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# Solving a linear equation in a set of integers II by

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(1)

i

1≤i≤k

1

1

k

k

1

k

1

k

1

k

1

k

1

k

1

k

i

i

[385]

(2)

−α

1

k

1

1

k

k

1

k

0

1

2

3

4

4

(3)

0

1

2

3

4

i

4

−1

−1

−k

4

1

k

1

k

i

i

4

(4)

2b j=1

j

j

1 0

j

1 0

1

k

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

1

k

k

k

k

k

−k

−k

(5)

0

1

i

1

k

0

4

0

i

1

k

1

k

1

k

1

k

1

1

k

k

i

i

i

0

i

i

0

i

i

0

0

0

1

1

2

2

k−1

k

k

k

(6)

k

k−1

i=0

i

i

i

i

i

i

i

i

k

0

+

ai>0

i

ai<0

i

+

+

1

+

+

1

1

k

k

i

1

1

0

1

+

+

(7)

2q+1

2i

i

1

2

1

0

1

2

2

2

3

4

2

1

4

2

3

4

i

0

1

2

3

4

(8)

Z

A∈H

x→∞

0

0

4

0

1

1

k

k

0

0

j

0

0

(9)

0

0

0

i

i

−1

−k

i

0

n∈B

n

n

1 0

0

0

(10)

j1

1

jk

k

ji

k

i=1

ji

1

k

ij

ij

i

j

i

0

i

j

k i=1

ji

0

2

i

ji

y∈A

y

y

(11)

0

1

k

i

i

i

i

i

i

i

1

k

1

k

ji

i

i

i

ji

i

ji

ji

i

i

ji

i

i

ji

ji

i

ji

i

ji

i

1

k

0

0

0

0

y

n∈B

n

n

1

0

n

1 0

1 0

y

(12)

y

y

y

φy6=0

y

2y

2

1 0

1 0

2

1 0

1 0

1 0

1 0

1 0

1 0

2

1

0

1 0

2

2

2

2

2

1 0

1 0

nn0

nn0

0

y

y

0

1 0

nn0

0

0

0

0

1 0

1 0

nn0

0

0

(13)

1 0

## R

1 0

References

### Hungar. 56, 155–158.

MATHEMATICAL INSTITUTE

HUNGARIAN ACADEMY OF SCIENCES BUDAPEST, PF. 127

H-1364 HUNGARY

E-mail: H1140RUZ@ELLA.HU

### and in revised form on 19.12.1994 (2681)

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