### 009: Diophantine Equations and Pell Equations

 Back to Index     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.       ◄