012: Cubic Polynomial as a kth Power



II. Cubic Polynomials as kth powers


A. Univariate: ax3+bx2+cx+d2 = tk

  1. ax3+bx2+cx+d2 = t2
  2. ax3+bx2+cx+d2 = t3

B. Bivariate: ax3+bx2y+cxy2+dy3 = tk

  1. x3+y3 = t2
  2. ax3+by3 = t2
  3. x3+y3 = nz2
  4. x3+ax2y+bxy2+cy3 = t2
  5. x3+ax2y+bxy2+cy3 = t3


Given an initial soln to the curve F(v):= av3+bv2+cv+d = t2, it is easy to do a small transformation F(v) → F(x) such that F(x) has a square constant term, a method discussed in the section on Quadratic Polynomials as kth powers.  Thus, for convenience, in the univariate case we will already assume that d is a perfect square.


A. Univariate: ax3+bx2+cx+d2 = tk


1. ax3+bx2+cx+d2 = t2


The general method is given by Fermat.  One can solve this in two ways as:


ax3+bx2+cx+d2 = (px+d)2, or,


ax3+bx2+cx+d2 = (px2+qx+d)2


where p,q are free variables. Expanding and subtracting one side from the other yields,


ax3 + (b-p2)x2 + (c-2dp)x = 0                                            (eq.1)


p2x4 + (-a+2pq)x3 + (-b+2dp+q2)x2 + (-c+2dq)x = 0        (eq.2)


For (eq.1), p can eliminate the x1 term.  For (eq.2),  p,q can do so for the x1 and x2 terms.  For both, one can then solve for x giving,


x = (c2-4bd2)/(4ad2)


x = 8d2(c3-4bcd2+8ad4)/(c2-4bd2)2


as two solns to ax3+bx2+cx+d2 = z2. 



2. ax3+bx2+cx+d = t3


For the monic case a = 1, what Fermat did was equate,


x3+3bx2+3cx+d = (x+b)3, 


then solved for x as 3x = (-b3+d)/(b2-c) though there are b,c,d such that the method is inapplicable. In fact, given an initial rational solution to ax3+bx2+cx+d = t3, one generally can find an infinite more. See also Form 5.



B. Bivariate: ax3+bx2y+cxy2+dy3 = tk


1. Form: x3+y3 = t2


Euler (also by R. Hoppe)


(3m4+6m2n2-n4)3 + (-3m4+6m2n2+n4)3 = (6mn(3m4+n4))2



2. Form: ax3+by3 = t2


Using the approach by Krafft, Lagrange discussed in Form 21 above, we can set one variable as zero, say, z = q2+2pr = 0, and solving for r, we can solve, 


x3+ny3 = t2


as {x,y} = {4p(p3-nq3), q(8p3+nq3)}, or more generally,


ax3+by3 = t2 


{x,y} = {4p(ap3-bq3), q(8ap3+bq3)}


However, for tk with k>2, it already involves polynomials in p,q,r of degree >1 so it is not so easy to set z=0 and find a general integral soln to ax3+by3 = tk for k>2.  Is there such a soln for other k?  An alternative soln to k=2, for a=b=1 and where x,y are relatively prime integers is,


R. Hoppe (also by Euler)


(p4+6p2q2-3q4)3 + (-p4+6p2q2+3q4)3 = (6pq)2(p4+3q4)2


This can be modified to solve x3+y3 = nz2.


 3. Form: x3+y3 = nz2


E. Fauquembergue


(p2+6pq-3q2)3 + (-p2+6pq+3q2)3 = pq (6(p2+3q2))2


so it suffices to find the factorization of n as n = pq.  Notice this is essentially the Euler-Hoppe identity. Finally, one soln implies a second,




(x3+4y3)3 + (-3x2y)3 = nz2(x3-8y3)2,  if x3+y3 = nz2



4. Form: x3+ax2y+bxy2+cy3 = t2


Lagrange, Legendre


This is the more general equation and has the beautiful soln,


{x,y}  = {u4-2bu2v2-8cuv3+(b2-4ac)v4,  4v(u3+au2v+buv2+cv3)}


for arbitrary variables u,v and where t is a sixth degree polynomial. 


