ON FERMAT’S LAST THEOREM AND THE FIRST
FACTOR OF THE CLASS NUMBER
OF THE CYCLOTOMIC FIELD
BYYosHIKAzu KARAMATSU AND SHulcHI ABE
1. Introduction. In the seventh of a series of papers conceming Fermat’s Last Theorem, Prof. T. Morishima[1]proved the fbllowing theorem. THEOREM. If(1) xρ+yP+zp−O
is satisfied by integers x,γ, z prime to the odd pr㎞e P, then the first factor of the class number of the cyclotomic丘eld k(ζ)(ζ=e2zilりis divisible by p12 provided〃 does not divide 75571・20579903. Afterwards D. H. Lehmer[9]showed that t与e proviso of above theorem was unnecessary・ In the present paper, using the similar method as Prof. T. Morishima attacked, we shall extend this result and prove the following Theorems and Lemma. LEMMA・ Ifρis an odd pr釦me and(1)is satisfied in Case 1, then fbr〃1=2,3,4,5,6,7and 8
D・一…一[d(2th”1)P log(1−e”t dv(2m−1)P)L。≠・(m・d・) but excepting p which is a factor of 3,547,114,323,481., where Di means differe皿tiat− ing匡一t口nes, and setting〃=0, also, 一’being any of the ratios x γ x z 夕 z γ’x’z’x’z’γ’ THEoR.EM A. If〃is an odd prime and xρ十γρ十zp=O is satis丘ed in Case 1, then fbr each m=2,3,4,5,6,7,8 b(P.2m)♪.、≡0 (modρ2), but excepting p which is a factor of 3,547,114,323,481, where bo=1, b,=−1/2, b2:一(−1)i+IBi, b2i.、=0, i>0, Bi are Bernou1H numbers. THEOREM B. If xp+YP十zp−O is satisfied in rational integers x, y, z prime to the odd prime P, then the 1]rst fac− [1]2 Y.KARAMATSU AND S. ABE
tor・of the class number of the cycloto皿ic field k(ζ)is divisible by ρ14 provided 」ク does not divide 3」547,114,323,481. 2. SupPose that(1)is satis丘ed in rational integers x,アand z,、none zero and each pr丘ne to the odd prhneク. Then(1)g輌ves xp十YP =(x+ζン)(x+ζ『γ)… (x+ζρ一1y)=up, x十ア whereζ=ε2πψ, and u is a rational血teger. It is well known that the idea1(x十ζッ) is p−th power of an idea19t in the field k(ζ), that is, (2) (x十ζ1ρ)=班♪ (i,ク)−1. Let p be a pr㎞e ideal in the field k(ζ)andωan integer in the field with p and ωpri血e to P, then we set ・…P・一・・lp≡(!P)(m・dP)・ where N(p)is the noml of p and(ω/p)is some powet ofζ..It fbllows that (・) (‘一)一ζ・・…一・・∼・一ζ((͡+・・一・・閣・ Ku㎜er[1 r,12]proved that, if Em(ζ)一・(ζ)・(ひ)・−2m・(ζ・2)・一繊…(ζ・−1)・−2”‘Pt−1’, whereμ=(P−1)/2, r is a primitive root ofクand ・(・一[(1嵩;呂≡i;≡}]憂then
(・)・・d・Em(・≡、僻.裟老+1),…}・d°♪一潔4(ev)・(m・d・)・
where
(Em(ζ P))−c・i・d・E・・(C・・ ais a rational integer,0<a<」ρ一1, ψ。(ζ)=Σζ一(a+助+i皿d・(gb’1), gis a pr丘nitive root of P,乃ranges over the integers O,1,2,…, gt−2, excepting (gt−1)/2, and g is selected so that (gt−1) 9P≡ζ (modク);whence
