PART 5. Pell Equations
I. Complete Solution
II. Transformations
III. Polynomial Parametrizations
IV. Diophantine Equations needing Pell Eqns
I. Complete Solution
C. Hermite
Given an initial soln {p,q} to p2-dq2 = ±1, one can find an infinite number of solns {x,y} by equating,
x2-dy2 = (p2-dq2)k
Using the same technique of equating factors,
x+yÖd = (p+qÖd)k, x-yÖd = (p-qÖd)k,
we easily find {x,y} as x = (ak +a-k)/2, y = (ak -a-k)/(2Öd), with a = p+qÖd, where for the negative case p2-dq2 = -1 only odd powers k should be used. Note that the contribution of a-k diminishes as k increases so effectively {x,y} are approximated by {x,y} ≈ {ak/2, ak /(2Öd)}.
P.Paoli, S. Realis (all rational solns)
Given an initial soln {p,q} to p2-nq2 = c for some constant c, then x2-ny2 = c where,
x = (pu2+2nquv+npv2)/(u2-nv2), y = (qu2+2puv+nqv2)/(u2-nv2)
for arbitrary {u,v}. However, if integral x,y is desired, then one can choose {u,v} to solve the Pell eqn u2-nv2 = ±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. Note that the formulas for x,y without their denominators satisfy,
x2-ny2 = (p2-nq2)(u2-nv2)2
Euler
More generally, given an initial soln {p,q} to mp2-nq2 = c, then mx2-ny2 = c where,
x = (pu2+2nquv+mnpv2)/(u2-mnv2), y = (qu2+2mpuv+mnqv2)/(u2-mnv2)
and an infinite number of integral x,y can be found by solving u2-mnv2 = ±1. Note that the Paoli-Realis soln is just the special case m=1. Again, without the denominators, the x,y satisfy,
mx2-ny2 = (mp2-nq2)(u2-mnv2)2
Solns for special forms are given by,
S.Realis
(n+2)x2-ny2 = (n+2)p2-nq2
{x, y} = {(n+1)p+nq, (n+2)p+(n+1)q}
x2+nxy-ny2 = p2+npq-nq2
{x, y} = {(n+1)p-nq, (n+2)p-(n+1)q}
II. Transformations
1. W. Brouncker, J. Wallis
Brouncker-Wallis theorem: “If x2-dy2 = 1, then (2x2-1)2-d(2xy)2 = 1.”
This is a very useful theorem for Pell equations. First, it shows that an initial soln (x,y) can generate some of the other solns (which yields even y’). Second,
Theorem 1a: “The fundamental soln to x2-dy2 = 1 yields the one for x2-4dy2 = 1.”
If y is even, one can simply use x2-4d(y/2)2 = 1. But if y is odd, use the Brounker-Wallis identity. Example:
x2-19y2 = 1; {x,y} = {170, 39), so,
p2-4(19)q2 = 1; {p,q}= {2(1702)-1, 170(39)}.
Third, let {x,y} = {u√-1, v√-1}, and we get the important variant, “If u2-dv2 = -1, then (2u2+1)2-d(2uv)2 = 1.”
Theorem 1b: “The fundamental soln to u2-dv2 = -1 leads to the one for x2-dy2 = 1.” Example:
x2-13y2 = -1; {x,y} = {18, 5), then,
p2-13q2 = 1; {p,q}= {2(182)+1, 2(18)(5)}.
This implies solns to the so-called negative Pell equation are generally smaller than those of the latter. After a challenge by Frenicle de Bessy, Brouncker solved x2-313y2 = 1 by solving the relatively easier negative case.
2. A. Cayley
Cayley’s theorem: “If m2-dn2 = 4, and {x,y} = {(m2-3)m/2, (m2-1)n/2}, then x2-dy2 = 1.” A variant can be obtained by using the same trick,{m,n} = {u√-1, v√-1} to get, “If u2-dv2 = -4, and {x,y} = {(u2+3)u/2, (u2+1)v/2}, then x2-dy2 = -1.”
Theorem: “If there is an odd fundamental soln {m,n} to m2-dn2 = ±4, then it leads to the one for x2-dy2 = ±1.” For example:
Given m2-dn2 = 4 and x2-dy2 = 1. Let d = 5, then {m,n} = {3,1}, and {x,y} = {9,4}.
Given u2-dv2 = -4 and x2-dy2 = -1. 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 the fundamental {x,y} = {6,1}. (In fact, the latter generated the former.)
Note 1: For x2-dy2 = ±4 to have a solution in odd integers a necessary (but not sufficient) condition is that d is odd number of form 8n+5. (Though there are exceptions like d = 37, 101, etc.)
Note 2: For odd d, if x2-dy2 = -4 is solvable, then so is x2-dy2 = -1. (For even d, this does not necessarily apply. For example, x2-20y2 = -4 has the soln {4,1}, but x2-20y2 = -1 has none.)
3. Euler
Theorem: “If m2-dn2 = -4, and {x,y} = {(m4+4m2+1)(m2+2)/2, (m2+3)(m2+1)mn/2}, then x2-dy2 = 1. If fundamental {m,n} are odd, then it leads to fundamental {x,y}.”
