zp - 1 = 0 (1) sums Ys = 2: Uk' l+s ' k=o The method of finding the va\ues of these periods is based upon the consideration of the sum

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1 Mathematics. -"On the binomial equation o{ prime degree". By Prof. J. C. KLUYVER. (Communicated at the meeting of April ). 1. The roots of the equation zp - 1 = 0 (1) where p is prime. can be expressed by radicals and the index of the radical signs involved is p- l. Likewise. putting p- I equal to )' )' 1' the equation of degree )'1' the roots of which are the soca lied GAUSSIAN periods of order )' I), admits of an algebraic solution. and the radical signs invo\ved have the index )'1' These GAUSSIAN periods are defined as follows. Let g be a primitive root of the prime number p. n. any complex root of the equation (1). then. putting the periods of order )' are the )'1 sums k='-i Ys = 2: Uk' l+s ' k=o (5 = O )'1-1). The method of finding the va\ues of these periods is based upon the consideration of the sum "= p- 2 {(n.. y"") = 2: a" )'''''. "=0 where fl is any integer and i' represents a primitive root of the new binomia\ equation Evidently we have and we can prove Z'I - 1 = o. (2) {(u. )'" 1) = {(n.. )',,"2). if fl l fl2' I ( d ). (ma )' 1 {(a. )'" ) = - 1. If fl 0., {(n.h. yl') = y-h:, {(a. y""). if fl =1= O. (mod )' 1)' I) Disquisitiones arithmeticae (deutsch von H. MASER). Art Proceedings Royal Acad. Amsterdam. Vol. XXVIII. 32

2 476 It is th en readily seen that the expression f (a y"'l) f (a y"'2) f(a y"'i+"'2) is a symmetrie function of the roots ak. depending only on ftl' ft2 and y. and that. when we write f(a y"'l) f(a y"'2)_ f(a. y"'i+"'2} -11''''1.'''2 (y). the righthand side will be in general a polynomial in y. Indeed. when we have ftl =1= O. ft2 =1= O. ftl + ft2 =1= O. (mod VI)' it is possible to show that where rl and r2 take all the values p-2. always remaining connected by the one to one relation.. (3) Hence. in the general case. 11''''1.'''2 (y) is a sum of p-2 terms. each of them being a certain power of the root y. and from the definition it follows that at the same time we have 111''''1''''2 (y) 1 = vp. However. when ftl and ft2 are not quite arbitrary. other results are obtained. and in particular we must note the equations 'P"'1."'2 (y) = - 1 if ftl - 0 of als ft2 Ol ( d ) ()-( 1}"1 '-1 -( 1)u2.-1 'f O( mo VI' 11''''1,'''2 Y - - P - - p. I ftl==. ft2== ftl ft2 = J Now. supposing one of the two indices to be. equal to unity. we have in partieular in which formulae rl and r2 are still connected by the relation (3). and the special sum G",(y) will satisfy the relations G"'I(y} = G"2 (y) if ftl - ft2' G",(y}=-l. if ft - 0. G",(y}=(-l}V- l p if ft+1=0. IG",(v)I=Vp. G ", (y) = (-1)' G _",_I(y). if ft :=I= O.ft+ 1=/= 0.

3 477 Pinally. it should be noticed that the sums f{a.1'm) can be expressed by means of GI'{1') and that we can establish the equations f{a '1')m = GI (1') G 2 (1') G 3 (1')... G m - I (1') f{a.1'm)., f{a 1')'\ = Go (1') GI (1') G 2 (1')... G,\_I (1'). ~. (7) Hence. as soon as the sums GI{1'). G 2{1')... GY\- l{1') are calculated. we find the value of f{a.1') by extracting a vi-th root. and subsequently we are able to determine f{a,1'm) for any value of m. Now. putting or.= '\-1 f (a. 1'). vi Y + 1 = f{a 1') + I G ( ) G () G ()'. (8). = 2 I l' 2 l' l' and introducing successively into this formula the I I I values of f{a, 1'). we can show that the V I va lues found for y are precisely the V I GAUSSIAN periods ys of the order v. It should be noted that if we replace g by another primitive root of p. or l' by another primitive root of the equation (2) the formula (8) still gives the same results, only the order in which the 1' 1 values of the periods Ys present themselves will be altered. 2. Por a given large prime number p the calculation of the sums GI'{ 1') by means of (4) is very laborious and al most impracticable. but when V I is smali. it is sometimes possible to form the equation of the "I-th degree whose roots are the 1'1 periods. To this end. we must observe that this equation. the leading coefficient being unity. necessarily has real and integer coefficients. and we must have recourse to the theory of numbers. In this note I will show how to deduce the equation of the periods when " 1 is 5 or 6. and how to solve it algebraically. In the first place. let p = 10 w + I and v I = 5. then we have according to (5). (7) and (8) Go (1') = - 1'1 + 1'3 + 1,2 + l' + 1 = 0 1. G 2 (l') = GI (1'2). G 3 (1')= GI (1'). G 1 (,,) = - p. Î GI (1'- I) = G~{,, )' G 2 (1'- I) = G~ { 1' )' ). (9) or f{a. 1' )5 = p Gi (1') G 2 (1'), _ f{a. 1')2 f{a 1')3 f{a, 1')" 5y f{a 1') + G I (1') + (;1 (1') G ;-{1') + Gi (1') G 2 (1')' 5y + I =f{a, 1') + --p- - + f(a, 1') f{a. 1')2 G;Trf P G I (1') + f(a. 1')2 ' 32*

