### 006: Quadratic Polynomial as a kth Power

A. Univariate: ax2+bx+c = z2

B. Bivariate: ax2+bxy+cy2 = zk
1. x2+cy2 = zk
2. ax2+cy2 = zk, k odd
3. x2+2bxy+cy2 = zk
4. ax2+2bxy+cy2 = zk, k odd

C. Bivariate: ax2+bxy+cy2 = dz2

1. ax2+bxy+cy2 = dz2
2. ax2+by2+cz2+dxy+exz+fyz = 0
3. 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)

## 2. Form: ax2+cy2 = zk, k odd

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}

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

and D = b2-4ac. See post 1 and post 2.

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,

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.

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?