Derivation: This can be found by generalizing f3.  To recall,


f3:= Resultant[x+wy+w2z, w3-n, w]


However, if we extend w3-n to the more general cubic v3-av2+bv-c = 0, this results in a long trivariate polynomial, or f31,


f31: = Resultant[x+vy+v2z, v3-av2+bv-c, v]


rather tedious to explicitly write down.  For a=b=0 and c=n, this just reduces to f3.  Lagrange noted that if the variable z = 0, the equation f31 = tk naturally reduces to the bivariate form x3+ax2y+bxy2+cy3 =  tk.  So using the system,


x+v1y+v12z = (p+v1q+v12r)k

x+v2y+v22z = (p+v2q+v22r)k

x+v3y+v32z = (p+v3q+v32r)k


with the vi as the three roots of the cubic v3-av2+bv-c = 0, one first solves for x,y,z.  For k=2, a soln to f31 = tk, after much simplication, is given by,


{x,y,z} = {p2+cr(2q+ar),  2pq-2bqr-(ab-c)r2,  q2+2(p+aq)r+(a2-b)r2}


One can then set z = 0, solve for p to get a soln purely in {x,y}.  By using a small transformation, Legendre found a more aesthetic version (given at the start of this section) as,


{x,y}  = {u4-2bu2v2-8cuv3+(b2-4ac)v4,  4v(u3+au2v+buv2+cv3)}


which solves x3+ax2y+bxy2+cy3 = t2.



5. Form: x3+ax2y+bxy2+cy3 = t3


For k>2, unfortunately it involves polynomials in p,q,r of degree >1 so it is harder to set z=0.  Using another method, one can still find k=3. 


Theorem: There is generally an infinite family of rational solutions {x,y} to x3+ax2y+bxy2+cy3 = t3.


For simplicity’s sake, we can transform u3+au2v+buv2+cv3 = t3 into one with a=0 using the transformation u = x-ay, v = 3y to get,


x3 - 3(b2-3c)xy2 + (2b3-9bc+27d)y3 = t3


or simply x3+pxy2+qy3 = t3.  To solve this, using a birational transformation, we get,


P. von Schaewen, J. von Sz.Nagy


x3+pxy2+qy3 = z3


{x,y,z} = {pm2-3n2±t,  6mn,  pm2+3n2±t}, if {m,n,t} satisfy the elliptic curve E,


E: = p2m4+12qm3n-6pm2n2-3n4 = t2


One soln of which is,


{m,n,t} = {u2-3q2v,  p2qv,  pu(u3-3q2v2)}, where {u,v} = {p3+9q2,  3(p3+6q2)}.


Since we can use ±t, this yields two solns x,y,z.  Using the negative case, after removing common factors this gives,


{x,y} = {-q(u2-6q2v),  p(u2-3q2v)}


with {u,v} as defined above, and so on for the positive case.  From this initial {m,n}, one can then come up with an infinite number of rational solns.  As a last point, we can extrapolate Lagrange’s method to quartics (and higher).  Forming,


f41:= Resultant[x+vy+v2z+v3w, v4-av3+bv2-cv+d, v]


one can easily solve f41 = tk for any positive integer k by solving the system,


x+viy+vi2z+vi3w = (p+viq+vi2r+vi3s)k


for {x,y,z,w} and where the vi are the four roots of v4-av3+bv2-cv+d = 0.  To reduce f41 to the bivariate form by setting two variables z = w = 0 unfortunately entails solving an equation of degree >1 even for just k=2, hence to find a bivariate or univariate polynomial soln to x4+ax3y+bx2y2+cxy3+dy4 = t2 is a more difficult problem than for the cubic case.  A limited soln is possible though, as we’ll see in the section on fourth powers.



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