A. Univariate: ax^{2}+bx+c^{ }= z^{2}

B. Bivariate: ax^{2}+bxy+cy^{2} = z^{k}

- x
^{2}+cy^{2} = z^{k}
- ax
^{2}+cy^{2} = z^{k}, k odd
- x
^{2}+2bxy+cy^{2} = z^{k}
- ax
^{2}+2bxy+cy^{2} = z^{k}, k odd

C. Bivariate: ax^{2}+bxy+cy^{2} = dz^{2}

- ax
^{2}+bxy+cy^{2} = dz^{2}
- ax
^{2}+by^{2}+cz^{2}+dxy+exz+fyz = 0
- ax
^{2}+cy^{2} = dz^{k}, k > 2

* *

Given an initial solution, the problem of finding an indefinite number of subsequent rational solns to *P*(x) = z^{2} 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: ax**^{2}+bx+c^{ }= z^{2}

If P(x) has an initial rational soln, we can assume we are dealing with the form,

ax^{2}+bx+c^{2} = z^{2}

To illustrate, given P(x):= px^{2}+qx+r. Change variables from *x* to *v* and let x = v+n (for some indefinite *n*). Expanding, we get,

P(v): = pv^{2 }+ (2pn+q)v + pn^{2}+qn+r

Thus, if *n* is a soln to pn^{2}+qn+r = y^{2}, then equivalently,

P(v): = pv^{2 }+ (2pn+q)v + y^{2}

which is the desired form. In general, given a polynomial P(x) of *any* degree, if *n* is a soln to P(x) = y^{2} then the substitution x = v+n will yield a new polynomial P(v) with the constant term y^{2}. Following Fermat, we then assume that,

ax^{2}+bx+c^{2} = (p/qx-c)^{2}

Subtracting one side from the other and collecting the variable x,

(p^{2}-aq^{2})x^{2 }- (2cpq+bq^{2})x = 0

One can then easily solve for x as,

x = (2cpq+bq^{2})/(p^{2}-aq^{2})

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 p^{2}-aq^{2} = ±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) = dz^{2}

{x, z} = {p^{2}-u, pq}, if p^{2}-dq^{2} = u-v

{x, z} = {p^{2}-v, pq}, if p^{2}-dq^{2} = -(u-v)

involves the solving the Pell-like equation p^{2}-dq^{2} = k. (Euler)

**B. Bivariate form: ax**^{2}+bxy+cy^{2} = z^{k}

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: x**^{2}+cy^{2} = z^{k}

Euler:

x^{2}+cy^{2} = (p^{2}+cq^{2})^{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,

(p^{2}-cq^{2})^{2} + c(2pq)^{2} = (p^{2}+cq^{2})^{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:

(13u^{3}+60u^{2}v-168uv^{2}-144v^{3})^{2} + 47(u^{3}-12u^{2}v-24uv^{2}+16v^{3})^{2} = 2^{3}(3u^{2}+2uv+16v^{2})^{3}

*Q: How then to find the complete soln of x*^{2}+cy^{2} = z^{k} 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,

x^{2}-dy^{2} = (p^{2}-dq^{2})^{k}(r^{2}-ds^{2})

where r^{2}-ds^{2} = 1 and solve for x,y using

x+yÖd = (p+qÖd)^{k}(r+sÖd)

x-yÖd = (p-qÖd)^{k}(r-sÖd)

## 2. Form: ax^{2}+cy^{2} = z^{k}, k odd

To solve ax^{2}+cy^{2} = (ap^{2}+cq^{2})^{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 = (a^{k}+b^{k})/(2Öa), y = (a^{k}-b^{k})/(2Ö-c), where a = (pÖa + qÖ-c), b = (pÖa - qÖ-c).

Example for k = 3,

{x,y} = {p(ap^{2}-3cq^{2}), q(3ap^{2}-cq^{2}) }

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: x**^{2}+2bxy+cy^{2} = z^{k}

Euler, Lagrange:

x^{2}+2bxy+cy^{2} = (p^{2}+2bpq+cq^{2})^{k}

Equating factors,

x+(b+d)y = (p+(b+d)q)^{k}
x+(b-d)y = (p+(b-d)q)^{k}

where d = Ö(b^{2}-c). Then solve for {x,y} giving,

x = ((-b+d)a^{k}+(b+d)b^{k})/(2d) y = (a^{k}-b^{k})/(2d), where a = (p+(b+d)q), b = (p+(b-d)q)

**4. Form: ax**^{2}+2bxy+cy^{2} = z^{k}, k odd

Fauquembergue:

ax^{2}+2bxy+cy^{2} = a^{k}(x^{2}+2bxy+acy^{2})^{k}

Equating factors,

x+((b+d)/a)y = a^{m}(p+(b+d)q)^{k}

x+((b-d)/a)y = a^{m}(p+(b-d)q)^{k}

Solving for {x,y},

x = a^{m}((-b+d)a^{k}+(b+d)b^{k})/(2d) y = a^{m+1}(a^{k}-b^{k})/(2d),

where a = (p+(b+d)q), b = (p+(b-d)q), d = Ö(b^{2}-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 ax^{2}+2bxy+cy^{2} = z^{k} 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: ax**^{2}+bxy+cy^{2} = dz^{2}

**5. Form: ax**^{2}+bxy+cy^{2} = dz^{2}

This form is merely a special case of the general theorem given by Desboves in the next section.