4 478 and since If(a. y)1 = IG(y)1 = v p. it appears that everyone of the five roots y is real. The equation of the periods F(y) = 0 presents itself in the form 0= 5y+ 1 - Vp G I (y) G 2 (y) -Vp 2 5y+ 1 - Vp GI (y) Vp- Gî(y) G 2 (y) GI (y) G 2(y) --V p 3 V p 2 5y+ 1 - Vp-- and expanding the determinant. we get G I (y-i) G 2 (y-i) Gi (y-i) G 2 (y-i) -- V p 2- - V p 3 G I (y-i) G I (y-i) G 2 (y-i) V p V p 2 5y+ 1 - Vp- GI(y-l) Vp- 5y+ 1 V p 0= (5y+ 1 )5-1 Op (5y+ 1 )3-5p (5y + 1)2 [GI (y) + GI (y-i) + G 2 (y) + Gh- I )] + + 5p (5y + 1) [p -I GI (y) + GI (y-i) I 1 G 2 (y) + G 2 (y-i) I] - - p [G~ (y) G 2 (y) + G~ (y-i) G 2 (y-i) + G~ (y) G I (y-i) + G~ (y-i)gi (y)]. Now we put G I (y)=vpei~l. G2(y)=Vpei~). SI =GI (y)+ GI(y-l)+ G 2 (y)+ G 2 (y-i)= = 2 Vp (cos cos 8 2 ), S2 = I GI (y) + G I (y- I) I 1 G 2 (y) + G 2 (y-i) I = 'tp cos 8 1 cos 82, whence we have the inequalities 1] = 8p V p sin 8 1 sin 8 2 (cos fi l - cos 8 2 ), Transforming the equation of the periods F(y) = 0 and introducing SI' S2 and 1] we get 2 1 0= y5 + y1_ 5 (p-l) y y2 (2-6p-pSI) y (1-6p + p2-2p SI - p S2) + (11) in which equation. the prime number p being g~ven. the th ree numbers SI' S2 and '7 must be evaluated. To this end I proceed as follows.

5 479 The equation (4) can be reduced to G J (1') = I 1'q+~ rt " G J (1') = QJ l' + Q2 1'2 + Q3 1'3 + Qi 1'f, and as we have G 2 (1') = G J (1'2). the sums G 2 (1'). G J (1' - J) and G 2 (1'-t) are expressible by the same coefficients Qh that necessarily are integers. Addition of G t (1'). G t (1'- t). G 2 (1') en G 2 (1'-t) gives at once hénce St is an integer. Now writing th = 1'h+ 1' - h, and noticing that we have tt = ti' t2 = t3 tj + t2 = - 1. tt t2 = - 1. (tt - t2)2 = 5. the condition IG t (1') I = Vj) gives (12) But we have also I G 2 (1') I = V p. therefore we must conclude to and to lf we now put Qt Q2 + Q2 Q3 + Q3 Qi = QJ Q3 + Q2 Qi + Qi Qt. i 2 2p = 2 I Qh - I Qh Q. t h.k we have and consequently (13) a formula showing that S2 is an integer and that St and Vare both odd or both even. Concerning the product St V = ~ (cos 2 (Jt - cos 2 ( 2) t l -t 2

