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Diophantine equation x2 + 2&quot; = yn
LE Maohua
Department of Mathematics, Zhanjiang Teachers College, Zhanjiang 524048, China

Keywords:

exponential Diophantine equation, positive integer solution, Baker's method.

be the sets of integers, positive integers and rational numbers, respectively. The
LET 2, N,
solutions ( x , y , m , n ) of the exponential Diophantine equation
x 2 + 2 I n = y n , X , y , rn, n E Pd, 2 + y , n &gt; 2
(1
are connected with many questions in number theory and combinatorial theory. In the recent
fifty years, there were many papers concerned with the equation written by Ljunggren, Nagell, Brown, Toyoizumi and Cohn. In 1986, ref. [ I ] claimed that all solutions of ( 1 ) had been
determined. However, we have not seen the proof so far. Therefore, the solution of ( 1 ) has
not been found yet. In this note, using Baker's method, we prove the following result.
Theorem. Eq. ( 1 ) has no solutions ( x ,y , m , n ) satisfying 21 m a n d m &gt; 2 .
Since ~ o h n ' used
~ ] algebraic number theory method to prove that eq. ( 1 ) has only the solutions ( x , ~ rn,
, n ) = ( 5 , 3 , 1 , 3 ) and ( 7 , 3 , 5 , 4 ) with 2+m, ~ e ' used
~ ' elementary number
theory method to prove that eq. ( 1 ) has only the solution ( x , y , m , n ) = ( l l , 5 , 2 , 3 ) with
771 = 2. Thus, combining the above-mentioned results, we can determine all the solutions of
( 1 ) as follows.
Corollary. Eq. (1) hasonly thesolutions ( x , ~ m, , n ) = ( 5 , 3 , 1 , 3 ) , ( 7 , 3 , 5 , 4 ) and
(11,5,2,3).

1 Lemmas
Lemma 1 . Let YZ be an odd prime such that n &gt; 3 and n =3(mod 4 ) . Then the cyclotomic polynorrliul ( X&quot; - Y&quot; ) / ( X - Y ) cun be expressed us
x
7
1 - y&quot;
X - Y = ( A ( X , y ) I 2 + n ( B ( X , I'll2,
where A ( X , Y ) , B ( X , Y ) ure homogeneous polynomials of degree ( n - 1)/2 satisfying
A(X, y ) =

( n

1112

I:

,

-

u , ~ ( n - 1 ) / 2Y

E ?N[X,

Y], a , =

-

a(n-1)/2-i, i =O,

1,

11

(a-3liZ

B(X, y ) =

b , ~ ( &quot; - l ) &quot; - jYj EIP\'[X,

Y ] , b,=b(n-I)12-j, j = l , 2,

'

...'

n-1
2 '

n -3

2 '
Proof. See sec. 357 of reference [ 4 ] .
Let a be an algebraic number of degree d with the defining polynomial a o z d + a l z J - ' +
+ u', . Then
J

I

denotes Weil' s height of a , where a l , a 2 , ... , ad are all conjugates of a .
Lemma 2 . Lrt a be an ulgebruic number o j degree d with I a I = 1. Let b
tive integers, und Irt A = blloga

-

,, b2 be posi-

b2x c l . If a is not u root of unity, then we huve

loglA 1 2 - 8 . 8 7 A 2 B 2 , w h r r e A = m a x ( 2 0 , 10.98110ga l + s h ( a ) ) , L3=max(17, 6 / 1 0 ,
5.03+2.35s+slog(b1/68.9+ b 2 / 2 A ) ) uncl s = d / 2 .
Proof. See Theorem 3 of reference [ 5 ] .
Chinese Science Bulletin

Vol .42 No. 18

September 1997

a

Let II be a non-square positive integer, and let u + v l
denote the fundamental solution of Pell' s equation:
u2 - n v 2 = 1, U , v E 2.
(2)
We record some further lemmas that will be needed later. Idernma 3 is proved by elementary
number theory method. Lemmas 4 and 5 are verified by the use of computers. These computations
were done using W H E M I C A E 2 . 2 system on a 486 PC and extended about 150 h.
Lemma 3. Ifeq . ( 2 ) hus a positive integer solution ( u , v ) satisfying 2k 1 a , where k is
u positive integer, then we have 2' / u .
Lemma 4. If D is urr odd prime satisfjring D &lt; 3 816 a n d D ~ 7 ( m o 8d ) , then the f i n damental solution of q . ( 2 ) satisfies 2 1 0 2 2 ~. 1
Lemma 5 . Let 7 1 be an odd prime satisfying n &lt; 43 782 a n d n --7(mod 8 ) , and let k , r
be positive integers sutisfjring- k &lt;(n - 1) / 2 a n d rG33, respectively. Then there are no positive integers u satisfying 1 2&quot;ctg( kx/ n ) - u I &lt; a 2 - n .

