*II. Cubic Polynomials as kth powers*

A. Univariate: ax^{3}+bx^{2}+cx+d^{2} = t^{k}

- ax
^{3}+bx^{2}+cx+d^{2} = t^{2}
- ax
^{3}+bx^{2}+cx+d^{2} = t^{3}

B. Bivariate: ax^{3}+bx^{2}y+cxy^{2}+dy^{3} = t^{k}

- x
^{3}+y^{3} = t^{2}
- ax
^{3}+by^{3} = t^{2}
- x
^{3}+y^{3} = nz^{2}
- x
^{3}+ax^{2}y+bxy^{2}+cy^{3} = t^{2}
- x
^{3}+ax^{2}y+bxy^{2}+cy^{3} = t^{3}

Given an initial soln to the curve F(v):= av^{3}+bv^{2}+cv+d = t^{2}, 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: ax**^{3}+bx^{2}+cx+d^{2} = t^{k}

**1. ax**^{3}+bx^{2}+cx+d^{2} = t^{2}

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

ax^{3}+bx^{2}+cx+d^{2} = (px+d)^{2}, or,

ax^{3}+bx^{2}+cx+d^{2} = (px^{2}+qx+d)^{2}

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

ax^{3 }+ (b-p^{2})x^{2 }+ (c-2dp)x = 0 (eq.1)

p^{2}x^{4 }+ (-a+2pq)x^{3 }+ (-b+2dp+q^{2})x^{2} + (-c+2dq)x = 0 (eq.2)

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

x = (c^{2}-4bd^{2})/(4ad^{2})

x = 8d^{2}(c^{3}-4bcd^{2}+8ad^{4})/(c^{2}-4bd^{2})^{2}

as two solns to ax^{3}+bx^{2}+cx+d^{2} = z^{2}.

**2. ax**^{3}+bx^{2}+cx+d = t^{3}

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

x^{3}+3bx^{2}+3cx+d = (x+b)^{3},

then solved for *x* as 3*x* = (-b^{3}+d)/(b^{2}-c) though there are b,c,d such that the method is inapplicable. In fact, given an initial rational solution to ax^{3}+bx^{2}+cx+d = t^{3}, one generally can find an infinite more. See also **Form 5**.

**B. Bivariate: ax**^{3}+bx^{2}y+cxy^{2}+dy^{3} = t^{k}

**1. Form: x**^{3}+y^{3} = t^{2}

Euler (also by R. Hoppe)

(3m^{4}+6m^{2}n^{2}-n^{4})^{3} + (-3m^{4}+6m^{2}n^{2}+n^{4})^{3} = (6mn(3m^{4}+n^{4}))^{2}

**2. Form: ax**^{3}+by^{3} = t^{2}

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

x^{3}+ny^{3} = t^{2}

as {x,y} = {4p(p^{3}-nq^{3}), q(8p^{3}+nq^{3})}, or more generally,

ax^{3}+by^{3} = t^{2}

{x,y} = {4p(ap^{3}-bq^{3}), q(8ap^{3}+bq^{3})}

However, for t^{k }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 ax^{3}+by^{3} = t^{k} 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)

(p^{4}+6p^{2}q^{2}-3q^{4})^{3} + (-p^{4}+6p^{2}q^{2}+3q^{4})^{3} = (6pq)^{2}(p^{4}+3q^{4})^{2}

This can be modified to solve x^{3}+y^{3} = nz^{2}.

**3. Form: x**^{3}+y^{3} = nz^{2}

E. Fauquembergue

(p^{2}+6pq-3q^{2})^{3} + (-p^{2}+6pq+3q^{2})^{3} = pq (6(p^{2}+3q^{2}))^{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,

A.Gerardin

(x^{3}+4y^{3})^{3} + (-3x^{2}y)^{3} = nz^{2}(x^{3}-8y^{3})^{2}, if x^{3}+y^{3} = nz^{2}