6 480 we may notice the inequality I SI VI < 4p _ I S svs (IS) and prove the relation - SI V = X2 + 4XY- y2 = (X + 2Y)2 - Sy2 = SX2 - (2X-Yf. (16) Finally we can deduce the equation 16p = 10X2 + 10y2 + S~ + SV (17) showing th at for any prime number p = 10 w + 1 the number 16 p can be resolved into a peculiar sum of squares. a partition the possibility of which is not easily established in another ~ay. Now it is possible to show that the integer V is a multiple of S. In facto in equation (11) the coefficients of y2 and of y are integers. Hence we have in the flrst place or But we have also leading to psi = 2-6p SI_-2(p+l). SI = +1 S2 = - 1 (mod S) (mod 2S) and by combining the congruencies we will find or (mod 2S)..... (18) (mod S). (mod 12S) p2 (S~ - 4S2) = Sp2V2 = 4 (1 - p)3 = 0 (mod 2S) Therefore we have from (16) and as X2+ y2 S V= 0 (mod S). X + 2Y = 0. 2X-Y _ 0 (mod S) S (X2 + Y2) = (X + 2 y)2 + (2X - Yf. is also an integer. In order to show. how the number 17 is connected with the other numbers. we remark that 1 GI(y) - GI(y-l) 11 G 2(y) - G 2(y-l) 1 = - 4p sin Ol sin O2 = = - (tt - t 2 ) (X2 - Xy - y2) and multiplying by 2 V p (cos Ot - cos ( 2 ) = (tt - t 2 ) V,

7 481 we get t = S.y 1(2X - y)2 - S y2j ~ 8pV psin {}. sin (}2 (cos {}. - cos (}2) = 'Ij = SV(X2_ XY- P) = (19) and 50 it is evident that 2~7 V is an integer. An useful relation follows from the identity S (X2 + Y)2 - (X2 + 4 XY - Y2)2 = 4 (X2 - XY -P)2. X2 + y2 S.V 'Ij In facto it shows that the three integers - - S S- and 2S Vare connected by the equation (20) We must now pay attention to some results of LEGENDRE.) concerning the numbers of the form (I 6) - S. V = X2 + 4XY -Y2=(X + 2Y)2- SY2=SX2 - (2X-Y)2. The prime factors occurring in such a number can be arranged in two separate classes. The first class contains. besides a possible prime factor S. only prime factors of the form IOx ± I; all factors of this class may occur raised to any power. In the second class we may find the factor 2 and possibly certain odd prime factors q different from S and not of the form IOx ± I. Their exponents necessarily are even. Hence we will have in general -SI V = X2 + 4XY - y2 = 4" Sb H2(IOy ± I). wh ere the odd integer H is a product of prime numbers q. and loy ± I the product of the prime numbers IOx ± l. Moreover. we must observe that whenever a certain factor q occurs in - SI V both numbers X and Y must have this factor in common. A single factor 4 will be found in -SI V. wh en X and Y are both odd. and only when both numbers are even the exponent a can exceed unity. It will appear from (I7) that S. and V cannot have a common factor q and as we have V O. S = I (mod S). we must conclude that S. = 4" H; (loy. ± I). V = 4 82 Sb Hf (IOY2 ± I). where H. and H 2 are prime to each other. It is possible to prove th at the exponents a. and a2 cannot be both greater than unity. it being clear besides that the supposition al = 0 involves a2 = 0 and inversely. ) Essai sur la théorie des nombres. Table III.

8 The preceding considerations concerning the integers S. S2. V. 1]. X and Y enable us to ascertain their values and to calculate the coefficients of the equation of the periods for a given prime number p = 10 w + 1. as I now will proceed to show. Let the prime number be p = then we begin with examining the possible values of S. Prom (18) we have and from (10) IS. I< 133; (mod 25) therefore we must consider as possible values the numbers , , But any factor q and also the factor 2 occurring in S. must have an even exponent. and so we are lead to discard the numbers Por each of the remaining possible values of S. we examine the corresponding possible values of V. According to (13) and to (15) we have and remembering that S. and Vare both odd or both even. that they have no common factor q. and that a possible factor q in V and also the factor 2 always is raised to an even power. we evidently have to retain only the suppositions written down in the following scheme. that X2 + Y2 also indicates the corresponding va lues of --5- ' as given by (17). 109 I ~ I X 2+ y I X2 + P Examining the values found for 5. we eliminate again several 't' b b. th t X 2 + y 2 t t.. f t SUppOSI lons y 0 servmg a canno con alo a prime ac or of the form 4n - 1 with an odd exponent. Hence the numbers 217 = 7 X = 3 X 5 X = 5 X 3 X = 2" X = 11 X 23

