an easy way to solve the solvable quintic using two sextics
TRANSCRIPT
“An Easy Way To Solve The Solvable Quintic Using Two Sextics”
Titus Piezas III
ABSTRACT: Using a method initially developed by George Young (1819-1889), Arthur Cayley (1821-1895), and later by George Watson (1886-1965), an explicit quartic is constructed to enable the solution in radicals of a quintic when it is a solvable equation. Not one, but two sextic resolvents are derived which are important to forming the coefficients of this quartic. Certain difficulties and their solutions as well as a novel consequence to the method are also addressed in this paper. Mathematics Subject Classification. Primary: 12E12; Secondary: 12F10. It was the year 450 AH (c. 11th century AD). The poet-mathematician bent over the sand and scribbled some words. In the distance, a caravan of merchants and their camels were trudging on the dirt road going to the great city of Samarkand, but the poet paid them no heed. He was looking at the words he had written which modern mathematicians would understand as the cubic equation 032203 =++ xx . It was just yesterday that the poet solved a similar equation using conic sections. He frowned, as a thought occurred to him, and he made several scribbles. The equation became the fifth degree equation 032205 =++ xx . The poet stared at it for a moment. It was probably impossible to solve, he thought, as there were only the three dimensions of length, breath, and width. As he continued to ponder over the matter, an idea for a poem came to him. He erased the equation, and started to write on the sand, “The Moving finger writes, and having writ…” Little did he know that it would take almost a thousand years before such equations could be solved. I. Introduction The theory of equations and the 4000-year history of its progress, from its roots in ancient Babylonia (c. 2000 BC) to giving rise to Galois theory in modern times, is a very interesting and fruitful aspect of the study of mathematics. Fruitful in the sense that after the general solutions to the cubic and the quartic equations were discovered in the Renaissance (the Babylonians much earlier already knew how to solve quadratic equations), the natural question to ask was: Was the general quintic also solvable using only a finite number of the four arithmetic operations and root extractions? It was in trying to answer this question that eventually led, through the work of persons such as Joseph-Louis Lagrange (1736-1813), Paolo Ruffini (1765-1822), Niels Abel (1802-1829), and Evariste Galois (1811-1832), among others, to the development of modern group theory. In particular, Galois theory would provide an answer to the question and more: the general quintic and other higher general degrees were not solvable in radicals. However, that didn’t mean that there were no particular quintics solvable as such—the 2-parameter solvable deMoivre quintic is an example. Particular higher degree equations could be solvable in radicals but only when their Galois group was solvable.
How then can we find the expressions for the roots of the quintic when it is solvable in radicals? It is the author’s belief that there will be a perennial interest in this question as the general quintic is “special” in the sense that it is the first degree without a formula in radicals.
Thus, that is what this paper will address and answer in a very accessible manner.
Unconventionally, we will also present the summary of our results in this introduction, so that those who only want to know how to solve the solvable quintic need only read this chapter and no further, while those curious at the derivation can continue with the rest of the paper.
The method used here is indebted to the work of George Young (1819-1889), Arthur Cayley (1821-1895), and George Watson (1886-1965), though it seems there were things left undone. It is the author’s hope that he had made the method more rigorous, refined, and complete than when he first encountered it.
We start with the general quintic equation: 0FExDxCxBxx 2345 =+++++ which we can easily transform to its reduced or depressed form, a form without the 1−nx term, by
the substitution 5
Byx
−= , then expanding the expression to get a quintic of the form:
0fey5dy10cy10y 235 =++++ (a) Let the first sextic resolvent in the variable p be:
0)cfdef2dfc2eec11ecd28ec35d16dc40c25(
p)D(p)df2e3ec2cd8c15(25p)ec3(625p312522322244236
22224426
=+−−+−++−−−+
+−+−+++−
And the second in the variable t:
0)dffeefc8fcd2fc16fcde12
efd6defc80fdc40edc100ecd100d25(t)Df(t)cf2
def6dfc8e4ec20ecd40d15(25t)ce4d3(625t3125
3222222242
33322224622
232224426
=+−+−−−
++−−+−++−
++−+−+−−
where D is the discriminant:
4322222322225
32234533
542323424223
fcdf120fce160efd360efc1440fdc2640fc3456
fde640fdec4480efcd10080defc11520fd3456fdc5120
e256ec2560ecd5760ec6400ed2160edc3200D
+−++−++
−+−−++
+−++−−=
Let p be the root such that 2p is rational and t be the root such that 2t is rational. Define: m = c-p n = c+p v = d-t w =d+t
Case 1. If 0p ≠ . Let,
0nm
znm
v)wmmnemn(w)vnnmnem()wv)(nm(nm
znm
wvvwvnwmemwenv)wvw3v(mnnm
z))nm)(wv(5f(z
55
246424642222
233232322
2255
34
=+
−
+−−+−−−−−
−−+−+−−++++−
−−−+
Case 2. If 0p = . Let,
0cfzcz)cdf2ceed4ec5c2(fzz 1052223534 =+−++−−++ Then a real root of the solvable quintic (a) is given by: 5/1
45/1
35/1
25/1
11 zzzzy +++= where the iz are the roots of the quartic of the appropriate case. Or if expressed in the original variable x:
5
zzzzBx
5/14
5/13
5/12
5/11 ++++−
=
Example: 032y20y5 =++ c = 0 d = 0 e = 4 f = 32 5512D = The first sextic resolvent is: 064p)5512(p1200p2500p3125 246 =+++− The second is: 016384t)516384(t6400t3125 26 =−+−
which have roots, respectively:
5
52p −= ,
554
t −=
We then form m, n, v, w as defined above. Since 0p ≠ , we form the Case 1 quartic:
031251024
z125512
z5
64z 24 =−−−
which has the solutions:
+−= )5225(555)25/4(z1
−−−= )5225(555)25/4(z 2
−+−= )5225(555)25/4(z 3
++= )5225(555)25/4(z 4
thus, ...36396.1zzzzy 5/1
45/1
35/1
25/1
11 −=+++= And as was mentioned, the method is easy and simple. II. The Derivation After the solution to the cubic equation was found by Scipione del Ferro (1465-1526) and independently by Nicolo Tartaglia (1499-1557), and the solution to the quartic by Lodovico Ferrari (1522-1565), the next step obviously was to solve the general quintic. Despite the efforts of many mathematical minds of the period, the solution, if any, was far from obvious.
