PART 6: Third Powers
I. Sum / Sums of cubes
II. Cubic polynomials as kth powers
A. Univariate: ax3+bx2+cx+d2 = tk
B. Bivariate: ax3+bx2y+cxy2+dy3 = tk
I. Sum / Sums of cubes
1. Sum of two cubes: x3+y3 = z3
P.Tait
(x3+z3)y3 + (x3-y3)z3 = (y3+z3)x3, if x3+y3 = z3
Or course, by Fermat’s Last Theorem, x3+y3 = z3 has no non-trivial integer solns.
2. Form: x3+y3+z3+t3 = 0
F. Vieta
a3(a3-2b3) + (a3+b3)b3 + (2a3-b3)b3 = a3(a3+b3)3
P. Sondat
(ap-q)3 + (bp-q)3 + (cp+q)3 + (dp+q)3 = (a3+b3+c3+d3)(a+b+c+d)3
{p,q} = {a+b+c+d, a2+b2-c2-d2}
Thus, given an initial soln, subsequent ones can be found. A similar identity exists for a3+b3+2(c3+d3) = 0.
Euler (complete)
(p+q)3 + (p-q)3 = (r+s)3 + (r-s)3
{p,q} = {3(bc-ad)(c2+3d2), (a2+3b2)2 - (ac+3bd)(c2+3d2)} {r,s} = {3(bc-ad)(a2+3b2), -(c2+3d2)2 + (ac+3bd)(a2+3b2)}
For details on how Euler derived this, see “Euler’s Extended Conjecture and ak+bk+ck = dk for k>4” by this author.
J. Binet (complete)
a3+b3 + n(c3+d3) = 0
a = (1 - mn(p-3q))r b = (-1 + mn(p+3q))r c = (m2n - (p+3q))r d = (-m2n + (p-3q))r
where m = p2+3q2 and r is just a scaling factor.
Proof (Piezas): For any rational soln a,b,c,d, one can always find rational p,q,r, to get those particular values using the formulas,
{p,q,r} = {(bc+ad-2(bd+ac))/(2v), (bc-ad)/(2v), -v(c+d)/(3(bc-ad))}
where v = a2-ab+b2. For example, for n=1, to get the soln {3,5,4,-6}, the formulas give {p,q,r} = {1,1,1/3}. In fact, by permuting the {a,b,c,d}, one finds there can be six distinct {p,q,r} up to sign. Binet’s original soln only discussed the case n=1 but with a little tweaking by this author, one can generalize it and a derivation will be given later. It should be pointed out that the algebraic form a2+ab+b2, or equivalently x2+3y2, appears a lot when dealing with third and fourth powers. It may be of interest that while x2+y2 = 1 defines a circle, x2+xy+y2 = 1 yields an ellipse.
J. Steggall (complete)
(s4-rs(p+3q))3 + (-s4+rs(p-3q))3 = (r2-s3(p+3q))3 + (-r2+s3(p-3q))3
if r = p2+3q2. For s=1 reduces to Binet’s (with n=1). Note that Euler, Binet, and Stegall’s solns involve quartic polynomials. Noam Elkies' complete soln uses only cubic polynomials,
N. Elkies (complete)
a3+b3+c3+d3=0
av = s3-m-2r3+(r2-s2)t+(s-2r)t2 bv = -s3+m-r3+nt-(s+r)t2 cv = t3+m+r3+nt-(s+r)t2 dv = -t3+2rs2-r2s+2r3-nt+(s+r)t2
where {m,n} = {rs2-2r2s, s2+2r2} and v is just a scaling factor.
A.Werebrusow (complete)
If 3w2xy = a2+ab+b2, then,
(-ax+wy2)3 + (ay-wx2)3 = (-bx+wy2)3 + (by-wx2)3 = ((a+b)x+wy2)3 + (-(a+b)y-wx2)3,
and,
(ax-wy2)3 + (-bx+wy2)3 = (ay-wx2)3 + (-by+wx2)3
For the second formula, set {a,b,w,x,y} = {(p-3q)r, (p+3q)r, r, 1, p2+3q2} to get Binet’s soln. Variations of this were found by K. Schwering and Ramanujan.