9 483 X2 + Y2 are impossible as values of ' so that only five suppositions are left. These are tried by means of the equation (20) and we find that only two of them lead to a perfect square as value of 17 )2 ( 25 V. Hence we have only to consider the following suppositions 109 Ifl X 2 + Y (~ )2 25 V and the evaluation of the positive or negative integers X and Y will enable us to decide between these two. The first supposition leads to X 2 + Y2 = 545. X 2 + 4XY - y 2 = ± 545. Whatever sign we adopt in the second equation. there are no integer values of X and Y satisfying both equations and the first supposition is therefore to be rejected. The second supposition leads to X 2 + y 2= 225. X 2 + 4XY - y 2 = ± 495. From these equations integer values for X and Y can be found. Omitting not integer solutions. we find V = + 55 'X = ± 12. I Y = ± 9. or V = - 55,X=± 9. ( Y = =t= 12. but whatever system of solutions we make use of. we get from (14) and (19) SI = - 9. S 2 = = Hence the second supposition is to be retained. and the required quintic equation of the periods (11) becomes At the same time we are now able to give the complete algebraic solution of tbis equation. From (12) we have

10 484 and this equation is to be combined with + 55 = QI + Qi - Q2 - Q J'. ~ - 55 = QI + Q1 - Q2 - QJ' ± 12 = QI - Qi. or wlth, ± 9 = QI - Q1' 1 ± 9 = Q2 - QJ. ~ =F 12 = Q2 - QJ. In this way we find four systems of integer va lues for the numbers QI' Q2' Q 3 and Q1. corresponding with the four sums G I (1'). G I (1'2). G I (r 3 ) and G I (ri). In one of these sums for example in G I (ri). the numbers QI + Q1-Q2-Q3 and QI-Q1 will be positive. Selecting this sumo we will find and may th en write QI =22. Q 2= - 7. Q J = Q1= 10 G I (rk)=22r-7r2-16r3+ lori. where k is one of the numbers I or 4. Now r k as weil as I' itself is a root of (9). therefore we can solve equation (21) by putting and then by writing [(a. rk)5 = 1091 GI (r k )2 GI (r 2i ) [( i) [(a. yi ) Gdyi) y - r:t. I' [(a. yi) Gdyk) [(a. yi)2 From the method applied to the treatment of the particular case p = 1091 it will be evident that in general a finite number of trials will suffice to ascertain the numerical values of SI. S2 and 1]. to write down the equa~ ti on of the periods and to arrive at its algebraic solution. 4. In the case p = 6w + I. V I = 6 the primitive root I' of (2) is one of the roots of the equation y2 - Y + 1 =0. (22) and according to the general formulae (5). (7) and (8) we have Go(y)=-l). G J(y)= (- I)w G 2 ()'). G,,(y)= (- I) w G I (y). G 5(r)=(-lr- ' p. [(a. y)6 = (- l)'v p G~ (y) G ~ (y). or _ [(r:t. I,)2 [(r:t. )')3 (- l)'v [(a. y)" [(r:t. y)5 6y+ l - [(r:t. y) + G I (y) + GI(y) G 2 (y) + G~(rfG2(Yr + G,(y)2Gh), the latter equation showing that the six periods all are rea I wh en w is even and that they are complex when w is odd. Eliminating [(a. y) the equation of the periods presents itself in the form

11 485 i w Vp[ l08y2 + 18y 12 + G 2\y) + G 2(y-l) p (3 + 2 (-1)'v ) + \ + 3(- 1)'vp,G1(y) + G2(Y)~ + 31 G 2 (y) + G (y- I) 1 + (G 2 (y) G1(y)\ G1(y) G2(y) + G1(y-l) G 2(y- l) 1 J= 216y3 + l08y2 + 18y(1-p) + (23) + 1-3p+3(-1)wpl Gh)+ Gh- I ) 1-3pl G1(y) + G1(y-l) 1- _ (- l)w p ~G~(y) + G~(y- I) ~ (G1(y) Gt(y-t)) and the following considerations lead to the evaluation of the coefficients. Supposing k,u =1= O. k(,u+ 1) =1= 0 (mod 6) we have (4) and as IGf" (Yi)1 = V p. we get Gf" (yk) = I yk(q+i~") = A + By, p = A 2 + AB + B2 = (A + B)2 - AB = (A - B)2 + 3AB. ~ 4p = (2A + B)2 + 3B2 = (2B + A)2 + 3A2. ) from which equations we infer that the factor p does not occur in one of the integers A. B. A+B. A-B. 2A+B and 2B+A. Likewise we observe that in the equations (A + By)2 = (A2 - B2) + yb (2A + B) = U2 + V 2y, p2 = U~ + U 2 V 2 + vi the integers U 2 and V 2 in this respect are similar to A and Band that. if we write (A + By)i = Ui + ViY. again neither Ui nor Vi is a multiple of p. Now on the ellipse p=x2 +xy + y2 there always lie twelve points both coordinates of which are integers. and by a finite number of trials their coordinates can be found. Selecting one of these points with the coordinates A. B. we may asser I that. always in the supposition k,u =1= o. k (,u+ 1) =1= 0 (mod 6). any sum GI" (yk) can be written in the farm fjh (A + BfJ). wh ere fj is either y or y-i. if we only give the right value to the exponent h. 50 for in stance we have G 1(y2) = fjh (A + BfJ), G,(y) = ftl(a + BfJ,). Gh) = ~2(A + BfJ2) and we must prove that fj. fj, and fj2 rep re sent one and the same root of (22). For this purpose we consider the equation G1(y) G,(y2) = (-1)w Gh) (25) immediately derived from (6); which we write in the farm fj~ lfjh (A + BfJ,) (A + BfJ) = (-l)w fj~h2 (A + BfJ2)2. (24)