It took a few centuries before significant advances were made and among several, we can cite the work of Joseph-Louis Lagrange (1736-1813). He observed that the known methods for solving quadratics, cubics, and quartics had a common feature: it reduced the problem to solving an equation of a smaller degree, the resolvent.
For the general quadratic, 0cbxx 2 =++ its resolvent was given by the linear equation, 0)c4b(z 2 =−−
such that a solution was,
2
zbx
2/1+−=
For the cubic, 0dcxbxx 23 =+++ its resolvent was given by the quadratic, 0)c3b(z)d27bc9b2(z 3232 =−++−+ such that a solution was,
3
zzbx
3/12
3/11 ++−=
where the iz are the 2 roots of the resolvent. In general, we have the theorem due to Lagrange. Let the ix be the roots, V a rational fraction in the ix , ω any nth root of unity, and )(ωL the Lagrange resolvent: ( ) n
1n3
221 xxxxL −++++= ωωωω ...
Theorem: Let n be prime. If 1≠ω , then the nL )(ω is a root of an equation of degree n-1 whose coefficients are rational fractions in V and in the elementary symmetrical polynomials in the ix . Morever, V is a root of an equation of degree (n-2)! whose coefficients are rational fractions in the elementary symmetric polynomials.
To avoid confusion, let us denote the equation of degree n-1 as the Lagrange resolvent (polynomial) and the equation of degree (n-2)! as the auxiliary resolvent (polynomial).
Thus the quadratic has a Lagrange resolvent of degree 2-1=1, with coefficients determined by an auxiliary resolvent of degree (2-2)!=0!=1. For the cubic, Lagrange resolvent is of degree 3-1=2, coefficients determined by an auxiliary resolvent of degree (3-2)!=1!=1. These results are consistent with the solutions discussed earlier.
The situation for composite n is slightly more complicated but the quartic ends up with a resolvent of cubic degree.
For the quintic however, the situation is vastly different. It has a Lagrange resolvent of degree 5-1=4 but the auxiliary resolvent that determines this quartic’s coefficients is of degree (5-2)!=3!=6. Thus, to solve the general quintic we end up trying to solve a sextic. And for the general septic, the auxiliary resolvent is a staggering (7-2)!=5!=120-degree equation. No wonder Lagrange, and probably other persons doing work on the problem, was cautious about the
possibility of solving higher degrees in the radicals though it was the later work of Ruffini, Abel, and Galois that finally settled the matter.
However, as mentioned earlier, if the equation has a solvable Galois group, then it is expressible in radicals. Indeed there are some quintics, an infinite number in fact, that have a solvable group, though there is also an infinite number, much greater, that do not have a solvable group.
To find the exact expressions for these quintics solvable in radicals, thus we have to find the Lagrange and auxiliary resolvents. We start with the general quintic: 0FExDxCxBxx 2345 =+++++ Let x= (-B+y)/5 to get the reduced form. Since if expressed in the original coefficients, it will be unwieldy, we can assume it to be equal to: 0fey5dy10cy10y 235 =++++ with the explicit expressions for the variables c, d, e, f easily obtained by expanding the substitution.
Next, we use the theorem which just a variation of Lagrange’s theorem discussed earlier, that if an irreducible quintic with rational coefficients is solvable in radicals, then its roots iy are expressible in the form: 4
k43
k32
k21
k1k uuuuy ωωωω +++=+
where ω is any complex fifth root of unity, or )5/i2exp( π , for k = 0,1,2,3,4, and
54
53
52
51 u,u,u,u are the roots of a quartic, the Lagrange resolvent, also with rational coefficients.
If we construct the quintic with roots 54321 y,y,y,y,y as defined above, we get:
0))uuuuuuuu)(uuuu(5uuuu(
y)uuuuuuuuuuuuuuuu(5
y)uuuuuuuu(5y)uuuu(5y
22
442
312
232
132415
45
35
25
1
33
413
343
223
143212
32
22
42
1
22
244
231
223
21
33241
5
=+−−−−+++−
−−−−−++
+++−+−
which one can see has no 4y term, thus explaining why the solvable quintic must be reduced first. In general, if we form the n degree equation, for prime n, with roots y analogous to our iy ,
then the equation will have no 1−ny term.
The next steps we will take, or similar ones, were familiar to George Young (1819-1889) and George Watson (1886-1965) as described in Berndt, Spearman, and Williams’ paper “Commentary On An Unpublished Lecture By G.N. Watson On Solving The Quintic”. However, beyond a certain point, our approach will start to differ from theirs.
