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19. Form: ax3+by3+cz3 = N
Theorem 1: If ax3+by3+cz3 = N has a solution, then one can generally find a second. (A. Desboves).
Proof: ap3+bq3+cr3 = (ax3+by3+cz3)(ax3-by3)3
{p,q,r} = {(ax4+2bxy3), -(2ax3y+by4), z(ax3-by3)}
Special cases a=b=1 were found by J.Prestet and A. Legendre while a different version was given by Cauchy. (Desboves also found a rather similar identity for fourth powers.) Special cases of the general form are given by,
S. Realis (quadratic)
(2x2-4xy+9yz-9z2)3 + (2y2-xy+9xz-18z2)3 = 9(2x2-4xz-yz+y2)3, if x3+y3 = 9z3
Piezas
Let q = p3+1, then
(px2-p2xy+qyz-qz2)3 + (py2-xy+qxz-pqz2)3 = q(px2-p2xz-yz+y2)3, if x3+y3 = qz3
(Realis’s was the case p=2 of this identity.)
E.Lucas (nonic)
p3+q3+cr3 = 27 (x3+y3+cz3) (x7y+x4y4+xy7)3
{p,q,r} = {x9+6x6y3+3x3y6-y9, -x9+3x6y3+6x3y6+y9, 3xyz(x6+x3y3+y6)}
More generally,
Theorem 2: If ax13+a2x23+...+ amxm3 = N has a solution, then this generally leads to a second.
Proof: Desboves’ identity in Theorem 1 is essentially the same as the difference of cubes,
a(ax4+2bxy3)3 - b(2ax3y+by4)3 = (ax3+by3)(ax3-by3)3
If ax3+by3 is the sum of cubes c1z13+c2z23+…+ cmzm3 = N, then,
a(ax4+2bxy3)3 - b(2ax3y+by4)3 = (c1z13+c2z23+…+ cmzm3)(ax3-by3)3 = 0
so, by distribution, one gets a new identity of the form ax13+a2x23+...+ amxm3 = 0 if there are initial solns x,y, and zi. (End proof)
Form p3+q3 = nr3
Expressing a given number n as the sum of two rational cubes is simply the case a = b = 1, and c = 0, of Desboves' identity above. The fact that an initial soln leads to more can be seen in a better light by using the birational transformation, {p,q,r} = {36n-y, 36n+y, 6x}, to reduce the eqn to the elliptic curve,
y2 = x3-432n2
and can be reversed as,
{x,y} = {12n/(p+q), -36n(p-q)/(p+q)}
More generally though,
Theorem 3: If au3+bv3+cw3 = 0 has a solution, then so does y2 = x3-432a2b2c2.
Proof: y2-x3+432a2b2c2 = (au3+bv3+cw3)(au3-bv3-cw3)(432b2c2/u6)
{x,y} = {4(p2+q)/r2, 4(2p3+3pq)/r3}, where {p,q,r} = {bv3-cw3, 3bcv3w3, uvw}
Theorem 4: If au3+bv3+cw3 = 0 has a solution, then so does x3+y3+abcz3 = 0.
Proof: x3+y3+abcz3 = 27bcv3w3(au3+bv3+cw3)(p2+q)3
{x,y,z} = {p3+3bqv3, -p3+3cqw3, 3(p2+q)r}, where {p,q,r} as in theorem 3.
A generalization was found by J.Sylvester. (Who found theorems 3 and 4?). }
20. Form: ax3+by3+cz3 = dxyz
S. Realis (complete, other than the case x = y = z)
x3+y3+z3 = 3xyz
{x,y,z} = {(a-b)3+(a-c)3, (b-a)3+(b-c)3, (c-a)3+(c-b)3}
All positive integers N other than those div by 3 but not by 9 are representable as N = x3+y3+z3-3xyz with integral x,y,z => 0. The primes (other than 3) are representable in this manner in one and only one way (Carmichael).
