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K R Y S T Y N A B I A L E K A N D A L E K S A N D E R G R Y T C Z U K
T H E E Q U A T I O N O F F E R M A T IN G2C k ) A N D Q j V l T ]
1. I N T R O D U C T I O N
Let- G ^ C k ) b e t h e set. of m a t r i c e s of t h e form
[ l ; ]
w h e r e k is f i x e d i n t e g e r s u c h t h a t k ^ 0 and a r e a r b i t r a r y i n t e g e r s .
T h e p u r p o s e o f t h i s p a p e r is to g i v e a c o n n e c t i o n b e t w e e n t h e s o l u t i o n of F e r m a t e q u a t i o n in G^Ck> a n d t h e s o l u t i o n o f t h i s e q u a t i o n in Q ^V' lc j .
S o m e p a r t i a l r e s u l t s c o n c e r n i n g a b o v e p r o b l e m a r e g i v e n in LI], 12 J, C d ] (comp, tS3).
We p r o v e t h e f o l l o w i n g t h e o r e m s :
T H E O R E M 1.
T h e n e c e s s a r y and s u f f i c i e n t c o n d i t i o n f a r t h o e q u a t i o n C2> An + Bn = Gn ,
Cn ^ 2J) t o h a v e a s o l u t i o n in e l e m e n t s A , B , C <£ G ^ C k } i s t h e e x i s t e n c e o f t h e n u m b e r s a, ß, y e Q k j s u c h t h a t
C 3 ) an + ßn = r r' .
T H E O R E M 2.
Let K be a n u m b e r f i e l d . If a , b , c e IC a n d a2 m + b2 m = c2 m
w i t h m p o s i t i v e i n t e g e r t h e n
A4 m + B4 t o = C *w , where A,B,C are m a t r i c e s of the form
A - [ a Í ]' » " [ 2 O l C = [ c
S ]
2. L E M M A SIn the proofs of the t h e o r e m s we can use the f o l l o w i n g lemmas:
L E M M A 1 I f
I r s y m f R S I L k s r J [ k S R J for some n £ 2 then
C 4 ) R = \ [ ( r + s V l T jn + (r-s-/lT]n ] , and
1
CS) s
2 vir
[ ( r + s V í T j " - (r-s-/"kT]n ] .
PROOF
In c a s e n=2 the L e m m a can be seen directly and ana c a n c o m p l e t e the p r o o f by m a t h e m a t i c a l induction on n.
L E M M A 2 If
A
• [ c
aS ]
with integers a,b,c,d, then for e v e r y integer n 2: 2
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_ f f Ca) byj 1
~ [ c v f C d ) J»
where yj is an integer,
f C a ) - f Cd) = Ca - d ) yj
and fCa), f C d ) are p o l y n o m i a l s of degree n.
P R O O F
For n=2 we have
A =
a +bc b C a + d ) c C a + d ) d2+ b c
f C a ) by cyj f Cd)
where y = a+d . It is easy to verify that f C a ) - f C d ) = C a - d ) C a + d ) = C a - d ) yj .
Assume that the Lemma is true for n=k, C k £ 2 ) that is
Ak =
f t C a ) by>
cv, f C d ) l l
and f j C a ) - f C d ) = C a - d ) ^
First we have
Ak + 1 - Ak A =
f 1C a ) cy
b yi%
f tC d )
a b c d
f2C a ) b ^2
cyT f Cd)
where
f2C a ) = af 1 C a ) + be yjx , f2Cd;> = df1cd;) + fac H* x
C 7 )
= f C a ) + d w 2 i
On the other hand
^2 = 3 f l + flC d : >
C 8 ) A k + 1 = A Ak =
a b c d
flC a ) b yt cyj± f <d>
a f tC a ) + b (ay^-i-f ±C d ) j c ( d y ^ + f±C a ) J df iC d ) + b c v1
k L.