By equating the coefficients of the constructed quintic with the coefficients of the reduced quintic, we get,
c2uuuu 3241 −=+ (1)
d2uuuuuuuu 22
442
312
232
1 −=+++ (2)
euuuuuuuuuuuuuuuu 33
413
343
223
143212
32
22
42
1 =−−−−−+ (3)
f)uuuuuuuu)(uuuu(5
uuuu
22
442
312
232
13241
54
53
52
51
−=+−−−−
+++ (4)
Assume 2 more variables: p2uuuu 3241 =− (5)
t2uuuuuuuu 22
442
312
232
1 =+−− (6) From (1) and (5), pcuu 41 +−= , pcuu 32 −−= (7) From (2) and (6) tduuuu 2
243
21 +−=+ , tduuuu 4
231
22 −−=+ (8)
Using the algebraic identities: )uu()uu(4)uuuu()uuuu( 32
241
22
243
21
22
243
21 −+=−
)uu()uu(4)uuuu()uuuu( 412
322
42
312
22
42
312
2 −+=− we have, using (7) and (8),
)pc()pc(4)td(uuuuR 222
243
211 +−+−±=−= (9)
)pc()pc(4)td(uuuuR 224
231
222 −+++±=−= (10)
By adding, and also subtracting, (8a) and (9),
2
Rtduu 1
32
1++−= ,
2Rtd
uu 12
24
−+−= (11)
and (8b) and (10),
2
Rtduu 2
12
2+−−= ,
2Rtd
uu 24
23
−−−= (12)
The objective is to express (3) and (4) in terms of p, t, 1R and 2R . Using the identities:
32
12
232
12
31 uu
)uu)(uu(uu =
41
22
412
24
32 uu
)uu)(uu(uu =
41
32
142
31
33 uu
)uu)(uu(uu =
32
42
322
43
34 uu
)uu)(uu(uu =
and (7), (11), (12), we can express (3) as: 21
222222 RpR)ep3c)(pc()td(c =−+−+− (13) And using the identities:
2
32
12
22
32
151 )uu(
)uu()uu(u =
241
22
42
12
252
)uu()uu()uu(
u = (13.1)
241
32
12
42
353
)uu(
)uu()uu(u =
232
42
32
22
454 )uu(
)uu()uu(u =
together with (7), (8), (11), and (12), we can express (4) as:
212232224
4222422332332
RR)tptccdp2(cpt2)pc(dtp)fcd6(
)p11pc18cdc7(pt2p)fc2ddc4()fcddc2(c
++=−+−+−+
+−++−++++ (14)
combining (13) and (14) by eliminating 21RR we get the cubic in t:
0p)fpfccde2d(pt)p25eppc10dt(ct)cedct( 22332232 =+−+−+−++++−− (15) where the peculiar arrangement will be justified eventually.
While we ended up with one equation in 2 unknowns, we can rationalize either (13) or (14) to have another equation. Rationalizing (13) by using the squares of (9) and (10) we have the quartic in t:
0)eep6ec4(p
)ct14dpt16p25pc35cd2c11(p)cedct(2222
24222422232
=+−−
+−+−+−+−− (16)
and we now have 2 equations in 2 unknowns. This is the point where we start to differ with Young and Watson. For one thing, given similar equations to (15) and (16), they did not attempt to eliminate one unknown to get a single equation in the other unknown. In the words of Watson, “…[one] will probably realize that the solution by inspection of a pair of simultaneous equations of so high a degree is likely to be an extremely tedious task…” Young had to make do with inspection in numerical examples, while Watson obtained one unknown through another approach.
This is one aspect where we can have an advantage over pre-electronics mathematicians. In the more than 50 years since Watson wrote his paper in 1948, we’ve seen the development of computers, computerized algebra systems (CAS), and the Internet. What they shirked from doing due to the sheer amount of effort involved, we can let the computer do in seconds.
So, to eliminate t in (15) and (16) to get a single equation in p, we get their resultant using a computer. After a minute or so, it gives the result: a 22-degree equation. One can readily understand Watson’s sentiment.
However, it factors, several of which are spurious like for fixed values of p, like p = 0, but the important one is the 12-degree factor which factors further into 2 sextics over a square root extension:
0)cfdef2dfc2eec11ecd28ec35d16dc40c25(
p)D(p)df2e3ec2cd8c15(25p)ec3(625p312522322244236
22224426
=+−−+−++−−−+
+−+−+++− (17)
This is our first sextic resolvent. The radical D could either be D± , though for convention we will use the positive case. It turns out D is the discriminant of the quintic of form: 0fex5dx10cx10x 235 =++++ given by:
4322222322225
32234533
542323424223
fcdf120fce160efd360efc1440fdc2640fc3456
fde640fdec4480efcd10080defc11520fd3456fdc5120
e256ec2560ecd5760ec6400ed2160edc3200D
+−++−++
−+−−++
+−++−−=
Strictly speaking, the discriminant may be more 5D than D, though for purposes of this paper, D will be referred to as the discriminant. In the paper “Commentary…” cited earlier, the authors mention, to quote, “…we are not aware of any rigorous direct proof of the equivalence of ∈2φ Q to the original quintic being solvable.” The use of φ is analogous to the root of our sextic in p. We will provide such proof. Theorem 1. The irreducible quintic 0fex5dx10cx10x 235 =++++ where the coefficients are in the rationals is solvable in radicals if and only if (17) has a root p such that 2p is rational. If so, (17) factors into a linear factor and an irreducible quintic polynomial.