A.Gerardin

ax^{2}+bxy+cy^{2}-dz^{2} = (am^{2}+bmn+cn^{2}-dp^{2})(au^{2}+buv+cv^{2})^{2}

{x,y,z} = {-(am+bn)u^{2}-2cnuv+cmv^{2}, anu^{2}-2amuv-(bm+cn)v^{2}, p(au^{2}+buv+cv^{2})}, for arbitrary u,v.

Piezas

ax^{2}+bxy+cy^{2}-dz^{2} = (am^{2}+bmn+cn^{2}-dp^{2})(u^{2}-cdv^{2})^{2}

{x,y,z} = {mu^{2}-cdmv^{2}, nu^{2}+2pduv+d(bm+cn)v^{2}, pu^{2}+(bm+2cn)uv+cdpv^{2}}

*Update*: For a form to easily set *z* = 1, and given an initial solution
to am^{2}+bmn+cn^{2} =
d, then,

1) ax^{2}+bxy+cy^{2}-dz^{2} =
(am^{2}+bmn+cn^{2}-d)(x/m)^{2}

or suppressing *z*,

2) ax^{2}+bxy+cy^{2}-d = d(u^{2}-Dv^{2}-1)(
u^{2}-Dv^{2}+1)

where for both,

{x,y,z} = mu^{2}-2(bm+2cn)uv+Dmv^{2}, nu^{2}+2(2am+bn)uv+Dnv^{2}, u^{2}-Dv^{2}

and D = b^{2}-4ac. See post 1 and post 2.

For the particular case a=d=1, the complete soln as established by Desboves is

x^{2}+bxy+cy^{2} = z^{2}

{x,y,z} = {u^{2}-cv^{2}, 2uv+bv^{2}, u^{2}+buv+cv^{2}}

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,

x^{2}+mny^{2} = z^{2}

{x,y,z} = {mu^{2}-nv^{2}, 2uv, mu^{2}+nv^{2}}

**6. Form: ax**^{2}+by^{2}+cz^{2}+dxy+exz+fyz = 0

S. Realis (complete soln)

Given one soln to ax^{2}+by^{2}+cz^{2} = 0 then an infinite more can be found.

aX^{2}+bY^{2}+cZ^{2} = (ax^{2}+by^{2}+cz^{2})(ap^{2}+bq^{2}+cr^{2})^{2}

X = x(-ap^{2}+bq^{2}+cr^{2}) – 2p(bqy+crz),

Y = y(ap^{2}-bq^{2}+cr^{2}) – 2q(apx+crz),

Z = z(ap^{2}+bq^{2}-cr^{2}) – 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),

ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2} = (ap^{2}+bq^{2}+cr^{2})(ax^{2}+by^{2}+cz^{2})^{2}

{x_{1},^{ }x_{2}, x_{3}} = {pv_{1}-2xv_{2}, qv_{1}-2yv_{2}, rv_{1}-2zv_{2}} where {v_{1}, v_{2}} = {ax^{2}+by^{2}+cz^{2}, apx+bqy+crz}

for arbitrary {x,y,z}. In fact, a more general statement can be made.

*Theorem*: "Given one solution to a_{1}y_{1}^{2}+a_{2}y_{2}^{2}+…+ a_{n}y_{n}^{2} = 0, then an infinite more can be found."

*Proof*: We simply generalize the soln. For four addends,

ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2}+dx_{4}^{2} = (ap^{2}+bq^{2}+cr^{2}+ds^{2})(ax^{2}+by^{2}+cz^{2}+dt^{2})^{2}

{x_{1},^{ }x_{2}, x_{3}, x_{4}} = {pv_{1}-2xv_{2}, qv_{1}-2yv_{2}, rv_{1}-2zv_{2}, sv_{1}-2tv_{2}} where {v_{1}, v_{2}} = {ax^{2}+by^{2}+cz^{2}+dt^{2}, 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 x_{1}^{4}+x_{2}^{4} = x_{3}^{4}+ x_{4}^{4} such that an initial soln leads to subsequent ones.

A. Desboves (complete soln)

*Theorem*: “Given one integral soln to ax^{2}+by^{2}+cz^{2}+dxy+exz+fyz = 0, then an infinite more can be given by a polynomial identity.”

aX^{2}+bY^{2}+cZ^{2}+dXY+eXZ+fYZ = (ax^{2}+by^{2}+cz^{2}+dxy+exz+fyz)(bp^{2}+fpq+cq^{2})^{2}

X = -x(bp^{2}+fpq+cq^{2})

Y = (dx+by+fz)p^{2 }+ (ex+2cz)pq - cyq^{2}

Z = -bzp^{2 }+ (dx+2by)pq + (ex+fy+cz)q^{2}

for arbitrary p,q.

**7. Form: ax**^{2} + cy^{2 }= dz^{k}, k > 2

Pepin

3(pr+3qs)^{2}-(pr+9qs)^{2} = 2(r^{2}-3s^{2})^{3}, and

3(3pr-15qs)^{2}-(5pr-27qs)^{2} = 2(r^{2}-3s^{2})^{3},

where {p,q} = {r^{2}+9s^{2}, r^{2}+s^{2}}.

*Note*: Pepin’s result also solves *a*^{2}+mb^{2} = c^{2}+md^{2}, 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+m^{2}qs)^{2} + m(mpr-mnqs)^{2} = (npr-m^{3}qs)^{2} + m(pr+mqs)^{2}

and is true for all {p,q,r,s} only if {m,n} satisfies the elliptic curve, m^{3}-m+1 = n^{2}, one soln of which is {m,n} = {3, 5}.

Q: Other poly solns to ax^{2} + cy^{2 }= dz^{k} for d >1, k >2?