gind・ζ≡ζ (modク).The symbo1
[♂−2餌logψ。(e” dvp−2m)],.。 means that.the(p.−2m)−th derivative of llogψ。(のl is taken with resp㏄t to v and v=Osubstituted in the result,θbeing the Napierian base. H...S. Vandiver also derived the relation[8]ON FERMAT’S LAST THEOREM AND THE F】RST FACTOR 3
(・・) 血・・Em(・≡み‘2 +1’D芳2$””(「2拠一1)(m・d・) wh・・e b、一一1/2, b、i−(−1)…Bi,転.、−0, i>0,β、 a・e B・fn・u皿i・numbers.Using t垣s in co皿㏄tion with K㎜er’s輌a[13]
エ (・)・・d・(}≡讐)≡≒1i・d・ζ一・Σ’nd’Em(鱈d’Em(ζ)(m・d・)・ n= wh・・eμ一(P−1)/2, th・・ymb・I S・ep・・…t垣g th・・。b・tit。ti。n(ζ/4r)i。 the n。t。ti。n °fthe K「°peke「“Hilbg「t[22・23コsymb・1i・p・wers・ Now if ぷ≡〆 (lnodク), 1 then』because of the relation. Ernst一γ2励=η♪; ηaunit in k(ζ), where現s’correspOnding to the substitution(瓦,(ζ)/Em(ζ・5)). We may write (6) ind.五』(ζs)≡S2m ind. Em(ζ) (modク).From(5)and(6)we obtain
ユ(・)・・d・G三f)≡≒1・・d・ζ一・Σ(52≒碁半(°(m・d・)・[13,19コ
”’=1 Using these relat輌on together with(3),(4a)and(5)we obtain エ(・)…d・G三f)≡一Σ一b・’…2”一 1’鱈’(S ’)(m・d・・)・
ラ扮=1 We also have the fb皿owing rcsult[3](・) Σ・・d・(≒讐 1)≡・・(・−1・…・・)(m・d・)・
hニ1 Applying the relation(8)and(9), fbr i=1,…,μ, we have エ ΣΣ[(f+1)2m−1]・・…・・…D−・≡・(m・d・・)・ k=1勿=1 0n the other hand it is known that Σ・’一・≡(1芦ibξ(m・dp),[5,・15コ k=l then, using this, we also have n Σ(’−3:/:lin)b2n[Σヒ吉1)…一 ・・D・…一・・]…一・・≡・(m・d・・)・[・コ 銘=2 m=1 1f we put n=2,3,4,…,8, in turn, we obtain n (1・)(…−1)・・λΣ鱈元1)…一・D 一・・]・ip−・ ・(m・d・・)・ 抗=1Forπ=2,3,4,5,6,7,8and p>3617 we have
4
Y.KARAMATSU AND S. ABE
(22カー1)b2n≠0 (mad p), the relation(10)gives b(♪.2餌四・D凱2。.、}≡0, (〃2=2,3,…,8) (mad P2), that is (11)…一・・・・…[d(2郁一1謬i−ev°]“.。≡・(m・d・・)・[16コ(m−・……・・)・ For the solution of(1)in integers prime toρ, we see that each of six ratios (12) 一’≡y/x,xfy, z/y, y/z, x/z,ψ. (madρ), satiSfies the congruence(11), which.iS Well−knowh as Kum ner Creiteria. By using Kummer Criteria[12,13,16](in co皿ection with Fe頂at’s Last Theorem) it may be shown that b(」−2が)ρ+1≡0(mod p2), m=2,3,4,5,(D. M垣manoff.[20], H.S. Vandiver[7]);and similarly fb⇔n=6,7,(T. Mo亘shhna[1,2], D. H. tβh− mer[9])if(1)is satisfied in CaSe 1, that is (13)[d( 1}鵠皇一eVt)],.。≡[d2m−1票し一判“.。≠・(m・d・)・where
[d{ 湯li−e”t)]。.。≡[ごi,]::12≡一[1」..,]:rE.2(m・d・)・ In a s血nilqr way fbr m=8, to be more exact, we may prove the relation(13). We have eas且y P−1 [1:≒。,]≡Ση・一…+・・[1』θ,’]:1;2(m・d・)・ ”=o Using these relation together with(13)we have(14) [4 1謬…−e”t)]“。。≡一({警諺≧・(m・d・)・