**4. Form: x**^{3}+ax^{2}y+bxy^{2}+cy^{3} = t^{2}

Lagrange, Legendre

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

{x,y} = {u^{4}-2bu^{2}v^{2}-8cuv^{3}+(b^{2}-4ac)v^{4}, 4v(u^{3}+au^{2}v+buv^{2}+cv^{3})}

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

*Derivation*: This can be found by generalizing f_{3}. To recall,

f_{3}:= Resultant[x+wy+w^{2}z, w^{3}-n, w]

However, if we extend w^{3}-n to the more general cubic v^{3}-av^{2}+bv-c = 0, this results in a long trivariate polynomial, or f_{31},

f_{31}: = Resultant[x+vy+v^{2}z, v^{3}-av^{2}+bv-c, v]

rather tedious to explicitly write down. For a=b=0 and c=n, this just reduces to f_{3}. Lagrange noted that if the variable z = 0, the equation f_{31} = t^{k} naturally reduces to the bivariate form x^{3}+ax^{2}y+bxy^{2}+cy^{3} = t^{k}. So using the system,

x+v_{1}y+v_{1}^{2}z = (p+v_{1}q+v_{1}^{2}r)^{k}

x+v_{2}y+v_{2}^{2}z = (p+v_{2}q+v_{2}^{2}r)^{k}

x+v_{3}y+v_{3}^{2}z = (p+v_{3}q+v_{3}^{2}r)^{k}

with the v_{i} as the three roots of the cubic v^{3}-av^{2}+bv-c = 0, one first solves for x,y,z. For k=2, a soln to f_{31} = t^{k}, after much simplication, is given by,

{x,y,z} = {p^{2}+cr(2q+ar), 2pq-2bqr-(ab-c)r^{2}, q^{2}+2(p+aq)r+(a^{2}-b)r^{2}}

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} = {u^{4}-2bu^{2}v^{2}-8cuv^{3}+(b^{2}-4ac)v^{4}, 4v(u^{3}+au^{2}v+buv^{2}+cv^{3})}

which solves x^{3}+ax^{2}y+bxy^{2}+cy^{3} = t^{2}.

**5. Form: x**^{3}+ax^{2}y+bxy^{2}+cy^{3} = t^{3}

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 **x*^{3}+ax^{2}y+bxy^{2}+cy^{3} = t^{3}*.*

For simplicity’s sake, we can transform u^{3}+au^{2}v+buv^{2}+cv^{3} = t^{3} into one with a=0 using the transformation u = x-ay, v = 3y to get,

x^{3 }- 3(b^{2}-3c)xy^{2} + (2b^{3}-9bc+27d)y^{3} = t^{3}

or simply x^{3}+pxy^{2}+qy^{3} = t^{3}. To solve this, using a *birational transformation*, we get,

P. von Schaewen, J. von Sz.Nagy

x^{3}+pxy^{2}+qy^{3} = z^{3}

{x,y,z} = {pm^{2}-3n^{2}±t, 6mn, pm^{2}+3n^{2}±t}, if {m,n,t} satisfy the elliptic curve *E*,

*E*: = p^{2}m^{4}+12qm^{3}n-6pm^{2}n^{2}-3n^{4} = t^{2}

One soln of which is,

{m,n,t} = {u^{2}-3q^{2}v, p^{2}qv, pu(u^{3}-3q^{2}v^{2})}, where {u,v} = {p^{3}+9q^{2}, 3(p^{3}+6q^{2})}.

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(u^{2}-6q^{2}v), p(u^{2}-3q^{2}v)}

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,

f_{41}:= Resultant[x+vy+v^{2}z+v^{3}w, v^{4}-av^{3}+bv^{2}-cv+d, v]

one can easily solve f_{41} = t^{k} for any positive integer k by solving the system,

x+v_{i}y+v_{i}^{2}z+v_{i}^{3}w = (p+v_{i}q+v_{i}^{2}r+v_{i}^{3}s)^{k}

for {x,y,z,w} and where the v_{i} are the four roots of v^{4}-av^{3}+bv^{2}-cv+d = 0. To reduce f_{41} 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 x^{4}+ax^{3}y+bx^{2}y^{2}+cxy^{3}+dy^{4} = t^{2} 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.