**PART 5. Pell Equations**

*I. Complete Solution*

*II. Transformations*

*III. Polynomial Parametrizations*

*IV. Minor updates*

*I. Complete Solution*

C. Hermite

Given an initial soln {p,q} to p^{2}-dq^{2} = ±1, one can find an infinite number of solns {x,y} by equating,

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

Using the same technique of equating factors,

x+yÖd = (p+qÖd)^{k},

x-yÖd = (p-qÖd)^{k},

we easily solve it as {x, y} = { (a^{k }+a^{-k})/2, (a^{k }-a^{-k})/(2Öd) }, where a^{ }= p+qÖd, and for the negative case p^{2}-dq^{2} = -1 only odd powers k should be used. Note that the contribution of a^{-k} diminishes as k increases so effectively are approximated by {x,y} ≈ {a^{k}/2, a^{k }/(2Öd)}.

P.Paoli, S. Realis (all rational solns)

Given an initial soln {p,q} to p^{2}-nq^{2} = c for some constant *c*, then,

x^{2}-ny^{2} = c

where x = (pu^{2}+2nquv+npv^{2})/(u^{2}-nv^{2}), y = (qu^{2}+2puv+nqv^{2})/(u^{2}-nv^{2})

for *arbitrary* {u,v}. However, if integral x,y is desired, then one can choose {u,v} to solve the Pell eqn u^{2}-nv^{2} = ±1, with two solns given by (-p,q) and (p,q). Since there is an infinite number of {u,v}, then so for the x,y. The formulas for x,y without their denominators satisfy,

x^{2}-ny^{2} = (p^{2}-nq^{2})(u^{2}-nv^{2})^{2}

Euler

More generally, given an initial soln {p,q} to mp^{2}-nq^{2} = *c*, then mx^{2}-ny^{2} = *c* where,

x = (pu^{2}+2nquv+mnpv^{2})/(u^{2}-mnv^{2}), y = (qu^{2}+2mpuv+mnqv^{2})/(u^{2}-mnv^{2})

and an infinite number of integral x,y can be found by solving u^{2}-mnv^{2} = ±1. The Paoli-Realis soln is just the special case m=1. Again, without the denominators, the x,y satisfy,

mx^{2}-ny^{2} = (mp^{2}-nq^{2})(u^{2}-mnv^{2})^{2}

Solns for special forms are given by,

S.Realis

(n+2)x^{2}-ny^{2} = (n+2)p^{2}-nq^{2}

{x, y} = {(n+1)p+nq, (n+2)p+(n+1)q}

x^{2}+nxy-ny^{2} = p^{2}+npq-nq^{2}

{x, y} = {(n+1)p-nq, (n+2)p-(n+1)q}

*II. Transformations*

1. W. Brouncker, J. Wallis

*Brouncker-Wallis theorem*: “If x^{2}-dy^{2} = -1 (eq.1), then (2x^{2}+1)^{2}-d(2xy)^{2} = 1.”

*Proof*: Square both sides of eq.1,

(x^{2}-dy^{2})^{2} = (-1)^{2}

(x^{2}+dy^{2})^{2} - d(2xy)^{2} = 1

Since dy^{2} = x^{2}+1 from eq.1, then,

(2x^{2}+1)^{2}-d(2xy)^{2} = 1

Example:

x^{2}-13y^{2} = -1; {x,y} = {18, 5),

p^{2}-13q^{2} = 1; {p,q}= {2(18^{2})+1, 2(18)(5)} = {649, 180}

This shows that *fundamental* solns to the negative Pell equation, when they exist, are *smaller *than those for the positive case. After a challenge by Frenicle de Bessy, Brouncker solved x^{2}-313y^{2} = 1 by solving the relatively easier negative case. (*Note*: Letting {x,y} = {u√-1, v√-1} gives the purely positive variant: “If u^{2}-dv^{2} = 1, then (2u^{2}-1)^{2}-d(2uv)^{2} = 1”, which implies one soln leads to some of the other solns.)

2. A. Cayley

*Theorem*: “If there is an *odd* fundamental soln {u,v} to u^{2}-dv^{2} = ±4, then it leads to the fundamental soln for x^{2}-dy^{2} = ±1.”

*Case 1*: If u^{2}-dv^{2} = 4, and {x,y} = {(u^{2}-3)u/2, (u^{2}-1)v/2}, then x^{2}-dy^{2} = 1.

Ex. Let d = 5. Then {u,v} = {3,1}, and {x,y} = {9, 4}.