This was found by combining the variants of the Brouncker-Wallis and Cayley identities. As previously, if {m,n} is fundamental and odd, it leads to the fundamental soln {x,y}. This shows that fundamental solns of m2-dn2 = -4, when odd, are much smaller than those for x2-dy2 = 1 since x is a sixth degree polynomial in m. Historically, Euler used a variant of the identity above (this author modified to make it more aesthetic) to find x2-61y2 = 1 which has the largest solns for d < 100 as {x,y} = {1766319049, 226153980} by solving the much easier m2-61n2 = -4 which only has {m,n} = {39, 5}.
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,
(p2-1)2 - d(pq)2 = 1, if p2-dq2 = 2
(p2+1)2 - d(pq)2 = 1, if p2-dq2 = -2
If d = 8n+r is prime, then the ff always have solns:
8n+1 for u2-dv2 = -1;
8n+3 for p2-dq2 = -2;
8n+5 for u2-dv2 = -1;
8n+7 for p2-dq2 = 2.
Theorem: “Primes of form d = 4n+1 solve x2-dy2 = -1, while those of form d = 4n-1 solve one case of x2-dy2 = ±2 (Legendre). For such d, the fundamental soln to x2-dy2 = ±2 leads to the one for u2-dv2 = 1.”
5. F. Arndt
(2pr2-1)2 - pq(2rs)2 = 1, if pr2-qs2 = 1
(pr2+qs2)2 - pq(2rs)2 = 1, if pr2-qs2 = ±1
(pr2-1)2 - pq(rs)2 = 1, if pr2-qs2 = 2
(pr2+1)2 - pq(rs)2 = 1, if pr2-qs2 = -2
which are generalizations of the identities discussed previously.
Q: Any other transformations?
III. Polynomial Parametrizations
The degree of a polynomial soln to x2-dy2 = ±1 will be given by the P(n) defining the variable x.
A. Degrees 1, 2, 3, 4, 6
Deg 1
(y2n+x)2 – (y2n2+2xn+d)y2 = x2-dy2
Deg 2
(an2+2ny+x)2 – (n2+2bn+d) (an+y)2 = z2,
where a = (x-z)/d, b = (x+z)/y, and x2-dy2 = z2.
Deg 3
(2an3+3abn2+6ny+x)2 – (4n2+4bn+d) (an2+abn+y)2 = x2-dy2
where a = 2(3x-by)/(bd) and b is a root of the cubic b3y-3b2x+3bdy-dx = 0.
Deg 4 (Avanzi and Zannier)
(a2-2b(b2+2b-c))2 – (a2+8bc+16c) (b2+c)2 = -16c3, if a = b2+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+n3))2 – (5+6n+3n2+2n3+n4) (n2+1)2 = -4
The case n=0 gives the fundamental soln to x2-5y2 = -4 while n=2 gives Euler’s soln to x2-61y2 = -4. In general, if n is even then x’ is odd, so using Euler’s transformation this gives a parametric fundamental soln to x2-dy2 = 1 for some d. Alternatively, for {b,c} = {2n-1, 1} yields,
(-1+2n+2n2-4n3+4n4)2 – 2(1+2n+2n4) (1-2n+n2)2 = -1
Deg 6 (G. Ricalde)
(8(a3+b3)2+1)2 – ((a+b)2+4) (4(a3+b3)(a2+b2))2 = 1, if b = a+1
Avanzi and Zannier
(2n(1+2n-2n2+8n3-4n4+5n5))2 - (1+4n+4n2+4n4) (1-2n+6n2-4n3+4n4)2 = -1
B. Degrees 5, 7, 11, 13, 17
For these all known poly solns to x2-dy2 = 1 are of the form,
(2pr2-1)2 - pq(2rs)2 = 1,
or equivalently, (pr2+qs2)2 - pq(2rs)2 = 1, hence also satisfy pr2-qs2 = 1. Thus, the variables {p,q,r,s} of the entries below are understood to refer to these equations.
Deg 5 (Avanzi and Zannier)
{p,q,r,s} = {2+3m2+2(-m+m3)n-(1+4m2)n2+2mn3, -1+2mn, -n, 1+mn-n2}
{p,q,r,s} = {-2+3m2+2(m+m3)n-(1-4m2)n2+2mn3, -1+2mn, n, 1+mn+n2}
and,
(2+an2)2 – a(1+n)(n(2-n+n2))2 = 4, if a = 1+3n-n2+n3
(2+an2)2 – a(1+n)(n(-2+n+n2))2 = 4, if a = -7-n+3n2+n3
For any n, the {x,y} of the pair above are even so it reduces to x2-dy2 = 1. Setting a = 1+3n-n2+n3 = 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+n2+2n3)/2, 2(-3+2n), -1-2n+2n2, (-1-n+2n3)/2}
Piezas
p = n/4
q = -4 + (1+2a+3a2-6a3+a4)n - 2a2(1-2a2+a3)n2 + (-1+a)2a4n3
r = 3+3a-a2 + (-1-3a-4a2+3a3)n + a2(2+2a-3a2)n2 + (-1+a)a4n3
s = (1-(1+2a)n+a2n2
A deg-17 soln will result by setting a = bn for some rational b.
Deg 11 (Weber)
{p,q,r,s} = {-1+n2+n3, -5-4n+n2+n3, (-2+n)(1+n)(-2+n2)/2, (-1+n)(-2-2n+n3)/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-n2)n/4, -4(4-4n+4n3-n5+n7), 2+2n-n2+n4+n5, (1-n2-n3)/4}
This was derived by using a variant of the 2-variable deg-7 soln and likewise setting a = bn for b = -1.
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