IV. Diophantine Equations needing Pell Eqns
These equations are also found in the various chapters of this work but I thought it would be a good idea to bring some of them together in one section. For the basics, go to Pell Equations.
1. ad-bc = ±1
{a,b,c} = {y2, x+y, x-y}, if x2-(d+1)y2 = ±1
2. x2+y2 = z2
Euler
(v2+1)2 + (x2+1)2 = (v2+8y2+1)2, if v = (3y2+1)/(2y) and x2-10y2 = 1
3a. x2+ny3 = z4
H. Mathieu
(q2(p2-2))2 + (2q2)3 = (pq)4, if p2-2q2 = 1
(p2(p2-1)/2)2 + p6 = (pq)4, if p2-2q2 = -1
This can be generalized as,
Piezas
(4q2(p2-2))2 + d(4q2)3 = (2pq)4, if p2-dq2 = 1
(4p2(p2-1))2 + (2p)6 = d2(2pq)4, if p2-dq2 = -1
Just multiply the first eq by d8 and the second by d6.
3b. x4+y3 = nz2
K.Brown
p4 + (q2-1)3 = (q3+3q)2, if p2-3q2 = 1
More generally,
p4 + (dq2-1)3 = d(dq3+3q)2, if p2-3dq2 = 1
(See Brown’s “Miscellaneous Diophantine Equations”.)
4. x2+y2 = z2-1
E. Grigorief
((a2-b2+c2-d2)/2)2 + (ab+cd)2 = ((a2+b2+c2+d2)/2)2 - 1
{a,b,c} = {y2, x+y, x-y}, if x2-(d+1)y2 = ±1
5. x2+y2 = z2+1
A. Gerardin
(cx)2 + (c2y2-1)2 = (c2y2+c)2 + 1, where x2-2(c+1)y2 = 1
6. x2+y2 = z2+2k
Piezas
Let w = 2uv-2v2, then
(u2+2uv-kw)2 + (ku2+2kuv-w)2 = (k-1)2(u2+w)2 + 2k, if u2-2v2 = ±1
7. x2+y2+z2 = t2+1
(u2+v2)2 + (2duv)2 + (du2-dv2)2 = (du2-2uv-dv2)2 + (u2+2duv-v2)2
Set u2+2duv-v2 = ±1 by letting {u,v} = {x-dy, y} to get the Pell eqn x2-(d2+1)y2 = ±1.
A. Gerardin
(y-1)2 + y2 + (y+1)2 = x2 + 1, where x2-3y2 = 1
Any Pell equation for x2+y2+z2 = t2-1?
8. ax2+bx+c2 = z2
Fermat
x = (2cpq+bq2)/(p2-aq2), where for integral solns let p2-aq2 = ±1.
9. (n+u)(n+v) = dz2
Euler:
{n, z} = {p2-u, pq}, if p2-dq2 = u-v
{n, z} = {p2-v, pq}, if p2-dq2 = -(u-v)
10. mx2-ny2 = k
Euler
Given an initial soln mp2-nq2 = k for some constant k, then,
x = (pu2+2nquv+mnpv2)/(u2-mnv2), y = (qu2+2mpuv+mnqv2)/(u2-mnv2)
and an infinite more can be found by solving u2-mnv2 = ±1.
11. {x2+y2-1, x2-y2-1} both squares
A.Genocchi (complete rational soln in p,q,r,s)
{x,y} = {(2r2-t)/t, 4pqrs/t}, if t = r2 – (p4+4q4)s2 .
For integral solns, it suffices to solve the Pell equation r2 – (p4+4q4)s2 = ±1. Similarly,
T. Pepin
{x2+y2-1, x2-y2-1} = {(2pq(r2+2s2))2, (2pq(r2-2s2))2}
{x,y} = {p2+(r4+4s4)q2, 4pqrs}, if p2-(r4+4s4)q2 = ±1.
For the particular case {r,s} = {3,1}, this yields,
{x2+y2-1, x2-y2-1} = {(22pq)2, (14pq)2}
{x,y} = {p2+85q2, 12pq}, if p2-85q2 = ±1.
(The discriminant of this Pell equation appears in two other forms given below.)
12. x3+y3+z3 = 1
Piezas
1. (1-ac+bc)3 + (a+c2-ac3)3 + (ac3-b-c2)3 = 1
{a,b,c} = {12qrt, 3(q-r)(3q+r)t, 3s2t2}
2. (1-ac-bc)3 + (a+c2-ac3)3 + (ac3+b-c2)3 = 1
{a,b,c} = {12qrt, 3(q+r)(3q-r)t, 3s2t2}
where, for both, r = p-18qs3t3 and the condition p2-3(108s6t6-1)q2 = s. There is a complete soln when s = 1.
Piezas
(9+44pq-404q2)3 + (10+45pq-417q2)3 = (12+56pq-518q2)3 + 1, if p2-85q2 = -4
(9+44pq+404q2)3 + (10+45pq+417q2)3 = (12+56pq+518q2)3 + 1, if p2-85q2 = 4
Using another initial soln 63 + 83 = 93 – 1, we get,
(6+222pq+4014q2)3 + (8+270pq+4806q2)3 = (9+312pq+5616q2)3 - 1, if p2-321q2 = 1
Given one initial soln to x3+y3+z3 = 1 it is possible to find identities like these as discussed in the section on "Third Powers".
13. x3+y3+z3+x+y+z = 0
L.Aubry
{x,y,z} = {u+v, -u+v, -2(3n+1)v}, if u2-(1+12n+36n2+36n3)v2 = n
with n some constant. (For n=1, this entails solving u2-85v2 = 1.) In general, for square n one can always find integral u,v as the conditional equation has the form u2 = av2+n2 for some constant n and which has the soln, courtesy of Fermat as,
v = 2nxy/(x2-ay2)
where if integral v is desired one has to simply solve the Pell equation x2-ay2 = ±1.
14. x4+bx2y2+y4 = z2
Euler
x4+bx2y2+y4 = (x3+y2)2, where b = nx2+2x, and x2-ny2 = 1
More generally, let b = nx2+2v,
x4+bx2y2+y4 = (vx2+y2)2, if v2-ny2 = 1
15. x4+y4 = z2+1
E. Fauquembergue
(17p2-12pq-13q2)4 + (17p2+12pq-13q2)4 = (289p4+14p2q2-239q4)2 + (17p2-q2)4
where q2-17p2 = ±1. It was proven by Fermat that x4+y4 = z2 has no non-trivial solns so this is the next best thing.
P.S. This is the list I have come up so far. Surely there are others? If you know of one, pls send it.