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
6npn−3n2−2n+3
/2
1/pnand 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
αk x. Then from (1) we get
(2) z − y = p
αpn−nx
p0n,
(3) z
pi− y
piz
pi−1− y
pi−1= px
pin, i = 1, . . . , n ,
where x
0, x
1, . . . , x
nare positive integers satisfying p - x
0x
1. . . x
nand (4) x = p
αx
0x
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
piz
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.
228
M.-H. L Ewhence we get
(5) B
2i+ p A
ip
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
ip ≡ 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
ip
√ −p = (X
i+ Y
i√ −p)
pn, i = 1, . . . , n .
From (8), (9) A
ip = p
nY
i(pn−1)/2
X
j=0
(−1)
jp
n2j + 1
p
j−nX
ipn−2j−1Y
i2j, i = 1, . . . , n . Notice that if p > 3 and j > 0, then j > (log(2j + 1))/ log p and
p
n2j + 1
p
j−n= p
n− 1 2j
p
j2j + 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
i6= 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+αby (4). Notice that α ≥ 3 by [2]. We get x > p
6npn−3n2−2n+3by (11).
Using the same method, we can prove that y > p
6npn−3n2−2n+3and
z > p
6npn−3n2−2n+3correspond to p | y and p | z respectively. Thus, y >
FERMAT’S EQUATION
229
p
6npn−3n2−2n+3/2
1/pnsince 2
1/pny > z. Simultaneously, we have
z − x = y
pnz
pn−1+ xz
pn−2+ . . . + x
pn−1> y
pnp
nz
pn−1> y
pnp
n(2
1/pny)
pn−1= y
2
1−1/pnp
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