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(1)

C O L L O Q U I U M M A T H E M A T I C U M

VOL. LXV 1993 FASC. 2

LOWER BOUNDS FOR THE SOLUTIONS IN THE SECOND CASE OF FERMAT’S EQUATION WITH PRIME POWER EXPONENTS

BY

MAOHUA L E (CHANGSHA)

Let p be an odd prime, and let n be a positive integer. Further, let x, y, z be integers satisfying

(1) x

pn

+ y

pn

= z

pn

, p | xyz, 0 < x < y < z, gcd(x, y) = 1 .

Recently, Zhong [2] proved that y > p

3npn−n

/2 and z − x > p

3npn−n−1

/4.

In this note we partly improve the above result as follows:

Theorem. If p ≡ 3 (mod 4), then y > p

6np

n−3n2−2n+3

/2

1/pn

and z − x > p

6npn−3n2−3n+3

/2

1−1/pn

.

P r o o f. It is a well known fact that (1) is impossible for p = 3, so we may assume that p > 3.

We first deal with the case that p | x. Let p

α

k x. Then from (1) we get

(2) z − y = p

αpn−n

x

p0n

,

(3) z

pi

− y

pi

z

pi−1

− y

pi−1

= px

pin

, i = 1, . . . , n ,

where x

0

, x

1

, . . . , x

n

are positive integers satisfying p - x

0

x

1

. . . x

n

and (4) x = p

α

x

0

x

1

. . . x

n

.

For any coprime integers X, Y , by the proof of the Theorem in [1], we find that if p ≡ 3 (mod 4) then (X

p

− Y

p

)/(X − Y ) = A

2

+ pB

2

, where A, B are integers satisfying gcd(A, B) = 1 and A ≡ 0 (mod (X − Y )). Hence, by (3), we have

z

pi

− y

pi

z

pi−1

− y

pi−1

= A

2i

+ pB

2i

= px

pin

, i = 1, . . . , n ,

1991 Mathematics Subject Classification: Primary 11D41.

Supported by the National Natural Science Foundation of China.

(2)

228

M.-H. L E

whence we get

(5) B

2i

+ p  A

i

p



2

= x

pin

, i = 1, . . . , n ,

where A

i

, B

i

(i = 1, . . . , n) are integers satisfying gcd(A

i

, B

i

) = 1 and A

i

≡ 0 (mod (z

pi−1

− y

pi−1

)) , i = 1, . . . , n .

Further, by (2), A

i

/p (i = 1, . . . , n) are integers satisfying

(6) A

i

p ≡ 0 (mod p

αpn−n+i−2

) , i = 1, . . . , n .

Notice that p > 3 and the class number of the imaginary quadratic field Q(

√ −p) is less than p. By an argument similar to the proof of the Theorem in [1], we see from (5) that there exist integers X

i

, Y

i

(i = 1, . . . , n) satisfying (7) x

i

= X

i2

+ pY

i2

, gcd(X

i

, Y

i

) = 1 , i = 1, . . . , n ,

and

(8) B

i

+ A

i

p

√ −p = (X

i

+ Y

i

√ −p)

pn

, i = 1, . . . , n .

From (8), (9) A

i

p = p

n

Y

i

(pn−1)/2

X

j=0

(−1)

j

 p

n

2j + 1



p

j−n

X

ipn−2j−1

Y

i2j

, i = 1, . . . , n . Notice that if p > 3 and j > 0, then j > (log(2j + 1))/ log p and

 p

n

2j + 1



p

j−n

= p

n

− 1 2j

 p

j

2j + 1 ≡ 0 (mod p) .

Since p - x

i

(i = 1, . . . , n), we have p - X

i

(i = 1, . . . , n) by (7), and hence (10) Y

i

≡ 0 (mod p

αpn−2n+i−2

) , i = 1, . . . , n ,

by (6) and (9). Since x

i

> 1 (i = 1, . . . , n) , we have Y

i

6= 0 (i = 1, . . . , n) by (7). Thus, we obtain

x

i

> p

2αpn−4n+2i−3

, i = 1, . . . , n , by (7) and (10), and hence

(11) x > p

α+Σni=1(2αpn−4n+2i−3)

= p

2αnpn−3n2−2n+α

by (4). Notice that α ≥ 3 by [2]. We get x > p

6npn−3n2−2n+3

by (11).

Using the same method, we can prove that y > p

6npn−3n2−2n+3

and

z > p

6npn−3n2−2n+3

correspond to p | y and p | z respectively. Thus, y >

(3)

FERMAT’S EQUATION

229

p

6npn−3n2−2n+3

/2

1/pn

since 2

1/pn

y > z. Simultaneously, we have

z − x = y

pn

z

pn−1

+ xz

pn−2

+ . . . + x

pn−1

> y

pn

p

n

z

pn−1

> y

pn

p

n

(2

1/pn

y)

pn−1

= y

2

1−1/pn

p

n

> p

6npn−3n2−3n+3

/2

1−1/pn

. The theorem is proved.

REFERENCES

[1] M.-H. L e, Lower bounds for the solutions in the second case of Fermat’s last theorem, Proc. Amer. Math. Soc. 111 (1991), 921–923.

[2] C.-X. Z h o n g, On Fermat’s equation with prime power exponents, Acta Arith. 59 (1991), 83–86.

Current address:

RESEARCH DEPARTMENT DEPARTMENT OF MATHEMATICS

CHANGSHA RAILWAY INSTITUTE HUNAN NORMAL UNIVERSITY

CHANGSHA, HUNAN P.O. BOX 410081

P.R. CHINA CHANGSHA, HUNAN

P.R. CHINA

Re¸ cu par la R´ edaction le 10.12.1992

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