A. Desboves (complete soln)
Theorem: “Given one integral soln to ax2+by2+cz2+dxy+exz+fyz = 0, then an infinite more can be given by a polynomial identity.”
aX2+bY2+cZ2+dXY+eXZ+fYZ = (ax2+by2+cz2+dxy+exz+fyz)(bp2+fpq+cq2)2
X = -x(bp2+fpq+cq2)
Y = (dx+by+fz)p2 + (ex+2cz)pq - cyq2
Z = -bzp2 + (dx+2by)pq + (ex+fy+cz)q2
for arbitrary p,q.
7. Form: ax2 + cy2 = dzk, k > 2
Pepin
3(pr+3qs)2-(pr+9qs)2 = 2(r2-3s2)3, and
3(3pr-15qs)2-(5pr-27qs)2 = 2(r2-3s2)3,
where {p,q} = {r2+9s2, r2+s2}.
Note: Pepin’s result also solves a2+mb2 = c2+md2, but the complete soln of this is,
(pr+mqs)2 + m(ps-qr)2 = (pr-mqs)2 + m(ps+qr)2
for all {m,p,q,r,s}. However, Pepin’s variant form is,
(pr+m2qs)2 + m(mpr-mnqs)2 = (npr-m3qs)2 + m(pr+mqs)2
and is true for all {p,q,r,s} only if {m,n} satisfies the elliptic curve, m3-m+1 = n2, one soln of which is {m,n} = {3, 5}.
Q: Other poly solns to ax2 + cy2 = dzk for d >1, k >2?
ax12+bx22+cx32+dx42 = (ap2+bq2+cr2+ds2)(ax2+by2+cz2+dt2)2
{x1, x2, x3, x4} = {pv1-2xv2, qv1-2yv2, rv1-2zv2, sv1-2tv2} where {v1, v2} = {ax2+by2+cz2+dt2, apx+bqy+crz+dst}
for arbitrary {x,y,z,t}. And so on for any n addends. This was previously discussed in Section 003. A similar general identity exists for cubes while there is more limited one for fourth powers of the form x14+x24 = x34+ x44 such that an initial soln leads to subsequent ones.
For the particular case a=d=1, the complete soln as established by Desboves is
x2+bxy+cy2 = z2
{x,y,z} = {u2-cv2, 2uv+bv2, u2+buv+cv2}
which can be derived by using the initial soln {m,n,p} = {1,0,1} on either of the two previous general identities. If b=0, after some modification this becomes,
x2+mny2 = z2
{x,y,z} = {mu2-nv2, 2uv, mu2+nv2}
6. Form: ax2+by2+cz2+dxy+exz+fyz = 0
S. Realis (complete soln)
Given one soln to ax2+by2+cz2 = 0 then an infinite more can be found.
aX2+bY2+cZ2 = (ax2+by2+cz2)(ap2+bq2+cr2)2
X = x(-ap2+bq2+cr2) – 2p(bqy+crz),
Y = y(ap2-bq2+cr2) – 2q(apx+crz),
Z = z(ap2+bq2-cr2) – 2r(apx+bqy)
for arbitrary {p,q,r} (or x,y,z since if the equation is equal to zero, either factor is of the same form). Or alternatively, to show its “internal structure” (after a small exchange of variables),
ax12+bx22+cx32 = (ap2+bq2+cr2)(ax2+by2+cz2)2
{x1, x2, x3} = {pv1-2xv2, qv1-2yv2, rv1-2zv2} where {v1, v2} = {ax2+by2+cz2, apx+bqy+crz}
for arbitrary {x,y,z}. In fact, a more general statement can be made.
Theorem: "Given one solution to a1y12+a2y22+…+ anyn2 = 0, then an infinite more can be found."
Proof: We simply generalize the soln. For four addends,
Update: For a form to easily set z = 1, and given an initial solution to am2+bmn+cn2 = d, then,
1) ax2+bxy+cy2-dz2 = (am2+bmn+cn2-d)(x/m)2
or suppressing z,
2) ax2+bxy+cy2-d = d(u2-Dv2-1)( u2-Dv2+1)
where for both,
{x,y,z} = mu2-2(bm+2cn)uv+Dmv2, nu2+2(2am+bn)uv+Dnv2, u2-Dv2
A. Univariate: ax2+bx+c = z2
B. Bivariate: ax2+bxy+cy2 = zk
x2+cy2 = zk
ax2+cy2 = zk, k odd
x2+2bxy+cy2 = zk
ax2+2bxy+cy2 = zk, k odd
C. Bivariate: ax2+bxy+cy2 = dz2
ax2+bxy+cy2 = dz2
ax2+by2+cz2+dxy+exz+fyz = 0
ax2+cy2 = dzk, k > 2
Given an initial solution, the problem of finding an indefinite number of subsequent rational solns to P(x) = z2 where P(x) is the general univariate polynomial of degree k=2,3,4 was solved by Fermat using a simple method. The general principle can be illustrated by the quadratic case.