Proof: We rationalize (17) to get the form:
0Dp)cfdef2dfc2eec11ecd28ec35d16dc40
c25p)df2e3ec2cd8c15(25p)ec3(625p3125(222232224423
622224426
=−+−−+−++−−
−−+−+++−
Then let, c = k/10; d = q/10; e = r/5; f = s and a change of variable from p to x defined by,
5
r12kx4101
p2 ++
=
and we get:
0rs9375sk3125kqrs1250skr2000srq3250
srk1200rsqk725rsk99skq125sqk150sk27sqr1600
sqrk160srkq590sqrk196rsq124rsqk12qrsk18sqk12
sqk4r256rk192rkq16rk48rq17rqk128rk4
rqk65rqkrkq13q(x)s3125kqs625skr500rsq2750
rsk525sqk325sk108sqr2400sqrk260rskq105qrsk117
sq58sqk31r512rk256rkq76rk32rq3rqk51
rqk19kq2(x)krs500sq625sqr1400qrsk90skq50r400
rk136rkq76rk9rq8rqk6qk(x)ks125qrs400qsk15
r160rk40rkq21q2(x)qs50r40rk6kq2(rx8x
442323222
2232222524224274
322324533652
3565242443432336
24222568432222
232222532234
533542323424223
42622222234
32222442342322
32224422256
=−+−−+
+−−−−−−
−++−++−
−+−−++−−
++−+−+−+
+−−−+++
−−+−+++−
+−+++−+−+
−++−−++−−
+−+−+−+−++
(18)
which some might recognize as the resolvent sextic established by David Dummit in his paper “Solving Solvable Quintics” for the quintic: 0sryqykyy 235 =++++ Hence, the two sextic resolvents are related. The relationship can also be expressed as: x4r12kp500 22 =−− With k = 10c and r = 5e as rational coefficients, if 2p is rational it implies that x is also rational. Since Dummit has already proven that the irreducible quintic with rational coefficients is solvable in radicals if and only if (18) has a unique rational root x, then it is equally valid to say that it is solvable in radicals if and only if (17) has a root p such that 2p is rational. The rationality of x necessarily follows and Theorem 1 is proven.
What remains is to determine t. We can get the resultant of (15) and (16) again, this time, eliminating p. We get a 22-degree equation in t which factors with the important factor similar in form to (17). It is given by:
0)dffeefc8fcd2fc16fcde12
efd6defc80fdc40edc100ecd100d25(t)Df(t)cf2
def6dfc8e4ec20ecd40d15(25t)ce4d3(625t3125
3222222242
33322224622
232224426
=+−+−−−
++−−+−++−
++−+−+−−
(19)
with D as discriminant and still using the positive case. This is our second sextic resolvent. When the quintic is solvable, it factors similarly to (17) and we can come up with a second theorem complementary to the first. Theorem 2. The irreducible quintic 0fex5dx10cx10x 235 =++++ where the coefficients are in the rationals is solvable in radicals if and only if (19) has a root t such that 2t is rational. If so, (19) factors into a linear factor and an irreducible quintic polynomial. Proof. We can appeal to the rationality of the coefficients of both the solvable quintic, namely its constant f, and its resolvent quartic to prove the rationality of 2t . From (4), (5), and (6), we have: f)t2)(p2(5uuuu 5
45
35
25
1 −=−+++ Since the 5
iu are the roots of our resolvent quartic which will be in the variable z, by forming the quartic we have, 0)uz)(uz)(uz)(uz( 5
45
35
25
1 =−−−−
Expanding this, one can see that the coefficient of the 3z term, label it Q, is )uuuu( 5
45
35
25
1 +++− . Since for solvable quintics with rational coefficients this quartic also has rational coefficients, then Q is rational. fpt20Q −=−− Or,
2
22
p400)Qf(
t+−
=
Since f is rational and for solvable quintics 2p and Q are rational, then 2t must also be rational and we have proven the first part of Theorem 2.
To prove the second part of the theorem, that if and only if the quintic is solvable then
(19) has a unique linear factor t, one need only prove that f, 2p , and Q of the above relationship
has unique values that will uniquely set the value of 2t .
For f and 2p , it is easily seen that they have unique values. For the unique value of Q,
we recall a part of Lagrange’s theorem, “… nL )(ω is a root of an equation of degree n-1”. Thus,
for n=5, nL )(ω takes on four and only four values, whether zero or non-zero. Thus, for solvable equations of prime n, its Lagrange resolvent polynomial with rational coefficients is unique. And since Q is a coefficient of our quartic resolvent, then Q has a unique value. Thus 2t also has a unique value and we have proven the whole of Theorem 2.
We now know how to get p and t by using the sextics of (17) and (19). However, we can always ask how Young and Watson did it. Assume they already had p. What Watson did was substitute the value into (15) and have a cubic in t, or (16) and have a quartic in t. And while it was a cubic or quartic, one need not employ the third roots of unity since we know it should factor such that 2t is rational. What one may observe, and which will later be illustrated with a solvable parametric quintic, was that (15) or (16) sometimes factors completely into linear factors. Which t to use since we’ve seen that there is a unique t? To appeal to the common factor of (15) and (16) won’t do because they may have all factors in common, with one repetition in (16). The problem doesn’t seem to have been addressed by either Young or Watson. The solution then is to have an equation in t solely in the coefficients c, d, e, f of the quintic which determine the correct value of t. Similarly, if one already had t and substituted it into either (15) and (16) and have an equation in p, it may have several linear factors. So we also need an equation in p solely in the coefficients c, d, e, f . Our two sextic resolvents answer that problem. If that was how Young or Watson obtained t, they had to have p first. Since they didn’t have the convenience of a computer to calculate resultants of a large degree for them, Watson obtained p from another source, namely a sextic resolvent which had been constructed by Arthur Cayley (1821-1895) a quarter of a century earlier. We start by recalling again the theorem that the roots of our quintic can be expressed as: 4
k43
k32
k21
k1k uuuuy ωωωω +++=+
It is easy to show that if we compute the value of the expression, call it q, defined as, 14422553311554433221 yyyyyyyyyyyyyyyyyyyyq −−−−−++++= using the roots as defined in the theorem, then it reduces to, 55)uuuu(q 3241 −= Since from (5) we have,
p2uuuu 3241 =−
Then,
5p10q = Thus, all we need is to find the polynomial in q and once a linear factor is found, we can obtain p. Since Cayley’s sextic was in fact the polynomial in q, Watson could easily find the desired p. And since we already have a polynomial in p, we can reverse-engineer and reconstruct the polynomial in q to see if we’ll arrive at the same polynomial derived by Cayley.