Comparing the e n t r i e s of A A and A A we obtain f ( a ) + dv, = av« + f Cd),
l i l l hence by C 7 ) we get
v2 = y*2- From C 7 ) it f o l l o w s that
f Ca) - f C d ) = af C a ) - df C d )
2 2 I 1
but
f C a ) - f Cd) + Ca-d)»/' -
1 I 1
f2C a ) - f2C d ) = a jf % < d H C a - d ) ^ j - d f ± C d ) = C a - d ) [f} C d J + a ^ j . T h u s
From C 7 ) we have
ftC d ) - a ^ - - V/2>
thus
f C a ) - f C d ) = Ca—d)V-'
2 2 2
what e n d s the proof.
LEMMA 3,
If a matrix
A =
a b c d
rith nS:2 and i n t e g e r s a,b,c,d s a t i s f i e s R S
k S Fi
where k is fixed integer such that. kffl and integers, then
A e G Ck).
2
E , S ^ O a r o
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P R O O F
From the a s s u m p t i o n we h a v e
C 9 )
a b n
R S
c d k S R
for s o m e n £ 2 a n d By L e m m a 2 we h a v e
C I O )
a b c d
f4( a )
f Cd>
w h e r e
f C a ) - f j C d ) = C a — d ) From C 9 ) and C I O ) we o b t a i n
f4 C a ) f±C d )
R S k S R
T h u s
f C a ) = f4 C d ) = R, = kS, = S.
From t h i s w e h a v e
f ^ a ) - f 1 C d ) = 0.
S i n c e Si*Q, then we o b t a i n and h e n c e
c = kb.
On t h e o t h e r hand
C a - d ) ^i = f ^ a ) ~ f 1 C d ; > ~
By t h e f a c t that 1 - 0 w e g«t a=d and the p r o o f is c o m p l e t e .
3. PROOFS OF T H E THEOREMS.
P R O O F OF THEOREM 1.
Assume t-hat A, B, G e G 2 C k ; > a n d l e t*
r s I s„ r s ^
A = 1 1
k Si ri .
, B = 2 2
[ 2 2 .
, 0 = 3 3
ks^ r
3 3 .
such that
d l ) An + Bn = Gn. By Lemma 1 we obtain
A —
M N M N
1 1
, Bn- 2 2 3 3
kN M , Bn- kN M > ^ — kN M
1 1 2 2 3 3
where
ír +s V j r ]n+ f r - s V T " ] M m m J ^ in m J I C 1 2 )
N =- 2VjT Hence by ( 1 1 ) we have
i — ír + s V i P ]n- í r - s VT"1]" , m=l,2,:
H TO m J I. TO m J J
C 1 3 )
M. = M +
3 1 2
N„ = N4 + N .
3 1 2
From C 1 2 ) and C 1 3 ) we get
( r1+ s1V i r ]n+ ( r2+ s2V i r ]n = [ l y ^ V k - j V
Putting in t h e last e q u a l i t y
a = r1+ si VIT, ß « r2+ s2V j T , y = r3+ s3V T , we obtain
ar'+ ßn=
I
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where ex, ft,? e Q ^VIPj . Now, let a,ft,y e Q ^VlP j . Then we can write
and
a = r^s^/lT, ft = r2+s2VT, y = r^+s^VT
a = r1-siV)T, J5 - r2-s2-vr, f = r3- s3 V r \
wit-h integers r , s , m=l,2,3. ° m* m' ' From the assumption we have
a r i
+
It is easy to see that
C a ) r'+ C T h u s we obtain
c i 4 > J- + I - \ [ rn+ r " ) >
and
( I S )
D o n o t e
2V1P ^ ° ^ ^VTT"2 VIT11 ^ ^ J ^VT"' ^ ] * 2YjT
C16> Mx= I ( a W ] , M2= I [/3n+?T],M3- £ ( / V p ) .
C17> N =
1 2YT
From t h i s and from C 1 4 > , C 1 S ) we have C 1 8 ) M3 = ti± + M2,
Consider the m a t r i c e s A ^ »1 8! » ^ of* form N = N + N .