*Case 2*: If u^{2}-dv^{2} = -4, and {x,y} = {(u^{2}+3)u/2, (u^{2}+1)v/2}, then x^{2}-dy^{2} = -1.

Ex. Let d = 5. Then {u,v} = {1,1} and {x,y} = {2, 1}.

In contrast, let d = 37. Since the fundamental soln of the negative case is *even* {u,v} = {12, 2}, using the formula does *not* lead to fundamental {x,y} = {6,1}.

*Note*: The first case can easily be transformed into the second using the same trick, {u’,v’} = {u√-1, v√-1}.

3. Euler

*Theorem*: “If there is an *odd* fundamental soln {u,v} to u^{2}-dv^{2} = -4, then it leads to the fundamental soln for x^{2}-dy^{2} = 1 using the transformation {x, y} = {(u^{4}+4u^{2}+1)(u^{2}+2)/2, (u^{2}+3)(u^{2}+1)uv/2}.”

Ex. Let d = 61. Then {u,v} = {39, 5}, and {x,y} = {1766319049, 226153980}.

This was found by combining the Brouncker-Wallis and Cayley theorems, and shows that fundamental solns of u^{2}-dv^{2} = -4, when odd, are *much smaller* than those for x^{2}-dy^{2} = 1 since *x* is a *sixth* degree polynomial in *u*. Historically, Euler used a variant of the identity above (this author modified to make it more aesthetic) to find x^{2}-61y^{2} = 1 which has the largest solns for *d* < 100. Parametric fundamental solns to u^{2}-dv^{2} = -4 with odd {u,v} are,

(n)^{2} - (4+n^{2}) (1)^{2} = -4

((1+n)(1+2n+n^{3}))^{2} - (5+6n+3n^{2}+2n^{3}+n^{4}) (n^{2}+1)^{2} = -4

**with even ***n* for the second. (For ex., for n = 2, it yields d = 61.) *Q: Any others?*

Note 1: For x^{2}-dy^{2} = ±4 to have a solution in odd integers a necessary (but *not* sufficient) condition is that *d* is odd number of form 8n+5. (There are exceptions like d = 37, 101, etc. which have no odd solutions.)

Note 2: For odd *d*, if x^{2}-dy^{2} = -4 is solvable, then so is x^{2}-dy^{2} = -1. But for even *d*, this does not necessarily apply. For example, x^{2}-20y^{2} = -4 has the soln {4,1}, but x^{2}-20y^{2} = -1 has none.

4. A. Cunningham, R. Christie

By doing the transformations a) {x√2, y√2} = {p, q}, and b) {x√-2, y√-2} = {p, q}, on the Brouncker-Wallis theorem, we get the variants,

(p^{2}-1)^{2 }- d(pq)^{2} = 1, if p^{2}-dq^{2} = 2

(p^{2}+1)^{2 }- d(pq)^{2} = 1, if p^{2}-dq^{2} = -2

If d = 8n+r is *prime*, then the ff always have solns:

d = 8n+1 for u^{2}-dv^{2} = -1;

d = 8n+3 for p^{2}-dq^{2} = -2;

d = 8n+5 for u^{2}-dv^{2} = -1;

d = 8n+7 for p^{2}-dq^{2} = 2.

*Theorem*: “Primes of form *d = 4n+1* solve x^{2}-dy^{2} = -1, while those of form *d = 4n-1* solve one case of x^{2}-dy^{2} = ±2 (Legendre). For such d, the fundamental soln to x^{2}-dy^{2} = ±2 leads to the one for u^{2}-dv^{2} = 1.”

5. F. Arndt

(2pr^{2}-1)^{2} - pq(2rs)^{2} = 1, if pr^{2}-qs^{2} = 1

(pr^{2}+qs^{2})^{2} - pq(2rs)^{2} = 1, if pr^{2}-qs^{2} = ±1

(pr^{2}-1)^{2} - pq(rs)^{2} = 1, if pr^{2}-qs^{2} = 2

(pr^{2}+1)^{2} - pq(rs)^{2} = 1, if pr^{2}-qs^{2} = -2

which are generalizations of the identities discussed previously.

Q: *Any other transformations?*

*III. Polynomial Parametrizations*

The degree of a polynomial soln to x^{2}-dy^{2} = ±1 will be given by the *P*(n) defining the variable x.

**A. Degrees 1, 2, 3, 4, 6**

**Deg 1**

(y^{2}n+x)^{2} – (y^{2}n^{2}+2xn+d)y^{2} = x^{2}-dy^{2}

This identity also shows that for any soln *y,* parametric or otherwise, to x^{2}-dy^{2 }= c, then there are an infinite number of *d'* that uses the same *y*.