Using (17) and letting 510
qp = , we get,
0)cfdef2dfc2eec11ecd28ec35d16dc40c25(40000
q)D5(800q)df2e3ec2cd8c15(2000q)ec3(100q22322244236
22224426
=+−−+−++−−−+
±−+−+++−
which is precisely the same as Cayley’s resolvent sextic for the quintic: 0fex5dx10cx10bx5ax 2345 =+++++ for a=1, b=0, and D as the quintic discriminant. One can only marvel at the amount of patience Cayley must have had to derive that sextic by hand. We can also ask if our t is a combination of the roots iy similar to our p. Define 2q as,
315154143532421
2151545434323212
yyyyyyyyyyyyyyy
yyyyyyyyyyyyyyyq
−−−−−
++++=
and roots iy as above, and we get, 55)uuuuuuuu(q 2
244
231
223
212 +−−−=
From (6) we then have, 5t10q 2 −= a result similar for the polynomial in p. Now that we have our p and t, we can determine 1R and 2R in (9) and (10), namely,
)pc()pc(4)td(R 221 +−+−=
)pc()pc(4)td(R 222 −+++=
From (7), (11), and (12), we can also determine, pcuu 32 −−= pcuu 41 +−=
2
Rtduu 1
32
1++−= ,
2Rtd
uu 12
24
−+−=
2
Rtduu 2
12
2+−−= ,
2Rtd
uu 24
23
−−−=
which is enough information to find the 5
iu which are the roots of our resolvent quartic. Before we do so, we can exploit certain symmetries in our variables to simplify our notation. Let, d-t = v d+t = w c-p = m c+p = n Thus, we have,
221 vnm4R +=
222 wmn4R +=
Using the algebraic identities for the 5
iu found in (13.1), we then have,
2
222
22
51
n8
wmn4wvnm4vu
++−
++−
=
2
222
22
52
m8
vnm4vwmn4wu
+−−
++−
=
2
222
22
53
m8
vnm4vwmn4wu
++−
+−−
=
2
222
22
54
n8
wmn4wvnm4vu
+−−
+−−
=
and we now have the roots of our resolvent quartic! While these expressions can indeed enable us to express the quintic roots in radicals, one can easily see two immediate problems: 1st, it involves square roots, hence a consequential ambiguity in sign, and 2nd, and more serious, there is division by a variable, which potentially could be zero. We will have to address these problems to maintain the integrity of our method. Since the 5
iu are roots of a quartic, one solution is to form the actual quartic itself to see if there are any cancellations. Let, 0)uz)(uz)(uz)(uz( 5
45
35
25
1 =−−−− We get,
0nmz2
RR)wmvn(vwmwvnwnm2vnm2
zmn2
RR)nvmw(mvwwnvvwnm6mn2nm2
znm2
RR)wnvm(vwnwvmnwm2vmn2z
55213323234224
22122332266
322
2122222244
4
=+
−++++−
−+++++−
−+++++
(20)
with 21RR as defined above. Note that the constant term of the Lagrange resolvent of the quintic
is a fifth power of the rational number 22 pcmn −= . To recall, the constant term of the Lagrange resolvent of the cubic was the third power of a rational number. It shouldn’t be hard to guess the power of the constant term of the Lagrange resolvent of the solvable 7th degree equation. We can call (20) as the generic Lagrange resolvent quartic. While there are still variables as divisors, we can use the presence of 21RR to remove the ambiguity with regards to sign since we can express its unique determinate value using other expressions in two ways. Case 1. If 0p ≠ We can use either (13) or (14) to define the determinate value of 21RR . Expressing (13) in terms of m, n, v, w, we have, )enmnm(mn2)mn(vwRR)mn( 22
21 −+−++=− (21) and (14) becomes,
2222
44222221
22
vwnwvm
nwm2vmn2)nm)(nm)(wv(10nfm2RR)wnvm(
−−
−−−−−=− (22)
Substituting (21) into the coefficients of z and 2z , and (22) into the coefficient of 3z of our generic quartic, we get,
0nm
znm
v)wmmnemn(w)vnnmnem()wv)(nm(nm
znm
wvvwvnwmemwenv)wvw3v(mnnm
z))nm)(wv(5f(z
55
246424642222
233232322
2255
34
=+
−
+−−+−−−−−
−−+−+−−++++−
−−−+
(23)
Since m-n = -2p, thus the restriction 0p ≠ . Case 2. If 0p =
If p = 0, we can’t use (21). However, we can still use (22). Since m = c-p and n = c+p, if p = 0, then m = n = c. If we substitute (22) into our generic quartic taking into account m = n = c, we get,
0cfzc
z)c/)wvwv(vw)wvw5v(c)wv(cfc2(fzz105
222222534
=+−
+++++−++−++
which can be simplified further. If p = 0, we can immediately see from (15) and (16) that, 0cedct 232 =+−− which justifies the peculiar arrangement of (15) and (16). The unknown t is then,
cedct 23 −+±= where the choice of sign doesn’t matter since v = d-t and w = d+t are symmetric in the 2z term above. Expressing the resolvent in the original coefficients, it has the surprisingly simple form,
0cfzcz)cdf2ceed4ec5c2(fzz 1052223534 =+−++−−++ (24) with no denominators, so no conditions other than p = 0.