12 486 The supposition that (31 and (3 were different roots of equation (22) would lead to and that is impossible since neither U 2 nor V 2 has the factor p. Further, (31 and (3 being equal. they cannot differ from (32 because in that case we would arrive at (3h+h1 (A + B(3)2 = (- I)w (3~ h (A + B(32) 2, (3; p2 = (A + B(32)" = U" + V,,(32' and again that would be impossible. In the first place 1 proceed to the evaluation of h by deducing the p-i equation of the periods of order - 3- ' Remembering that I' still denotes one of the roots of (22), we have according to (5), (7) and (8) G O (y2) = - 1. Gh 2 ) = - p f(a, y2p = P G I (y2) f( 2) + f(a, 1'2)2 f( 2) + p y - a, I' G I (y2) - a, I' f(a, 1'2) and eliminating + f(a, 1'2) the required equation becomes y3 + y2 + y (1-p) [1-3p - p I G I (1'2) + GI (1'- 2) I ] = O. (26) Here the last coefficient is an integer, hence we have (mod 27)... (27) and this congruency will give the value of h as soon as p is a known prime number. In fact, we may write (3h = L + M (3 and the corresponding values of h, Land M must be those of the following scheme L M hl I.

13 In this way we get 487 G,(y2) + G,(y-2) = L (2A + B) + M (A - B) and th en from (27) the congruency 1-3p - pl(2a + B) - pm(a - B) = O. (mod 27) by which the exponent h is readily determined. A second congruency wijl determine the exponent h 2 In facto the equation of the periods of p-i order 6 (23). wh en rationalised. must have integer coefficients. and that requires or 2 + L (2A + B) + M(A - B) = O. (mod 6) (mod 6).. (28) This congruency in connexion with the preceding scheme will give the value of h 2 and finally h, is found from the congruency (mod 6). (29) directly derived from (25). Now all the coefficients of (23) are known. and we have only to p-i rationalise this equation. to obtain the equation of the periods of order As an example I will treat the case p = Equation (24) becomes (2A + B)2 = B2, and af ter a few trials we find ± 3. ± 31 and ± 34 as possible values of B. Giving. for instance. B the value 34 and adopting - 3 as one of the two corresponding values of A. we may write Hence the congruency (27) becomes IOL-8M= 0 (mod 27) and considering the scheme. we find that we have necessarily L = - I. M = + I. h = 2. Havïng thus established the equation we have also the re sult G,(y2) + G,(y-2) = - 65, and by the way we have found that the equation (26) of the periods p- I of order - 3- is y3 + y2-354y = 0.

14 488 Thecongruency (28) now takes the form L - 37 M _ 0 (mod 6). and we necessarily have L = 1. M = O. h 2 = o. G 2 (y) = ( fj). Hence from (29) we get hl = 1 and then successively Gl(y) = fj ( fj) Substituting these results in (23) we obtain - ( 1 5 ) ( ) i VI 063 "2 y2 + 2" y - 29 = y3 + "2 y2-2 y and th en by squaring we find 0= y6 + y y y y y p-l as the equation of the periods of order 6' an equation all roots of which are complex. Solving algebraically, we have at once {(a, y)6 = fj2 ( fj)1 = fj2 R {( ) _ (2(a. y )+1063 fjr+[3(a. y ) y - a. y {(a. y) fjr {2 (a, y) fjr2' It will be found that the equation of the periods of order p 4 1 may be derived in a quite similar way.