Theorem 1: If ax3+by3+cz3+dxyz = 0 has a solution, then one can generally find a second. (A. Cauchy)
Proof: ap3+bq3+cr3+dpqr = (ax3+by3+cz3+dxyz)(pqr)/(xyz)
{p,q,r} = {x(by3-cz3), y(-ax3+cz3), z(ax3-by3)}
This generalizes Desboves’ result in a certain way. Cauchy original statement was that given an initial soln {x,y,z}, a second soln {p,q,r} is,
p/(v1x) = q/(v2y) = r/(v3z)
{v1, v2, v3} = {by3-cz3, -ax3+cz3, ax3-by3}
After a little algebraic manipulation, one can get the above identity.
E. Lucas
Similarly, given an initial soln {x,y,z}, a second {p,q,r} satisfies the elegant relations p/x+q/y+r/z = 0 and apx2+bqy2+ crz2 = 0. One can then solve for {p,q} which gives essentially the same result as Cauchy’s.
Theorem 2: If ax3+by3+cz3+dxyz = 0 has a soln, then so does p3+q3+abcr3+dpqr = 0. (J. Sylvester)
Proof: {p,q,r} = {f2g+g2h+h2f-3fgh, fg2+gh2+hf2-3fgh, xyz(f2+g2+h2-fg-fh-gh)}, where {f,g,h} = {ax3, by3, cz3}.
Note that this generalizes theorem 4 of the previous section which is just the case d=0. For a=b=1, d=0, and after some simplification this also gives the nonic parametrization of Lucas.
C.Souillart, E. Mathieu
x3+ny3+n2z3-3nxyz = (p3+nq3+n2r3-3npqr)(u3+nv3+n2w3-3nuvw)
{x,y,z} = {pu+n(qw+rv), pv+qu+nrw, pw+qv+ru}
which proves that, like the quadratic form x2+y2, the product of two forms x3+ny3+n2z3-3nxyz is of like form. In general, the result extends to cyclic determinants of order m.
21. Form: x3+y3+z3-3xyz = tk
Krafft, Lagrange
To recall, the algebraic form we can designate as f2:= x2-dy2 factors over the square root extension d1/2 (which may involve the complex unit). For its cubic analogue, use the general form f3:= x3+ny3+n2z3-3nxyz which factors linearly over a complex cube root of unity w as,
f3:= x3+ny3+n2z3-3nxyz = (x+my+m2z)(x+mwy+m2w2z)(x+mw2y+m2wz)
where w3-1 = 0 and m = n1/3 for convenience. Or alternatively, if one has Mathematica,
f3:= Resultant[x+wy+w2z, w3-n, w]
Because it has three linear factors and variables, one can then easily solve the equation,
x3+ny3+n2z3-3nxyz = (p3+nq3+n2r3-3npqr)k
for any positive integer k (just like for f2) by factoring both sides. In fact, for the special case f3 = ±1, this is sometimes known as the cubic Pell equation, an analogue to the Pell equation f2 = ±1 since starting with one initial soln {p,q,r}, one can then find an infinite number of solutions {x,y,z}. Specifically, one solves the system,
x+my+m2z = (p+mq+m2r)k x+mwy+m2w2z = (p+mwq+m2w2r)k x+mw2y+m2wz = (p+mw2q+m2wr)k
for x,y,z and where m = n1/3. For k=2, this gives,
{x,y,z} = {p2+2nqr, nr2+2pq, q2+2pr}
and so on for other k.
II. Cubic Polynomials as kth powers
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+d2 = 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.
B. Bivariate: ax3+bx2y+cxy2+dy3 = tk
3. Form: x3+y3 = t2
Euler (also by R. Hoppe)
(3m4+6m2n2-n4)3 + (-3m4+6m2n2+n4)3 = (6mn(3m4+n4))2
4. 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.
5. 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,
A.Gerardin
(x3+4y3)3 + (-3x2y)3 = nz2(x3-8y3)2, if x3+y3 = nz2
6. 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]
By setting n=1 and using a root of the general cubic v3-av2+bv-c = 0 (instead of a complex cube root of unity) 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.
7. 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 polynomial 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)}
and so on for the positive case. From this initial {m,n}, one can then come up with an infinite number of polynomial 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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