### 018: Quartic Polynomials as kth Powers

 Back to Index     II. Quartic Polynomials as kth Powers   ax4+by4 = cz2 ax4+bx2y2+cy4 = dz2 au4+bu2v2+cv4 = ax4+bx2y2+cy4 ax4+bx3y+cx2y2+dxy3+ey4 = z2  1. Form:  ax4+by4 = cz2   Theorem:  "If the form ax4-y4 = z2 has a solution, then so does the form x4-ay4 = z2."   Proof:   1. If ax4-y4 = z2 has a soln, then so does x4+4ay4 = z2, since,     z4 + 4a(xy)4 = (ax4+y4)2,  if ax4-y4 = z2.   2. And if in turn x4+4ay4 = z2 has a soln, then so does x4-ay4 = z2, since,    z4 - a(2xy)4 = (x4-4ay4)2,  if x4+4ay4 = z2.   Combining the two,    (ax4+y4)4 - a(2xyz)4 = (4ax4y4-z4)2,  if ax4-y4 = z2.   For example, the eqn 2x4-y4 = z2 has smallest soln {x,y,z} = {13, 1, 239}.  The theorem implies that p4-2q4 = r2 has solns as well, yielding {p,q} = {57123, 6214}, which is not necessarily the smallest.  Furthermore, given a non-trivial integral soln to ax4+bx2y2+cy4 = dz2, one can generally find an infinite more as proven by an identity of Desboves.  For the special case when a = c = 1, we have   Lagrange   x4+by4 = z2   {x,y,z} = {p4-bq4,   2pqr,  4bp4q4+r4},  if p4+bq4 = r2   Desboves   More generally, if ap4+bq4 = cr2, then,   ax4+by4 = cz2   {x,y,z} = {p(4m2-3(m+n)2),  q(4n2-3(m+n)2),  r(4(m+n)4-3(m-n)4)},  where {m,n} = {ap4, bq4}   re-arranged by this author to make it more aesthetic.  (Note how the algebraic form 4u2-3v2 appears).  This can be generalized even more to cover the form ax4+bx2y2+cy4 = dz2.  The identity involves polynomials of high degree, 9th deg in the {p,q}, but special cases like a+b = c need only cubics,   Desboves   If ap4+bq4 = (a+b)r2, then,   ax4+by4 = (a+b)z2   where {x,y,z} = {-d2p2+4bcq2,  (2c2-d2)pq±2cdr,  4bdpq(4acp2+d2q2)±(2c2-d2)(d2p2+4bcq2)r}, where {c,d} = {a+b, a-b}.    with z as a cubic.  And since this involves the ± sign, for this particular form there are solns that share the same x variable.  For ex, the eqn 2x4-y4 = z2 has small solns {x,y} = {13, 1} and {1525, 1343}.  Using Desboves’ identity, the second yields a pair of {x,y} as,   {28145221, 2165017} {28145221, 30838067}   Euler, Lagrange   Let {u,v} be the legs of a triangle. The problem of making the sum of the legs and its squares as a square and a fourth power, respectively, or   u+v = y2, u2+v2 = x4   can be reduced to a single condition.  These simultaneous equations are true for,   {u,v} = {(y2+z)/2, (y2-z)/2},   if 2x4-y4 = z2   A soln given by Euler is,   {x,y,z} = {p3+2pq2-qr,  p3-4pq2+qr,  p6+p4q2-6p3qr+24p2q4-8q6},  if  p4+8q4 = r2   Piezas   2x4-y4 = z2   {x,y} = {12(p+q)3-(3p+2q)v,  12(p+q)3-(4p+3q)v},  if 2p4-q4 = r2   where v = 2p2+8pq+5q2+r, and z is a sixth deg polynomial.  Note that with the sign of p held constant, the other two variables q,r can come in four combinations: {+,+}, {-,+}, {+,-}, {-,-} generally giving four distinct values {x,y}.   S.Realis, A.Gerardin   In general, one soln to ax4+by4 = cz2 can lead to subsequent ones.  Let n,x be unknowns,   a(u+px)4 + b(v+qx)4 = c(w-nx+rx2)2   Collecting powers of x,   (ap4+bq4-cr2)x4 + (Poly1)x3 + (Poly2)x2 + (Poly3)x + (au4+bv4-cw2) = 0   Assume that (ap4+bq4-cr2) = (au4+bv4-cw2) = 0.  Since n is only linear in Poly1 (or Poly3), equate Poly1 = 0, then solve for n.  The whole equation then reduces to the form,   (Poly2)x2 + (Poly3)x = 0   which, after factoring, is simply a linear eqn in x so we have both our unknowns {n,x} in terms of {p,q,r} and {u,v,w}. These generally lead to addends that are cubic polynomials in terms of p,q,r as in the explicit example given by this author.      2. Form:  ax4+bx2y2+cy4 = dz2   Theorem 1: Given an initial non-trivial solution to ap4+bp2q2+cq4 = dr2, one can generally find an infinite number of solutions. (Desboves)  (See also no. 4 below)   Proof:   ax4+bx2y2+cy4+dz2 = (ap4+bp2q2+cq4+dr2)(z/r)2   {x,y,z} = {p(u2-4cq4w),  q(u2-4ap4w),  r((p4q4v-w2)2-4p4q4u2v)},  where {u,v,w} = {ap4-cq4,  b2-4ac,  ap4+bp2q2+cq4}   Alternatively, to show its affinity to the diagonal form (when b = 0),   {x,y,z} = {p(4m2-3(m+n)2-nt),  q(4n2-3(m+n)2-mt),  r(4(m+n)4-3(m-n)4+mnt(4m+4n+t)+t(m+n)3)},  where {m,t,n} = {ap4, 4bp2q2, cq4}.   