**Deg 2**

(an^{2}+2ny+x)^{2} – (n^{2}+2bn+d) (an+y)^{2} = z^{2},

where a = (x-z)/d, b = (x+z)/y, and x^{2}-dy^{2} = z^{2}.

**Deg 3**

(2an^{3}+3abn^{2}+6ny+x)^{2} – (4n^{2}+4bn+d) (an^{2}+abn+y)^{2} = x^{2}-dy^{2}

where a = 2(3x-by)/(bd) and b is a root of the cubic b^{3}y-3b^{2}x+3bdy-dx = 0.

**Deg 4** (Avanzi and Zannier)

(a^{2}-2b(b^{2}+2b-c))^{2} – (a^{2}+8bc+16c) (b^{2}+c)^{2} = -16c^{3}, if a = b^{2}+2b+c

Appropriate choice of b and ±c will result in conventional Pell equations. For {b,c} = {2n, 4} this yields, after removing common factors,

((1+n)(1+2n+n^{3}))^{2} – (5+6n+3n^{2}+2n^{3}+n^{4}) (n^{2}+1)^{2} = -4

The case n=0 gives the fundamental soln to x^{2}-5y^{2} = -4 while n=2 gives Euler’s soln to x^{2}-61y^{2} = -4. In general, if n is even then x’ is odd, so using Euler’s transformation this gives a parametric fundamental soln to x^{2}-dy^{2} = 1 for some d. Alternatively, for {b,c} = {2n-1, 1} yields,

(-1+2n+2n^{2}-4n^{3}+4n^{4})^{2} – 2(1+2n+2n^{4}) (1-2n+n^{2})^{2} = -1

**Deg 6** (G. Ricalde)

(8(a^{3}+b^{3})^{2}+1)^{2} – ((a+b)^{2}+4) (4(a^{3}+b^{3})(a^{2}+b^{2}))^{2} = 1, if b = a+1

Avanzi and Zannier

(2n(1+2n-2n^{2}+8n^{3}-4n^{4}+5n^{5}))^{2} - (1+4n+4n^{2}+4n^{4}) (1-2n+6n^{2}-4n^{3}+4n^{4})^{2} = -1

**B. Degrees 5, 7, 11, 13, 17**

Solutions to the three equivalent forms,

1. (2pr^{2}-1)^{2} - pq(2rs)^{2} = 1,

2. (pr^{2}+qs^{2})^{2 }- pq(2rs)^{2} = 1

3. pr^{2}-qs^{2} = 1

**Deg 5 **(Avanzi and Zannier)

{p,q,r,s} = {2+3m^{2}+2(-m+m^{3})n-(1+4m^{2})n^{2}+2mn^{3}, -1+2mn, -n, 1+mn-n^{2}}

{p,q,r,s} = {-2+3m^{2}+2(m+m^{3})n-(1-4m^{2})n^{2}+2mn^{3}, -1+2mn, n, 1+mn+n^{2}}

As well as,

(2+an^{2})^{2} – a(1+n)(n(2-n+n^{2}))^{2} = 4, if a = 1+3n-n^{2}+n^{3}

(2+an^{2})^{2} – a(1+n)(n(-2+n+n^{2}))^{2} = 4, if a = -7-n+3n^{2}+n^{3}

For any n, the {x,y} of the pair above are even so it reduces to x^{2}-dy^{2} = 1. Setting a = 1+3n-n^{2}+n^{3} = 0 involves a radical that can be expressed in terms of the Weber class poly for Ö-11, similar to the one for deg 11 which involves Ö-23.

**Deg 7** (Avanzi and Zannier)

{p,q,r,s} = {(-1+n^{2}+2n^{3})/2, 2(-3+2n), -1-2n+2n^{2}, (-1-n+2n^{3})/2}

Piezas

p = n/4

q = {-4 + (1+2a+3a^{2}-6a^{3}+a^{4})n - 2a^{2}(1-2a^{2}+a^{3})n^{2 }+ (-1+a)^{2}a^{4}n^{3}}/4

r = 3+3a-a^{2 }+ (-1-3a-4a^{2}+3a^{3})n + a^{2}(2+2a-3a^{2})n^{2 }+ (-1+a)a^{4}n^{3}

s = 1-(1+2a)n+a^{2}n^{2}

A deg-17 soln will result by setting a = bn for some rational b.