When〃1=8, the numerator in(14)can be written as fb皿ows; 1己5(’)=’十16,369’2十4,537,314t3十198,410,786t4十2,571,742,175’5十12,843,262,863’6 十27,971,176,092〆+27,971,176,092’8+12,843,262,863’9+2,571,742,175’10 十198,410,786tli十4,537,314’12十16,369t13十t14. Now let us consider the equation F15(’). In order to do this it is more convenient to dist姐guish three cases as fbllows; Case A. The six values of’in(12)are all incongnlent modulo p. Case B. The three values of t in(12)are ea¢h congruent modulo p. Case C. The two values of t in(12)are eadl congruent.moduloρ. First of all, by the meaning of Case I we have t≠0,1 (modρ): In Case A:Whence(12)are roots of『the congruence (15) g1(1)=〆−3t6十ゐ’5−(2b−5)〆十あ’3−3t2十’≡0 (modρ) where’≠−1 (mod P)and b is an integer. Hence we look f()r p such that satiSfy F15ω…≡0 (modク).ON FERMAT’S LAST THEOREM AND THE FlRST FACTOR
5
We put
F、5(’)≡一τ(1+’)F’、5(t), then F’15(t)is divisible by p only when、F15(’)is divisible by」ρand F’15ωcan be wntten as F’15(t)=ρω91(’)+R(t) WhereρωiS a qUatient. If F’15(’)is divisible by p, then Rωmust be divisible byク. Whence we obta垣 R(t)=Ast5+A4t4+A3t3+A2t2+Alt1+Ao; A5=:−16,374b3十216,887,491b2十23,307,494,728b十219,878,141,658, ∠44=ゐ4−4,570,065b3十2,806,598,822b2−37,851,528,221b−94,225,276,608, A3=−2ゐ4十9,189,252b3−6,263,909,239b2十146,276,072,107カー365,015,148,464, ノ12=b4−4,602,810b3十3,226,614,495b2−75,432,750,220ゐ十55,336,765,782, 111=−3b3十13,792,057b2−9,467,460,569ゐ十204,957,634,488, 1lo=b3−4,602,810b2十3,226,565,377b−74,800,629,967. If R(’)is divisible byρ, then fbr all i q A{≡0 (i=o,1,2,3,4,5) (modρ). By solving above six copgruences・we have three fb皿owing congruences; From the relation (A4+2A,+3A2+3A、)/−2we have
(16) 、. 8,187b十21,686,186,363≡0 (mhd p). Using the above relation together with (16,373A,一∠44−16,373/13−32,749A2−49,123ノ{1−49,122Alo),we have
(17) 2,513b−198,399,178,670,415,714≡0 (modρ), and moreover, from(16)and (A,+!14+A3+A2+A、), we obtain (18) 2,069b−115,461,529,129,395,018≡0 (modρ). ・ Using these relations∫ogether with(17),(18)and(16)we have the fb皿owing two numbers (19) 1,624,294,130,272,079,780,737≡0 (mad p),and
(20) 945,238,583,851,076,597,413≡0 (mod p). A㏄ordingly if the congruence F15(’)≡0 (modク), is satisfied in Case A, then」ρmust satisfy at least above two numbers(19)and(20). But we have a relation of (number(19), number(20))=1. Therefbre, we have the fb皿owing resUl’t A; Result A:Under the conditions of Case A, we have for m=8and fbr allρ1arger than 49,1236 、 . Y.KARAMATSU AND Sc ABE