When p = 0, it means the constant term of our sextic in p (17) is zero. Thus, a
parametrization of Case 2 can be given by setting the constant term of (17) to zero, which turns out to be just a quadratic in f. Letting the variables c, d, e as arbitrary, one can then easily solve for f.
So finally we have attained our objective, to find the explicit resolvents—the quartic
Lagrange and the two sextic auxiliary resolvents—to solve the reduced quintic, 0fey5dy10cy10y 235 =++++ such that a real root is given by, 5/1
45/1
35/1
25/1
1 zzzzy +++= or for our original quintic, 0FExDxCxBxx 2345 =+++++ with a real root given by,
5
zzzzBx
5/14
5/13
5/12
5/11 ++++−
=
in a manner analogous to the solution of the quadratic and cubic or, in general, of all solvable equations of prime degree.
In addition, the explicit quartic we have derived indicates it is possible a root or more is zero as in the case 022 =−= pcmn . Parametric examples will be provided later. III. Pair Quintics Before giving examples, it should be mentioned that there is a novel consequence to the method that we have used. To recall, the generic resolvent quartic involved square roots, namely
21RR . We circumvented that problem by finding an expression that gave its definite value, hence its correct sign. But we can ask: what if the opposite sign was used? Would it be the resolvent for some other quintic? It turns out the answer is yes and in fact we can easily find this other quintic. Since the resolvent quartic involves the variables m, n, v, w, or c, d, p, t, the two quintics must have identical values for those and differ only in the remaining free variables, namely e and f, which should indicate how to find the other quintic.
We can find e and f in (15) and (16) where they are linear in (15) and where e is a quadratic in (16). Expressing (15) and (16) in m, n, v, w and solving for e and f, and letting c = (m+n)/2, d = (v+w)/2 , identities easily derived from how we defined m, n, v, w, then we can have a third theorem. Theorem 3. The pair of quintics with 4 free parameters, 0fye5y)wv(5y)nm(5y 11
235 =++++++
0fye5y)wv(5y)nm(5y 22235 =++++++
where,
mn2RR)nm(vw)nm(
nmnme 21221
−++++−=
mn2RR)nm(vw)nm(
nmnme 21222
−−+++−=
and,
2221
222244
1nm2
RR)wnvm(vw)wnvm(nwm2vmn2)nm)(wv(5f
−+++++−−=
2221
222244
2nm2
RR)wnvm(vw)wnvm(nwm2vmn2)nm)(wv(5f
−−++++−−=
are solvable for all m, n, v, w where 0≠mn with the solution given by the resolvent quartic in (20),
0nmz2
RR)wmvn(vwmwvnwnm2vnm2
zmn2
RR)nvmw(mvwwnvvwnm6mn2nm2
znm2
RR)wnvm(vwnwvmnwm2vmn2z
55213323234224
22122332266
322
2122222244
4
=+
−++++−
−+++++−
−+++++
such that,
5/14
5/13
5/12
5/11 zzzzy +++=
for the positive and negative cases of )wmn4)(vnm4(RR 222221 ++= , appropriately
applied.
There are two ways to apply this theorem. First, one has a solvable family with 4 parameters that can be arbitrary. In fact, it is a near-complete parametrization. For all solvable quintics, one can find m, n, v, w, using the 2 sextic resolvents. However, if mn = 0, then it can’t be placed in this form. If otherwise, it belongs to this family.