**Deg 11** (Weber)

{p,q,r,s} = {-1+n^{2}+n^{3}, -5-4n+n^{2}+n^{3}, (-2+n)(1+n)(-2+n^{2})/2, (-1+n)(-2-2n+n^{3})/2}

This was found using a modular equation between two elliptic functions at arguments f(w) and f(23w). Note that p,q have discriminant d=23 and *p* in fact is the inverse of the Weber class poly for Ö-23.

**Deg 13 ** (Piezas)

{p,q,r,s} = {(1-n^{2})n/4, -4(4-4n+4n^{3}-n^{5}+n^{7}), 2+2n-n^{2}+n^{4}+n^{5}, (1-n^{2}-n^{3})/4}

This was derived by using a variant of the 2-variable deg-7 soln and likewise setting a = bn for b = -1.

*IV. Minor updates*

(Update, 12/14/09): Given the fundamental soln {p,q} to x^{2}-dy^{2} = 1, Gerry Martens defined the following functions:

2G[n] := (p+q√d)^{n} + (p-q√d)^{n}

A[i,j] := (G[i]-1) (G[i+2j+1]+1)/4

B[i,j] := (G[i]+1) (G[i+2j+1]-1)/4

2C[i,j] := (√A[i,j] - √B[i,j] )^{2} + 1

2D[i,j] := (√A[i,j] + √B[i,j] )^{2} + 1

then {A,B} and/or {C,D} are squares for certain d.

1. For {A,B} as squares, d = {2, 5, 8, 10, 13, 17, 20, etc}.

2. For {C,D} as squares, d = {2, 7, 8, 14, 23, 31, 34, etc}.

3. For {A,B,C,D} as squares, d = {2, 8, 50, 98, 200, etc}.

*Note*: Apparently, a necessary but *not sufficient* condition is that d = u^{2}+v^{2} for [1], while d = 2m^{2} for [3]. (*End update*.)

(Update, 12/21/09): (Piezas) This arose in trying to solve in the rationals,

y^{4 }+ 2(u^{2}+auv+bv^{2})y^{2} + (u^{2}+v^{2})^{4} = z^{2}

Note that this can be made a square if u^{2}+auv+bv^{2} = (u^{2}+v^{2})^{2}. Given the quadratic equation, x^{2}+ax+b = 0, and its roots {2x_{1}, 2x_{2}} = {-a+√d, -a-√d}, where d = a^{2}-4b, define the sequence, P_{n} = 1/d (x_{1}^{n}-x_{2}^{n}).

*Theorem*: “If b = -1, then an infinite number of solns to the Diophantine equation,

(u^{2}+v^{2})^{2} = (u^{2}+auv+bv^{2})w^{2}

can be given by,

{u,v,w} = {P_{2m+1}, P_{2m}, P_{4m+1}}, and,

{u,v,w} = {P_{2m+1}, -P_{2m+2}, P_{4m+3}}.”

This can be verified in *Mathematica* after a little manual tweaking.

Example 1. Given x^{2}-x-1 = 0, where {x_{1}, x_{2}} = {(1+√5)/2, (1-√5)/2}, then P(n): = 1/√5 (x_{1}^{n}-x_{2}^{n}) yields the *Fibonacci sequence*,

P(n): = {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,…}

so the eqn, (u^{2}+v^{2})^{2} = (u^{2}-uv-v^{2})w^{2}, has solns {u,v,w} = {2,1,5}, {2,-3, 13}, {5, 3, 34}, {5, -8, 89}, etc.

Example 2. Given x^{2}-2x-1 = 0, where {x_{1}, x_{2}} = {1+√2, 1-√2}, then P(n): = 1/√2 (x_{1}^{n}-x_{2}^{n}), after removing the common factor 2, gives the *Pell numbers*,

P(n): = {1, 2, 5, 12, 29, 70, 169,…}

so (u^{2}+v^{2})^{2} = (u^{2}-2uv-v^{2})w^{2}, has {u,v,w} = {5, 2, 29}, {5,-12, 169}, etc. (*End update*.)

(Update, 12/23/09): I realized my solns had u^{2}+v^{2} = w, or (P_{2m})^{2} + (P_{2m+1})^{2} = P_{4m+1}, which is a known attribute of Fibonacci numbers. Thus, in general, the Diophantine equation,

(u^{2}+v^{2})^{2} = (u^{2}+auv+bv^{2})w^{2}

can be solved by,

{u,v,w} = {p-aq, 2q, u^{2}+v^{2}}, where {p,q} are solns of the Pell equation,

p^{2}-(a^{2}-4b)q^{2} = 1

generalizing the theorem given in the previous update. (*End update*.)

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