Note how when b = t = 0, the identity reduces to the one given in the previous section.  Less general solns are also known,   1. J. Lagrange   x4+cy4 = z2   {x,y,z} = {p4-cq4,   2pqr,  4cp4q4+r4},  if p4+cq4 = r2   This was already given, but is a special case of the next one,   2. V. Lebesgue   x4+bx2y2+cy4 = z2   {x,y,z} = {p4-cq4,  2pqr,  (b2-4c)p4q4-r4},  if p4+bp2q2+cq4 = r2   For b = 0, this reduces to the previous.  This, in turn, in generalized by,   3. A. Desboves   x4+bdx2y2+acd2y4 = z2   {x,y,z} = {ap4-cq4,  2pqr,  (b2-4ac)p4q4-d2r4},  if ap4+bp2q2+cq4 = dr2   which for a=d=1 again reduces to the previous.   4. A. Desboves   ax4+bx2y2+cy4 = dz2   {x,y,z} = {p(u2-4cdq4r2),  q(u2-4adp4r2),  r(4p4q4u2v-(p4q4v-d2r4)2)}   where {u,v} = {ap4-cq4,  b2-4ac},  if ap4+bp2q2+cq4 = dr2   This is basically the identity discussed by Theorem 1 in its original form but takes too long to factor symbolically on a computer. Can the degree of the polynomials be reduced yet still apply to general {a,b,c,d}?   Euler   Let b = nx2+2x,   x4+bx2y2+y4 = (x3+y2)2,  if x2-ny2 = 1   One can then use solns of Pell equations to solve the above form for certain b.  More generally, let b = nx2+2v,   x4+bx2y2+y4 = (vx2+y2)2,  if v2-ny2 = 1     3. Form:  au4+bu2v2+cv4 = ax4+bx2y2+cy4   Piezas   (√p+√q)k + (√p-√q)k = (√r+√s)k + (√r-√s)k     (eq.1)   Poly solns to eq.1 have been found this author for k=5,6,8, with k=7 found by D.Rusin.  For k=8, this becomes,   p4+28p3q+70p2q2+28pq3+q4 = r4+28r3s+70r2s2+28rs3+s4   with one soln,   {p,q,r,s} = {(n-1)(n2-n-1),  (n+1)(n2+n-1),  n3-n-1,  n3-n+1}   Since the above is a symmetric polynomial, a second transformation {p,q,r,s} = {u/2+v, u/2-v, x/2+y, x/2-y} gives it the simpler form,   u4-4u2v2+2v4 = x4-4x2y2+2y4   Any other soln for this as well as for the general case? Also, any non-trivial soln to eq.1 for k>8?   (Update, 11/10/09):  In "Parametric Solutions of the Quartic Diophantine Equation f(x,y) = f(u,v)", Choudhry discusses how, if an initial soln is known, then one can find another soln in general.  He also provides an explicit identity:   Choudhry   x14+2nx12x22+x24 = y14+2ny12y22+y24   {x1, x2} = {1+n2t+st2-rt3+qt4+3t5+pt6+nt7,  -n+pt-3t2+qt3+rt4+st5-n2t6+t7} {y1, y2} = {1-n2t+st2+rt3+qt4-3t5+pt6-nt7,    n+pt+3t2+qt3-rt4+st5+n2t6+t7}  where {p,q,r,s} = {1-n+n2,  -2+4n+n2,  -4n+n3, 1+n2+n3}.   Note how, for {x1,y1} the signs are different with odd powers t, while for {x2,y2}, they are different for even powers t.  For the case n = 0, this reduces to the formula found by Gerardin, while n = 1 is trivial.  (End update.)   4. Form:  ax4+bx3y+cx2y2+dxy3+ey4 = z2   For the 4th power univariate case to be made a square, Fermat's method still applies. If it has an initial soln, we can use the form with the square constant term,   ax4+bx3+cx2+dx+e2 = z2   and assume it as equal to,   ax4+bx3+cx2+dx+e2 = (px2+qx+e)2   Using the same approach, the unknowns {p,q} are enough to knock out enough terms such that one can solve for x, given by,    x = (be-dp)/(e(p2-a)),  where p = (4ce2-d2)/(2e)3   Unfortunately, for the quintic function f(x) = z2, not enough terms can be removed so the method stops here.  Interestingly, this is similar to the situation of solving f(x) = 0 where x is in radicals since there are general formulas only for degree 2,3,4 but none for higher.  Has it been explicitly proven there is no general formula to find rational and non-zero x that solves f(x) = z2 where f(x) is a quintic or higher degree with rational coefficients and a square constant term?       ◄