023: Sixth Powers (4 or 6 terms)

 Return to Index     PART 9.  Sum / Sums of Sixth Powers      6.1  Four terms   As was already discussed in a previous section, the equation (√p+√q)6 + (√p-√q)6 = (√r+√s)6 + (√r-√s)6 has a polynomial soln.  It is conjectured that,   x16 + x26 = y16 + y26   is only trivially solvable. However, if we relax it a bit, there is an infinite number of solutions to,   x16 + x23 = y16 + y23       (1)   Proof:   We use the form,   (c-x)6 + (x+y)3 = (c+x)6 + (-x+y)3   Expanding this, one must solve,   6cx4 + (20c3-1)x2 + 6c5 = 3y2   which, for an appropriate constant c, can be treated as an elliptic curve. For example, for c = 7/2, one gets {x,y} = {1/2, 67/2} which yields,   36 + 343 = 46 + 333   From this initial point {x,y}, one can then easily get subsequent ones, ad infinitum.  Surprisingly, there is also a small solution to, x16 + x26 = y16 + y23        (2)   as, 156 + 186 = 196 + (-118)3 It can also be shown there is an infinite number of primitive solutions to (2) using the well-known identity,   (3t2)6 + (3t-9t4)3 = 1 + (9t3-1)3   by making the second term a square,   3t-9t4 = y2   This can be treated as an elliptic curve. One such point is t = 3/13 which yields,   276 + 1386 = 1312 + (-25402)3   Another is t = 208098704151/634343459737 (though presumably there may be points of smaller height), and so on.  A different approach is used to solve the analogously relaxed,   x18 + x28 + x34 = y18 + y28 + y34   discussed in Eighth Powers.      6.2  Six Terms: Equal sum of three sixth powers, x16+x26+x36 = x46+x56+x66   Duncan Moore has done an exhaustive search within a radius of 17,800.  Out of about 400 solns, it turns out that 92% (!) are good for k = 2,6.  Why that is the case, no one knows, though it may have to do with the algebraic form x2+xy-y2.  (See Bremner and Kuwata's work below.)  Data results are here.  In fact, the smallest (6.3.3), (6.3.4), (6.4.4) are multi-grade for k = 2,6, [1]   23k + 10k + 15k = 3k + 19k + 22k [2]   73k + 58k + 41k = 70k + 65k + 32k + 15k [3]   10k + 6k + 5k + 3k = 9k + 9k + 2k + 2k The situation can be contrasted to the smallest known (8.4.4), (8.4.5), (8.5.5), (8.5.6), (8.6.6), none of which is multi-grade.  [1] and [3] belong to parameterizations as, (ad+b)k + (c-2d)k + (de+f)k = (de-f)k + (c+2d)k + (ad-b)k,    where {a,b,c,d,e,f} = {-4x+11y,  y,  2-5(2x-5y)y,  x-y,  2x-3y,  2x-5y},  and {x,y} satisfies x2-6y2 = 1, with {x,y} = {7/5, -2/5} yielding [1] while, (a+d)k + (-a+d)k + (b+c)k + (b-c)k = 2(c+d)k + 2(-c+d)k  where 5a2+2b2 = 7c2, and 2a2+5b2 = 7d2, with {a,b,c,d}= {13, 1, 11, 7} giving [3], after removing common factors for both examples.  It is not known whether [2] belongs to a family.  For 6th powers, identities, whether in terms of a polynomial or an elliptic curve, are known only for multi-grades k = 2,6, or 1,2,6:   1. x1k+x2k+x3k = y1k+y2k+y3k,  for k = 2,6, and x1y1(x12-y12)+x2y2(x22-y22)+x3y3(x32-y32) = 0.  Additional side conditions include,     a) 3x1+x2+x3 = y1+y2+3y3     b) 2x2 = y2+y3, 2x3 = y2-y3    2. x1k+x2k+x3k = y1k+y2k+y3k,  for k = 1,2,6.  Additional conditions include,     a) nx1+x2-x3 = ny1+y2-y3,  for n = 12, 15, 21, etc.   If you know of another class, pls send it.  The second condition of the first class can be expressed as a set of three simultaneous eqns given below.  As was mentioned, small solns to (k.3.3) for k = 4 or 6 also turn out to be valid for k = 2.  