A. Univariate form: ax2+bx+c = z2
If P(x) has an initial rational soln, we can assume we are dealing with the form,
ax2+bx+c2 = z2
To illustrate, given P(x):= px2+qx+r. Change variables from x to v and let x = v+n (for some indefinite n). Expanding, we get,
P(v): = pv2 + (2pn+q)v + pn2+qn+r
Thus, if n is a soln to pn2+qn+r = y2, then equivalently,
P(v): = pv2 + (2pn+q)v + y2
which is the desired form. In general, given a polynomial P(x) of any degree, if n is a soln to P(x) = y2 then the substitution x = v+n will yield a new polynomial P(v) with the constant term y2. Following Fermat, we then assume that,
ax2+bx+c2 = (p/qx-c)2
Subtracting one side from the other and collecting the variable x,
(p2-aq2)x2 - (2cpq+bq2)x = 0
One can then easily solve for x as,
x = (2cpq+bq2)/(p2-aq2)
for arbitrary p,q. (The method then is to remove enough terms so solving for x involves only a linear equation. This can easily be extended to other powers as shall be seen later.) If integral x is desired, one can set p2-aq2 = ±1 and solve this Pell equation. For non-square, positive integer a, this then provides an infinite number of integral solns x. For the related quadratic form,
(x+u)(x+v) = dz2
{x, z} = {p2-u, pq}, if p2-dq2 = u-v
{x, z} = {p2-v, pq}, if p2-dq2 = -(u-v)
involves the solving the Pell-like equation p2-dq2 = k. (Euler)
B. Bivariate form: ax2+bxy+cy2 = zk
The problem of making the general binary quadratic form into a kth power can be divided into two classes: the monic form a=1 where generic solns are known for all k and the non-monic form where generic solns are known only for odd k.
1. Form: x2+cy2 = zk
Euler:
x2+cy2 = (p2+cq2)k
Same technique of equating factors
{x+yÖ-c, x-yÖ-c} = {(p+qÖ-c)k, (p-qÖ-c)k}
and easily solving for x,y. Example, for k=2,
(p2-cq2)2 + c(2pq)2 = (p2+cq2)2
which for c=1 gives the familiar Pythagorean triples, and so on for other k, though it should be pointed out that for k > 2 this does not generally give all solns (Pepin). For example, for the particular case c = 47, all relatively prime solns with odd z are given by the method. But a class of even z is given by,
Pepin:
(13u3+60u2v-168uv2-144v3)2 + 47(u3-12u2v-24uv2+16v3)2 = 23(3u2+2uv+16v2)3
Q: How then to find the complete soln of x2+cy2 = zk for k>2?
For the negative case, or c = -d, one can use a variation of the method above by employing a Pell equation and equate,
x2-dy2 = (p2-dq2)k(r2-ds2)
where r2-ds2 = 1 and solve for x,y using
x+yÖd = (p+qÖd)k(r+sÖd)
x-yÖd = (p-qÖd)k(r-sÖd)
To solve ax2+cy2 = (ap2+cq2)k, as before, equate factors,
xÖa + yÖ-c = (pÖa + qÖ-c)k
xÖa - yÖ-c = (pÖa - qÖ-c)k
then solve for x,y.
x = (ak+bk)/(2Öa), y = (ak-bk)/(2Ö-c), where a = (pÖa + qÖ-c), b = (pÖa - qÖ-c).
Example for k = 3,
{x,y} = {p(ap2-3cq2), q(3ap2-cq2) }
and so on for all odd k. For even k there is the problem of {x,y} containing the radical Öa, though it disappears in the monic case a=1.
3. Form: x2+2bxy+cy2 = zk
Euler, Lagrange:
x2+2bxy+cy2 = (p2+2bpq+cq2)k
Equating factors,
x+(b+d)y = (p+(b+d)q)k
x+(b-d)y = (p+(b-d)q)k
where d = Ö(b2-c). Then solve for {x,y} giving,
x = ((-b+d)ak+(b+d)bk)/(2d) y = (ak-bk)/(2d), where a = (p+(b+d)q), b = (p+(b-d)q)
4. Form: ax2+2bxy+cy2 = zk, k odd
Fauquembergue:
ax2+2bxy+cy2 = ak(x2+2bxy+acy2)k
Equating factors,
x+((b+d)/a)y = am(p+(b+d)q)k
x+((b-d)/a)y = am(p+(b-d)q)k
Solving for {x,y},
x = am((-b+d)ak+(b+d)bk)/(2d) y = am+1(ak-bk)/(2d),
where a = (p+(b+d)q), b = (p+(b-d)q), d = Ö(b2-ac), and k = 2m+1. (Note that for a=1, this reduces to the monic case discussed by Euler and Lagrange.)
Q: There are certain {a,b,c} such that ax2+2bxy+cy2 = zk for even k has no soln for rational x,y,z. For what {a,b,c}, with a ¹ 1, can we find integral polynomial solutions for all k?
C. Bivariate form: ax2+bxy+cy2 = dz2
5. Form: ax2+bxy+cy2 = dz2
This form is merely a special case of the general theorem given by Desboves in the next section.
A.Gerardin
ax2+bxy+cy2-dz2 = (am2+bmn+cn2-dp2)(au2+buv+cv2)2
{x,y,z} = {-(am+bn)u2-2cnuv+cmv2, anu2-2amuv-(bm+cn)v2, p(au2+buv+cv2)}, for arbitrary u,v.
Piezas
ax2+bxy+cy2-dz2 = (am2+bmn+cn2-dp2)(u2-cdv2)2
{x,y,z} = {mu2-cdmv2, nu2+2pduv+d(bm+cn)v2, pu2+(bm+2cn)uv+cdpv2}