

^6m

(mod p

^2k−2m

) . Proceeding as in Lemma 12, these two congruences combine to produce (24) Q

²

(M

₅²

+ M

₆²

+ M

₇²

− 2(M

5

M

6

+ M

5

M

7

+ M

6

M

7

))

≡ M

₅

M

₆

M

₇

(2Q(2M

₄

−M

₅

−M

₆

−M

₇

)−M

₅

M

₆

M

₇

) (mod p

min(2k+4m,µ+3m)

) . Define x by

(25) p

^x

k (M

₅²

+ M

₆²

+ M

₇²

− 2(M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

), p

^µ

) .

Lemma 15. If H

12

= {S

⁰

: (25) holds, x ≥ 2k > ε = 2m and 2k + m >

γ

₅

+ γ

₆

+ γ

₇

} then #H

₁₂

µ

⁵

A.

(14)

P r o o f. Since p

^2m

k Q (25) implies that p

min(2k+4m,µ+3m)

| RHS of (24) or, by (7),

4Q(2M

4

−M

5

−M

6

−M

7

) ≡ 2M

5

M

6

M

7

(mod p

min(2k+4m,µ+3m)−γ5−γ6−γ7

) . We know that min(2k + 4m, µ + 3m) − γ

₅

− γ

₆

− γ

₇

> 3m. Since p

^3m

k LHS and p

^γ⁵^+γ⁶^+γ⁷

k RHS of the above we conclude that

(26) µ −

µ 4

≤ 3m = γ

5

+ γ

6

+ γ

7

< µ .

As 4Q = (2M

4

−M

5

−M

6

−M

7

)

²

−M

₅²

−M

₆²

−M

₇²

+2(M

5

M

6

+M

5

M

7

+M

6

M

7

) we can rewrite the above as

(2M

4

− M

5

− M

6

− M

7

)((2M

4

− M

5

− M

6

− M

7

)

²

− M

₅²

− M

₆²

− M

₇²

+ 2(M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

))

≡ 2M

₅

M

₆

M

₇

(mod p

min(2k+m,µ)

) . By (10) and (25) it follows that

p

^2m

k 6(2M

4

− M

5

− M

6

− M

7

)

²

−2(M

₅²

+ M

₆²

+ M

₇²

− 2(M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

)) . This is the derivative of the above expression with respect to M

₄

and so there are h/p

min(2k−m,µ−2m)

+ 1 choices for M

₄

given M

₅

, M

₆

, M

₇

. Therefore

#H

₁₂

h

p

^µ

+ 1 X

k,m

h

p

min(2k−m,µ−2m)

+ 1

×

h

p

max(2k−m,µ−2m)

+ 1

h p

^µ−k

+ 1

× X

γ5,γ6,γ7

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

h p

^γ⁷

+ 1

. As (26) holds, the result follows by comparison with Lemma 12.

Lemma 16. If H

₁₃

= {S

⁰

: (25) holds, 2k > x ≥ µ − m, 2k > ε = 2m and 2k + m > γ

₅

+ γ

₆

+ γ

₇

} then #H

₁₃

µ

⁵

A. P r o o f. There are h/p

^m

+1 choices for M

₄

from (10) given M

₅

, M

₆

, M

₇

and h/p

^[(x+1)/2]

+ 1 choices for M

₇

from (25) given M

₅

, M

₆

. By (7) and (23),

#H

13

h

h p

^µ

+ 1

× X

k,m,x

h p

^µ−k

+ 1

h p

^m

+ 1

h p

^2k−m

+ 1

h

p

^[(x+1)/2]

+ 1

(15)

× X

γ5,γ6

h p

^γ⁵

+ 1

h p

^γ⁶

+ 1

µh

h

p

^µ

+ 1 X

γk,m5,γ6

h

⁶

p

^D¹³

+ h

⁵

p

^D¹⁴

+ h

⁴

p

^D¹⁵

+ h

³

,

where

D

₁₃

= µ + k + γ

₅

+ γ

₆

+

µ − m + 1 2

≥ µ + 2

µ − m + 1 2

+ max

2m, µ −

µ 2

,

D

14

=

µ − m + 1 2

+ min(µ + k, γ

5

+ γ

6

+ min(µ − k + m, 2k))

≥ µ +

µ − m + 1 2

> µ +

µ 2

−

µ 4

,

D

₁₅

= min

µ + k,

µ − m + 1 2

+ min(2k, γ

₅

+ γ

₆

+ m, µ − k + m)

> µ −

µ 2

.

If m > [µ/4] then D

₁₃

≥ 2µ + m ≥ 2µ + [µ/2] − [µ/4] and if m ≤ [µ/4] then D

13

≥ 3µ − m − [µ/2] ≥ 3µ − [µ/2] − [µ/4].

It remains to consider min(2k, µ − m) > x. From (8), (14) and (25) we see that p

^x+4m

k LHS of (24) and thus

p

^x+4m

k M

₅

M

₆

M

₇

(2Q(2M

₄

− M

₅

− M

₆

− M

₇

) − M

₅

M

₆

M

₇

) . This together with (23) implies that x + 4m ≥ γ

5

+ γ

6

+ γ

7

+ 3m, or (27) x ≥ γ

₅

+ γ

₆

+ γ

₇

− m .

We now look at the derivatives of (24) with respect to M

₄

. Define T, U and V by

(28) p

^T

k (Q(2M

₄

− M

₅

− M

₆

− M

₇

)

× (M

₅²

+ M

₆²

+ M

₇²

− 2(M

₅

M

₆

+ M

₅

M

₇

+ M

₆

M

₇

))

− M

5

M

6

M

7

(2Q + (2M

4

− M

5

− M

6

− M

7

)

²

), p

min(2k+4m,µ+3m)

) , (29) p

^U

k ((2Q + (2M

₄

− M

₅

− M

₆

− M

₇

)

²

)

× (M

₅²

+ M

₆²

+ M

₇²

− 2(M

5

M

6

+ M

5

M

7

+ M

6

M

7

))

− 6M

₅

M

₆

M

₇

(2M

₄

− M

₅

− M

₆

− M

₇

), p

min(2k+4m,µ+3m)

)

An upper bound for the number of solutions of a system of congruences

LXVI.4 (1994)