009: Diophantine Equations and Pell Equations

 
 

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+q2(r4+4s4),  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.

 

 

 

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