A.Werebrusow:
(-a2d+bc2)3 + (ad2-b2c)3 = c3(ab-cd)3 + d3(ab-cd)3, if a3+b3 = c3+d3
Thus, this is another identity such that one soln to a3+b3+c3+d3 = 0 leads to a second. Related to the identity below,
A. Desboves
(b3-d3)(a2d-bc2)3 + (a3-c3)(ad2-b2c)3 = (ab-cd)3(-b3c3+a3d3)
C. Hermite
((a+2b)c-1)3 + (-a-2b+c2)3 = ((a-b)c-1)3 + (-a+b+c2)3, if c = a2+ab+b2
S. Baba
(pq)3 + (p+6q3)3 = (p-6q3)3 + ((p+12)q)3, if p = q6-4
J. Young
(a2+6ab3-3c)3 – (a2-6ab3-3c)3 = b3(a2+6a+3c)3 – b3(a2-6a+3c)3, if c = b6-1
The above generalizes Baba’s. Also useful for finding Martin triples. There are also quadratic parametrizations though these are no longer complete.
J. Young
(p2+16pq-21q2)3 + (-p2+16pq+21q2)3 + (2p2-4pq+42q2)3 = (2p2+4pq+42q2)3
Ramanujan
(3x2+5xy-5y2)3 + (4x2-4xy+6y2)3 + (5x2-5xy-3y2)3 = (6x2-4xy+4y2)3
Other binary quadratic form solns have been found by other authors such as Gerardin, Womack, etc. A generalization has been found by this author as,
Piezas
(ax2-v1xy+bwy2)3 + (bx2+v1xy+awy2)3 + (cx2+v2xy+dwy2)3 + (dx2-v2xy+cwy2)3 = (a3+b3+c3+d3)3(x2+wy2)3
where {v1, v2, w} = {c2-d2, a2-b2, (a+b)(c+d)}
thus for any soln to a3+b3+c3+d3 = N where N is zero or any number of cubes, then one can always find a quadratic parametrization. It is also the case that,
(ax-v1y)k + (bx+v1y)k + (cx+v2y)k + (dx-v2y)k = (ax+v1y)k + (bx-v1y)k + (cx-v2y)k + (dx+v2y)k, for k =1,3
Piezas
If a+b = c+d. Let {u1, u2} = {c-d, a-b}, then,
(ax2+u1xy+by2)3 + (bx2-u1xy+ay2)3 + (cx2-u2xy+dy2)3 + (dx2+u2xy+cy2)3 = (a3+b3+c3+d3)3(x2+y2)3
There are many particular cubic equations with this property, one of which is 93+133+193+233 = 283, (9+23 = 13+19) as well as those in a nice arithmetic progression like,
113+123+133+143 = 203 313+333+353+373+393+413 = 663
the latter found by D. Rusin in the context of a certain elliptic curve. There is also a quadratic identity, similar to the general one,
J. Nicholson
(au+pv)3 + (-au+qv)3 + (bu+nv)3 + (-bu+mv)3 = ((m+n)b2+(p+q)a2)3 (m3+n3+p3+q3)
where {u, v} = {(m2-n2)b-(p2+q2)a, (m+n)b2+(p+q)a2}
3. Form: x3+y3+z3 = 1
It turns out solving this completely in the integers may involve a certain elliptic curve. The complete integral soln to this equation is given by Werebrusow’s identity with one term set equal to 1.
(1-ac+bc)3 + (a+c2-ac3)3 + (ac3-b-c2)3 = 1
if a2+ab+b2 = 3c(ac-1)2.