3 1 2
M1 M
".1 M N
Al = kN
l ' Bi = kN 2 > Cl = kN M 3 3
l l 2 2 . 3 3
where N m=l,2,3.
By ( 1 8 ) we h a v e
A + B = C . l l l
F r o m the a b o v e e q u a l i t y a n d f r o m L e m m a 1 and L e m m a 3 we o b t a i n that t h e r e e x i s t t h e m a t r i c e s A , B , G s u c h t h a t
A± = A , B= An i = B " , Ct = G' a n d t h e r e f o r e w e h a v e
An + Bn = Gn. T h u s A,B,C a r e m a t r i c e s o f the f o r m
A =
r s r
l l
k s r , B =
2 2
k s r , G =
3 3
ks_ r l 2 2_ 3 3_
h e n c e A , B , C <£ G 2 C k ) , w h a t g i v e s t h e proof o f t h e Theorem.
P R O O F OF T H E O R E M 2.
Let
C19> A =
r s a s r t h e n by L e m m a 1 we h a v e
( 2 0 )
r s n
R S
a s r a S R
w h e r e
( 2 1 )
R = I j ^ [ r + s V ^ Jn+ ^ r - s V S r Jn| ,
P u t t i n g in C 2 1 ) r=0, s=l we g e t
= 4 -
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I ^ F [ M - M " ] -
F o r n=2k
( 2 2 ) R = a2 and S=0.
follows. By C 2 0 ) and ( 2 2 ) we get-
An= 0 1
n n a2 0
n
n
1 0
a 0 . o a2. . 0 1
S i m i l a r l y we o b t a i n
Bn = b2.
For n=dm we have
1 0 0 1
Cn = c2.
A4 m .+ti =a D4 m 2 m
1 0 0 1
1 0 0 1
1 0 0 1
= (a2Tri+b2r"] .
1 0 0 1
1 0 0 1
= c
4 mand t h e p r o o f i s c o m p l e t e .
From T h e o r e m 2 we get t h e f o l l o w i n g C o r o l l a r y : C O R O L L A R Y C R . Z . D o m i a t y [ 3 ] )
If K = Q and a , b , c <s Z then t h e e q u a t i o n A4 + B? = C4
a b e have i n f i n i t e l y s o l u t i o n s of the f o r m
A . - [ 2 h l K - i i £ ] >
c* = [ ° b ] -
w h e r e
a = ^ m2- n2] . l , b = 2 m n l , c= | m2+ n2] . 1 , m > n , C m , n ) = l , 12:1
R E F E R E N C E S
tl3 E . D . B o l k e r - " S o l u t i o n s of Ak+ Bk= Ck in nXn i n t e g r a l m a t r i c e s " - A m e r . Math. M o n t h l y , 7 3 , 1 9 6 8 , 7 3 9 - 7 6 0 .
[23 J . I . B r e n n e r and J . d e P i l l i s — " F e r m a t ' s e q u a t i o n Ap Bp = CP f o r m a t r i c e s o f i n t e g e r s " — Math. M a g . , 4 3 , 1 9 7 2 , 1 2 - 1 3 .
133 R . Z . D o m i a t y - " S o l u t i o n s o f x4+ y4= z4 in 2 x 2 i n t e g r a l m a t r i c e s " - A m e r . Math. M o n t h l y , 7 3 , 1 9 6 6 , 6 3 1 .
113 R . Z . D o m i a t y - " L ö s u n g e n d e r G l e i c h u n g xn+ yn= zn m i t n=2™
im R i n g g e w i s s e r g a n z z a h l i g e r M a t r i z e n " — Elem.
M a t h . 2 1 , 1 9 6 6 , 3 - 7 .
[33 P . E i b e n b o i m — "13 l e c t u r e s on F e r m a t ' s last t h e o r e m "
S p r i n g e r — V e r l a g • N e w Y o r k — H e i d e l b e r g — B e r l i n , 1 9 8 0 .
a