Second, given a known quintic with rational coefficients, and with m, n, v, w solved for, one can then find a second solvable quintic also with rational coefficients and whose discriminant seems to be related to the first, as will be shown later in the examples. For lack of a name, they can be called pair quintics, being two quintics having the same variables m, n, v, w, as defined in this paper, namely, d-t = v d+t = w c-p = m c+p = n It may be possible though, with 0RR 21 = , that the two quintics are identical. IV. Examples Because the method is quite straightforward, we can easily use parametric examples. All parametric families cited here, unless indicated otherwise, have been found by the author. Example # 1 Consider a version of the quintic that Ramanujan (1887-1920) submitted as a challenge, namely, 01r3r2rrr 2345 =+++++
a quintic associated with an elliptic function of period 79− . Such polynomials are also known as Weber class polynomials. Another, for period 127− , is, 01r3rr2rr 2345 =++−−+ With more data from Kluners and Malle’s website, and enough pen and paper doodling, one can eventually observe that the two are members of a 2-parameter solvable family, namely, 0ar)aa2(r)ba2a(r)1ba3(r)2a(r 2222345 =+−+++++++++ where the above are both a = -1, and b = 3 and 0, respectively. We can also mention another 2-parameter family found by Kondo and Brumer, 0abxx)b21aa(x)3ab(x)3a(x 22345 =++−−−++−+−+ which has a suspicious similarity to the previous family not only in form, but in its connection to Weber class polynomials. Let a = 1, and b = 0, 1, we get, 01xx2x2x 2345 =+−+−
01xx3x3x2x 2345 =++−+−
which are Weber polynomials, for periods 47− and 103− . The relation between the 2 families turn out to be quite simple. Let, r = x-1 and the first family transforms into the second! In fact, the reduced form of each family converges to identical forms. Thus the Weber class polynomials of the discriminants 47, 79, 103, and 127 for quadratic imaginary fields with class number 5 belong to the same parametric family. However, considering Theorem 3, it shouldn’t be so surprising. We’ll solve the second family and leave the first to the interested reader. For simplicity as this is just an example, we’ll let a =1. 01bxx)b21(x)2b(x2x 2345 =+++−++− To get the reduced form let x = (2+y)/5
0)2897b450(y)b75140(y)b10015(y)b2510(y 235 =+++−−+++ so, c = (10+25b)/10 d = (15-100b)/10 e = -(140+75b)/5 f = (450b+2897)
Discriminant D = 22315 )47b24b28b4(5 +++ From (17) and (19),
2
5)4b(p
−= ,
55)b823(
t+−
=
and thus, ( ) 2/5)4b()b52(pcm −−+=−=
( ) 2/5)4b()b52(pcn −++=+=
( ) 2/5)b823()b203(tdv ++−=−=
( ) 2/5)b823()b203(tdw +−−=+= This is a case 1 quintic so we form its resolvent quartic given by (23) and we get,
0)b5b1519(z)b6250
b13125b382500b432125b3362750b4063850b49850502614747(
z)b625b375b25250b13075b3035026919(z)b200b225597(z
527
65432
25432324
=++−++
−−−−−−−+
−+−−−++++
and we have a parametric quartic to solve our parametric quintic. For a specific example, let b = 0,
01xx2x2x 2345 =+−+− with discriminant 215475D = . Resolvent is, 019z2614747z26919z597z 5234 =−−++ where,
4/)5202530(4755215597z 4&1
−±+−=
4/)5202530(4755215597z 3&2
+±−−=
so,
...576471.05
zzzz2x
5/14
5/13
5/12
5/11 −=
++++=
Its partner, unreduced from its reduced form, is given by,
090252772
x95
188xx2x2x 2345 =−+−+−
with discriminant 8
229
1922320648475
D = .
Thus the discriminant numerators have the common factor 47, the powers of 5 being trivial. It seems the discriminants of pair quintics, or their numerators if they are not whole numbers, always have a common factor, though no rigorous proof will be provided in this paper. For this particular family, by forming the parametric pair quintics from the m, n, v, w, it is easy to see that the common factor is 223 )4724284( +++ bbb . Example # 2 It was mentioned that the resolvent quartic may in fact have a root equal to zero and we can provide a parametric example. In addition, this family has parameters such that with p known and t unknown, making (15) as a cubic in t and (16) as a quartic in t, (15) and (16) factors completely into linear factors and results would be ambiguous were it not for the sextic resolvent in t, which can only have one linear factor.
0b
1b5b4bx5x)b1(5x5x
23235 =
−+−+++++
so, c = 5/10 d = 5(1+b)/10 e = 1 b/)1b5b4b(f 23 −+−=
4223456 b/)1b5b8b31b76b31b(D −−++−+= From (17) and (19) p = 1/2, t = (1-b)/2 If we used (15) we would have had,
t = (1-b)/2, t = (1+b)/2, t = (-3-b)/2 making things ambiguous, and thus showing the advantage of using (19). Knowing the correct t, we then have, v = b, w = 1, m = 0, n = 1 Since this is a case 1 quintic, we use (23) again, but this time it factors 0)bz1)(zb)(zb(z 2 =+−++ and we have our resolvent. Specific example, let b = 3, 03/5x5x20x5x 235 =++++
Then, ...99483.1)3/1(33x 5/15/25/1 −=+−−= Example # 3 It is possible that the resolvent quartic may have not one but two roots equal to zero and effectively the quintic is solvable only by a quadratic. We can use the deMoivre quintic as an example. Let, 0fxb5bx5x 235 =+++ So, c = b/2, d = 0, 2be = , f = f, 225 )fc4(D += From (17) and (19), p = -b/2, t = 0 and, v = 0, w = 0, m = b, n = 0 Quartic from (23) reduces to: 0)bfzz(z 522 =−+ Specific example, let b = 2, f = 3, 03x20x10x 35 =+++
...14836.02
13732
1373x
5/15/1
−=
−−+
+−=
Example # 4 We will give an example for a Case 2 quintic, as well as for a quintic whose discriminant is not a square.