2 Proof of Theorem
Let ( x , y , m , n ) be a solution of eq. ( 1) satisfying 2 1 m and m &gt; 2. By the proof of
Theorem 1 of ref. [ 6 ] , n must be an odd prime satisfying n &gt;23 and n=7(mod 8 ) and there
exists an odd positive integer u such that
(3)
y = a2+2&quot;,
and

From ( 3 ) and ( 4 ) , we have

(5)

310, 3 1 7 ~ .
Let
E

= a

-

TI
=,
a - 2m'2

+ 28~1/2

(6)

E

We see from ( 3 ) and ( 6 ) that there exists a real number 8 such that 0 &lt; 8&lt; x/2 and
= JyeO&amp;l,

T

= ,J'-- 0 c l

Ye

(7)

From ( 4 ) , (6) and (7), we get

E.

We find from (7) and (8) that there exist some positive integers t satisfying t
L
&lt; n and I a n - 1 1 3 - I n l o g a - t x F 1 I .
Let A = ) n l o g a - t x c l I . By ( 7 ) and ( 8 ) ,
x
we get
Let a = E /

log*&amp; - log I A I &gt; nlog&amp;.
(9)
Since a satisfies y r 2 - 2( u2 - 2&quot; ) a + y = 0 by (3) and (6), a is not a root of unity and its degree is two. Further, we see from (7) that 1 a I = 1, 1 loga 1 = 20&lt; x and h ( a ) = l o g r y . Furthermore, by (3) and (5 ), we get &gt;3' + 2' = 73. Therefore, by Idernma 2, we get from (9) that
n &lt; 43 782.
(10)
From ( 7 ) and ( 8 ) , we have

Chinese Science Bulletin

Vol. 42 No. 18 September 1997

On combining (11) with ( 7 ) , there exist a positive integer k and a real number 6 such that

and
sin6
s i n s = y'n-~)/2'

(13)

since t g ( 6 / n ) &lt;( t g s ) / n =2&quot;1'2/n3. by ( 7 ) , ( 8 ) and (111, we get from ( 6 ) , ( 7 ) , (12)
and (13)
'X
k~
x
x
1
2&quot;&quot;ctg - - u
z c t g
-ctg - &lt;
n
n
Since 3 1 m by ( 5 ) , on applying Lemma 5, we see from (10) that if m/2&lt;99, then (14) is
impossible. So we have nz /2&gt;102, and by ( 3 ) , we get y &gt;2&quot; 3 2 2 0 4 .On applying Lemma 2
again, we obtain
n &lt; 3 816.
(15)
Since n is an odd prime with n ~7 (mod 8 ) , by Lemma 1, we get from (6) and ( 8 )
( 2 ~ / 2 ~ -) 2ns2 = 1 ,
(16)
where r , s are positive integers. We see from (16) that ( u , v ) = (2&quot;I2r, s ) is a positive integer solution of Pell' s equation :
u 2 - nu2 = 1 , u , v E Z
(17)

,&lt;

I&lt;

3.

with 2&quot;&quot; 1 u . Therefore, by Lemma 3, we see that the fundamental solution u 1 + vl
(17) satisfies
2&quot;/2
Recalling that rn /2&gt;102,
The theorem is proved.

&amp; of

IUI(18)
and applying Lemma 4, we see from (15) that (18) is impossible.

References
I Cao, Z . I;., The etl~zation.r2+ 2&quot;' = y&quot; and Hugh Edgar problem, Chinese Sci . Bull., 1986, 31 ( 7 ) : 555.
2 Cohn, J. H. E . , The Diophantine equation sZ+ 2&quot; y n . , Arch. Math. (Basel), 1992, 59(4): 341.
3 I,?, M . H . , O n the lliophi~ntinrequation a,'+ hv c', J . Chn?lgchun Teachers College (in Chinese), 1985, 2 ( 2 ) : 50.
4 Gauss, C. F. , l)isqui.~itio~rrs
Aritlrmrticut., Fleischer: I,eipzig, 1801.
5 1-aurent, M . , Mignotre, M . , Ncsterenko, Y.. Formes 1inCaires en deux logarithmes et dbterminants d'interpolation, J .
Numbpr Tlzmry, 1995, 5 5 ( 3 ) : 285.
6 l x , M. H . , On the Iliophantinr cquation s2f 2'&quot;=y&quot;, Adv. M a t h . (in Chinese), 1996, 25,(4): 328,

-

d to thank Prof. Xu Guangshan for his valuable discussion on the note. This work was
Acknowledgement The ;luthor w o ~ ~ llike
supported by the National Natural Scicnce Foundation of China (Grant No. 19571069) and the Guangdong Provincial Natural Science Foundation (Grant No. 950750).

Chinese Science Bulletin

Vol .42 No. 18 September 1997   