Proof (Piezas): To find any non-trivial soln x,y,z, use the formulas for {a,b,c} as,
{a,b,c} = {(y-c2)/(1-c3), (-c2x-z)/(1-c3), (1-x)/(y+z)}
The conditional equation above is a quadratic in b and to make its discriminant a square gives,
y2 = 3(4a2c3-8ac2+4c-a2)
which is an elliptic curve in c, or equivalently,
y2 = 3(4c3-1)a2-24c2a+12c
a quadratic curve in a. This is easier made a square if its constant term is also a square. Let c = 3v2,
y2 = 3(108v6-1)a2-216v4a+36v2
This is of the form y2 = ax2+bx+c2 where an infinite number of integral solns can be found by solving the Pell equation p2-aq2 = ±1. (See the relevant section in Part 2, Quadratic Polynomial as a kth power.) Sparing the reader some algebra, this gives,
Piezas
1. (1-ac+bc)3 + (a+c2-ac3)3 + (ac3-b-c2)3 = 1
{a,b,c} = {12qrt, 3(q-r)(3q+r)t, 3s2t2}
2. (1-ac-bc)3 + (a+c2-ac3)3 + (ac3+b-c2)3 = 1
{a,b,c} = {12qrt, 3(q+r)(3q-r)t, 3s2t2}
where, for both, r = p-18qs3t3 and the condition p2-3(108s6t6-1)q2 = s. By setting s = ±1, this is of course a Pell equation. Note that the second addend is unchanged, implying a pair of solns which share a term thus solving the system,
x13+x23 = x33+x43 = x53+1
Example, for the case s=t=1, this gives p2-321q2 = 1, with fundamental {p,q} = {215, 12} yielding the pair,
{x,y,z} = {4528, 3753, -5262} {x,y,z} = {-3230, 3753, -2676}
It turns out that for s = 1, the Pell equation,
p2-3(108t6-1)q2 = 1
has a parametrization for its fundamental soln given by {p1, q1} = {216t3-1, ±12t3}. From this one can then generate an infinite family. Using trivial {p0, q0} = {1, 0} on the expressions for x,y,z yields the known 4-deg identity,
(1-9t3)3 + (9t4)3 + (3t-9t4)3 = 1
while {p1, ±q1} with one sign yields a 10-deg,
(1+9t3+648t6-3888t9)3 + (-135t4+3888t10)3 + (-3t-81t4+1296t7-3888t10)3 = 1
while the other sign gives a 16-deg. The next, {p2, ±q2} gives a 22 and 28-deg, {p3, ±q3} a 34 and 40-deg, and so on for an infinite sequence with degree 6n+4. One of the terms is always a polynomial with only even powers, unchanged for ±t, so a term is shared by two distinct solns to x3+y3+z3 = 1. (This in fact is the same infinite family that can be derived by a recursion found by D.H.Lehmer.)
Note: Aside from s = 1, I do not know of any other non-trivial integral s such that p2-3(108s6t6-1)q2 = s is solvable in the integers {p,q} for a given t.
One can also use the binary quadratic forms together with Pell equations to find an infinite number of integral solns if an initial one is known. To recall,
(ax2-v1xy+bwy2)3 + (bx2+v1xy+awy2)3 + (cx2+v2xy+dwy2)3 + (dx2-v2xy+cwy2)3 = (a3+b3+c3+d3)3(x2+wy2)3
{v1, v2, w} = {c2-d2, a2-b2, (a+b)(c+d)}
so it suffices to find one soln to a3+b3+c3+1 = 0. It is easy to set {x,y} = {p+v2q, 2q} to transform the fourth addend x2-v2xy+cwy2 to the form p2-nq2. (The discriminant n is a function of a,b,c with 3! = 6 possible values and it is the experience of this author that at least one has n > 0). Assume d = ±1 and one can then solve the Pell equation p2-nq2 = ±1. For example, using the famous taxicab number 1728 = 93+103 = 123+13, and after a little modification,
(9+44pq-404q2)3 + (10+45pq-417q2)3 = (12+56pq-518q2)3 + 1, if p2-85q2 = -4
(9+44pq+404q2)3 + (10+45pq+417q2)3 = (12+56pq+518q2)3 + 1, if p2-85q2 = 4
Using another initial soln, 63 + 83 = 93 – 1, we get,
(6+222pq+4014q2)3 + (8+270pq+4806q2)3 = (9+312pq+5616q2)3 - 1, if p2-321q2 = 1
and so on.