0)441n37nn(x)14n15nn(5x)18n3n(5x10x)x(F 2342322351 =++−+−++−++++=
1c = 2/)18n3n(d 2 ++=
)14n15nn(e 23 −++−=
)441n37nn(f 234 ++−=
2432322 )nn3n26n20225()nn217()n21(D +++++++−=
While the previous examples had a square discriminant, it is possible that the discriminant of a solvable quintic may not be a square. As Dummit pointed out in his Theorem 2, the Galois group of a solvable quintic is the Frobenius group of order 20 if and only if the discriminant is not a square. From (17), p = 0 Since now this is a Case 2 quintic, we need not derive t and can just directly form the resolvent quartic. From (24),
01z)441n37nn(z)3530n4326
n2623n1123n312n70n9n(z)441n37nn(z2342
23456732344
=+++−−++
++++++++−+
Specific example, let n = 1, 0478x15x110x10x 235 =+−++ Quartic is, 01z478z11994z478z 234 =+−++ where,
2/)5454110225(10595239z 4&1
−±+−=
2/)5454110225(10595239z 3&2
+±−−=
Thus, ...50991.4zzzzx 5/1
45/1
35/1
25/1
1 −=+++= We can also compare )(1 xF and its partner. The other quintic is given by,
0)171n54n31n4n)(1n(
x)14n15nn(5x)18n3n(5x10x)x(F234
2322352
=+++++−
−++−++++=
While the discriminant of )(1 xF is,
2432322 )nn3n26n20225()nn217()n21(D +++++++−= the discriminant of )(2 xF is,
243222322 )nn5n19n4357()nn321()nn217()n1(D +++++++++−=
with a common factor which interestingly enough is the cube. We can also cite the situation with the quintic studied by Emma Lehmer (1906- ) which turns out to be a Case 2 quintic,
01x)10n10n4n(
x)5n15n11n5n(x)10n10n6n2(xnx)x(F23
22343234251
=+++++
++++++++−+=
and its partner,
025/)5n10n5n)(20n10n5n(x)10n10n4n(
x)5n15n11n5n(x)10n10n6n2(xnx)x(F232323
22343234252
=++++++−++++
++++++++−+=
which was derived by reducing the first quintic, getting m, n, v, w, forming the second quintic, and un-reducing it. The discriminants, respectively, are:
4432232151 )nn5n15n2525()nn5n107(5D +++++++=
44322322329
2 )nn5n15n2525()n4n10n2525()n3n20n5050(5D ++++++++++= again with a common factor, this time, the one with the 4th power. We can ask the question: Do the numerators of the discriminants of pair quintics always have a common factor, other than the trivial powers of 5? Though it won’t be explored any further in this paper, there seems to be something interesting going on. V. Beyond The Quintic The construction of the explicit Lagrange resolvent quartic was motivated by several things. First, was simply the aesthetic: the desire to see the quartic with rational or even integral coefficients that would somehow result from weaving together the irrational roots of the solvable quintic. Second, if it was done symbolically, whether it could be done with an economy of symbols, since there is the possibility that, starting with complicated expressions, by combining them, they may reduce into something simpler. And third, these simpler expressions might allow one to circumvent awkward situations like division by zero. The most obvious drawback of the approach is that the labeling of the quartic roots as
54
53
52
51 uuuu ,,, disappears, and thus to express the other four roots of the quintic in radicals
would entail additional effort. However, since the two forms of the quartic resolvent allows one to avoid the possibility of division by zero, a possibility present in the expressions for the 5
iu ,
then perhaps the loss is acceptable. And since the 5iu come in conjugate pairs, it would take only
a little judicious permutation to recover the labeling. Beyond the quintic, the next prime degree is the septic. And while in older texts, like in Burnside’s classic book on the theory of equations, it was sometimes referred to as the septemic, with its present name a websearch for “septic equation” can give one interesting references to sewage treatment. Not much seems to be known about solving the solvable septic, other than the complete list of transitive groups of that degree. It seems to be understandable since, while the quintic
needs a sextic auxiliary resolvent, the septic needs a 120-degree auxiliary resolvent! However, given the Lagrange resolvents of a solvable septic, one can construct a solvable sextic in analogy to what was done for the quintic. To give a concrete example, consider the septic Weber class polynomial, 01xxxxxx2x 234567 =−−+++−− which has the solution,
...13061.27
zzzzzz2x
7/16
7/15
7/14
7/13
7/12
7/11 =
++++++=
where the iz are the roots of the sextic,
0461z614282391563429731
z965914675972280z3928463233514z718991529z120671z7
23456
=−+
−−+−
with the constant term a 7th power, as expected. Not much is also known about the sextic resolvent of the septic. It might be interesting to know how many of its roots can be zero, or its factorizations if it is reducible. The above sextic, however, is irreducible but is solvable. And how to solve the solvable sextic? That should be a good question to answer. Almost a thousand years later and several thousand kilometers away from Samarkand, a young mathematician with an interest in poetry sat down on the powdery white sand of a beach on a tropical island. He too was working on the theory of equations. He had an algorithm to solve solvable equations and was te sting it on quintics, beginning with the solvable 032205 =++ xx . He pressed some keys on his laptop and the solution, which was so elusive centuries ago, appeared on the screen, namely, 5/1
45/1
35/1
25/1
1 zzzzx +++= where,
+−= )5225(555)25/4(1z
−−−= )5225(555)25/4(2z
−+−= )5225(555)25/4(3z
++= )5225(555)25/4(4z
which was approximately –1.36396.
After testing it several more times, and satisfied that the programming had no bugs, he turned off his laptop and took out a volume of poetry from his knapsack. He opened it to where he last finished and started to read. “The Moving Finger writes, and having writ, Moves on; nor all thy Piety nor Wit Shall lure it back to cancel half a Line, Nor all thy tears wash out a Word of it.” He paused, as a thought came to him. He idly traced circles with his finger on the fine white sand. Interesting, he wondered, that the word “calculus” meant “pebble” in Latin. Maybe the Moving Finger also writes the laws of Nature in the sandbox of mathematics. After a moment, he got out of his reverie, placed the book on top of his other things, and stood up to take a dip in the warm inviting water. On the cover of the book was the title: The Rubaiyat of Omar Khayyam…
-End- Copyright 2003, 2004 by Titus Piezas III December 2003 Revised: August 25, 2004 [email protected]
The author pleads artistic license with the story. While Omar Khayyam (1048-1131) did solve particular cubics prior to del Ferro, there is no evidence that he, nor other mathematicians of his time, considered higher degree equations.
This paper is dedicated to the author’s family, especially to his nephews and nieces. Optime te habeas te desidero. Have the best in life.
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