[d2m二1109(1−eVt dv2・・−1)],.。≠・伽・d・)・ In Case B:Whence(12)are roots of the congru㎝㏄ (21) 92(’)一τ2−’+1≡0 (mod P), where’≠0,1,−1(mod、ρ)and it is only type p=6h十1 that(21)is satisfied withP. We have
F’、5(の=0(t)92(t)+R(t) where Oωis a quotient and R(’)=25,201,039,501. This number R(t)is split口1 the f()皿owing pr㎞e factors; R(t)=.11・331.491・16,519 wherβ331 and 16,519 are the 6h十1 type primp. Clearly we must except these prhne numbers. T垣{允re, we have the follow血g result B; Result B:Under the conditions of Case B, we have for m=8.and for all p larger ’than 16,519 [42楊}1109(1一ε㍗ 吻2餌一1)L。≠・(m・d・)・ In Case C:We have t≡2,1/2,−1 (modρ). Hence 1「15(−1)is identically zero and 1㌦5(2)=−Fls(1/2)when㏄
Fl5(2)=2・3・3,547,114,323,481. Aocordingly if the.cQngmence FL5(’)≡0 (modク), is satis6ed in Case C, then 3,547,114,323,481 is divisible by p. So that, it remains to show that we sp五t in some prime factors of 3,547,114,323,481. Therefore, we have the fb皿owing result C; Result C: Under the conditions of Case C, we have for m=8, and for all p larger than 3 [42椀一110g(1一θ拶’ 吻2抗一1)]。.。≠・(血・・)・ but excepting p which is a factor of 3,547,114,323,481.From above ResUlts A, B孤d C we h的e the制ow㎞g Le㎜a;
LEMMA. Ifクis an odd prime and(1)is satisfied五1 Case 1, then fbr〃1=2,3,4,5,6,7and 8
‘ D・・一一[d伽一謬誓i一θ”’)],一。≠・(m・a・),
but exceptingクwhich is a factor of 3,547,114,323,481., where Di means differentiat− ing i−t註nes, and settmg v=0, also −t being any bf theτatios x y x z. ア z ’ γ’x’z’x’z’y’ ’ON FERMAT,S LAST THEOREM AND THE FIRST FACTOR
Now applying this Lemma to(11), we easiy obtain the fbllowing theorem. THEoREM A. if P is an odd p血1e and(1)is satisfied』in Case 1, then fbr 〃2=2,3,4・,5,6,7,8 ・ (22) 『 ゐ(p−2抗}ラ+1≡0 (mod p2), but fbr m ・8 ex㏄pting p which is a factor of 3,547,114,323,481., where b,=−1/2, b2i ==(−1)i+1」Bi, b2i+1=0, i>0, Bi are Bernoulli numbers. By H. S. Vandiver’s[5,6]result we also have i−1 カ i(・・) 〃‘;1Σ・』ΣΣ・/自(与)sゾ・・
a=1 a=1 S=1where
da≡−a/P (mod n), 0<da<〃, (n,ク)=1, whence fbr 7=(クー2m)〆十1, c>0, トエ P−1(24) 〃’;1Σ・・≡・Σ伽・・(m・d・・)・
a=1 a=1 0n the other hand it iS known that i−ユ(25) 撒Σ・・≡・・(m・d・・)・[15コ
a=1Hen㏄
P−ユ(26) ”≒1・・≡Σ・・al(m・d・・)・
a=1 For c=1 and 14, this yields (28)Hence we have
therefore n(クー2nt)P+1−1 (29) (P−2m)P十1For each m=2,3,4,5,6,7,8we have
whence from(29)and(22)we have
(30) We also have from(23)and(24) P ユ農編・・一・・≡Σda・・一(m・d・・)・
α=1ρ_1(芸蒜1・・一・4+・≡Σda・…一・碗1・(m・d・・)・
a=1 − a《♪−2繊≡a{♪−2碗14 (mOdρ), ・・一・・≡(:器是i…一 ・4t・(m・d・・)・ η(♪二2勿)ρ14十1≠1 (modρ) ゐ{ρ.2拐}ヵ・4.、≡0 (mOPρ2). 7each
8