A. Gerardin
(p4+9pq3)3 + (3q2)6 = (3p3q+9q4)3 + p12
For p=1 this gives a parametrization to a3+b3+c3 = 1.
4. Form: x3+y3+z3 = 2
(1+ax3)3 + (1-ax3)3 = 6a2x6 + 2
It suffices to make a = 6 to satisfy the equation, with x arbitrary so there are parametrizations for x13+x23+x33 = n for n = 1,2. It turns out there is a fifth power version,
x15+x25+x35+x45+x55+x65+x75 = n,
also for n =1,2, with n=1 to be discussed in a later chapter. For n=2 by Seiji Tomita, surprisingly this is dependent on Pythagorean triples and is analogous to the cubic case,
(1+ax5)5 + (1-ax5)5 + (1+bx5)5 + (1-bx5)5 + (-1+cx5)5 + (-1-cx5)5 + (dx4)5 = 2
where {a,b,c,d} should satisfy the two conditions a2+b2 = c2 and 20a2b2 = d5, one soln of which is {a,b,c,d} = {270, 360, 450, 180}.
5. Form: x3+y3+z3 = (z+m)3
J. Jandasek (m =1)
n3 + (3n2+2n+1)3 + (3n3+3n2+2n)3 = (3n3+3n2+2n+1)3
This implies that any integer n appears in a cubic quadruple at least once. A similar identity exists for second powers as was already discussed.
G. Ampon (m =1)
(3n2)3 + (6n2+3n+1)3 + (9n3+6n2+3n)3 = (9n3+6n2+3n+1)3
Q: Any soln for some other constant m?
6. Form: p(p2+bq2) = r(r2+bs2)
Derivation: To derive Binet’s version, for convenience, set,
(p+q)3 + (p-q)3 = (r+s)3 + (r-s)3
Expanding, we get p(p2+3q2) = r(r2+3s2). This can generalized to,
p(p2+bq2) = r(r2+bs2) (eq.1)
Assume that,
r2+bs2 = (p2+bq2)(u2+bv2) (eq.2)
Factor over √-b and equate factors,
r+s√-b = (p+q√-b)(u+v√-b) r-s√-b = (p-q√-b)(u-v√-b)
Solve for {r,s},
r = pu-bqv (eq.3) s = pv+qu (eq.4)
Substitute eq.2 into eq.1 (by eliminating p2+bq2),
p = r(u2+bv2) (eq.5)
Substitute r from eq.3 into eq.5,
p = (pu-bqv)(u2+bv2)
Collect {p,q},
p(-1+u(u2+bv2)) = qbv(u2+bv2)
And the equation is true if,
p = bv(u2+bv2), q = -1+u(u2+bv2)
Substitute these into eq.3 and eq.4 to get,
r = bv, s = -u+(u2+bv2)2
and we have all unknowns {p,q,r,s} for any b. But this can be generalized even further.
7. Form: (x+c1y)(x2+c2xy+c3y2)k = (z+c1t)(z2+c2zt+c3t2)k
Since by using the transformation {x,y} = {p-c1q, q} on the left hand side (and a similar one for the right) will transform this into the form p(p2+apq+bq2)k, one can simply assume c1 = 0 without loss of generality,
p(p2+apq+bq2)k = r(r2+ars+bs2)k
Piezas
{p,q,r,s} = {bvwk, -1+uwk, bv, -(u+av)+wk+1}, if w = u2+auv+bv2
This is easily be proven for k=1,2. For k=1 and a=0, this reduces to the formula for third powers given previously. For k=2, this is relevant to equal sums of fifth powers to be discussed later. Using computer algebra one can see it is also true for other k>2, but I have no proof it is the case for all positive integer k.
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