The smallest for k = 6 is,   23k + 10k + 15k = 3k + 19k + 22k   which, in fact, is for k = 2,6.  This deceptively simple-looking eqn has a lot of structure.  If expressed as,   {-23, 10, 15} = {3, 19, -22}   labelled {a,b,c,d,e,f} respectively, then,   a2+ad-d2 = -(b2-be-e2) b2+be-e2 = -(c2-cf-f2) c2+cf-f2  = -(a2-ad-d2) 3a+b+c = d+e+3f   Not surprisingly, this is just the smallest instance of a parametric soln.  Cubic and 4th powers have already been discussed and it was seen that a lot of identities involved equivalent forms of F3:= x2+xy+y2.  For 5th and 6th powers, it seems now it is the form F5:= x2+xy-y2 that is implicit in some identities.  For example, recall that,   (√p+√q)5 + (√p-√q)5 = (√r+√s)5 + (√r-√s)5   {p,q,r,s} = {5vw2,  -1+uw2,  5v,  -(u+10v)+w3}   where w = u2+10uv+5v2 and it takes only a small change of variables {u,v} = {x-2y, x+2y} to transform this to the form x2+xy-y2.  Though F3 and F5 are similar-looking, they are quite different since the former has discriminant d = -3 and hence factors over the imaginary field √-3, while the latter has d = 5 and factors over the real field √5.  It has been shown that solns to a2+ab+b2 = c2+cd+d2 also solve ak+bk+(a+b)k = ck+dk+(c+d)k  for k = 2,4, and vice versa.  For k = 2,6, one now needs a system of three equations,   Theorem 1:  (Bremner, Kuwata)  If there are {a,b,c,d,e,f} such that, call this system S1,   a2+ad-d2 + (b2-be-e2) = 0 b2+be-e2 + (c2-cf-f2) = 0 c2+cf -f2 + (a2-ad-d2) = 0   is true, then so is system S2,   ak+bk+ck = dk+ek+fk,  for k = 2,6   Proof:  Simply add the expressions together to get the identically true statement,   (a2+ad-d2)k + (b2-be-e2)k + (b2+be-e2)k  + (c2-cf-f2)k + (c2+cf-f2)k + (a2-ad-d2)k  = 2(a2k+b2k+c2k-d2k-e2k-f2k),  for k = 1,3   If the LHS is zero, then so is the RHS, which proves the theorem.  (Masato Kuwata, Equal Sums of Sixth Powers and Quadratic Line Complexes, 2007. End proof.)  (Note:  This probably contributes to the reason why more than 90% of solns are multi-grade for k = 2,6 since being valid for one power means being valid for the other.  Unfortunately, no similar expressions are yet known for 8th powers!)    Corollary 1: (Piezas) Solutions to S1 imply the ff sums are squares,   4(a2+ad-d2) + 5b2 = (b+2e)2,   4(b2+be-e2) + 5c2 = (c+2f)2,   4(c2+cf-f2) + 5a2 = (a+2d)2   4(a2-ad-d2) + 5c2 = (c-2f)2,      4(b2-be-e2) + 5a2 = (a-2d)2,   4(c2-cf-f2) + 5b2 = (b-2e)2   In Unsolved Problems in Number Theory, R. Guy asked if the reverse was true: does S2 imply S1?  This was answered in the negative by Delorme who gave one more condition, and Choudhry who gave a parametric example that solved S2 but not S1.   Theorem 2: (Delorme) The system S2 implies S1 if it is also the case that,   ad(a2-d2) + be(b2-e2) + cf(c2-f2) = 0       S. Brudno (1970); S. Brudno, I. Kaplansky (1974)   This is a simple example that solves both S1 and S2 ,   ak + bk + ck = dk + (b+c)k + (b-c)k,   k = 2,6   Note that for k = 2, one must solve a2 = b2+c2+d2.  The complete soln for k = 2,6 is then,   {c,d} = {ax/(a2+9b2),  3bx/(a2+9b2)}   where {a,b,x} satisfies the elliptic curve,   (a2-b2)(a2+9b2) = x2   Labelled as xi, yi, the terms also satisfy x1y1(x12-y12) + x2y2(x22-y22) + x3y3(x32-y32) = 0.  One small soln, among infinitely many, is {a,b} = {5/4, 1} which yields,   65k + 52k + 15k = 36k + 67k + 37k    Piezas   This is another simple example that solves both systems,   (ad+b)k + (c-2d)k + (de+f)k = (de-f)k + (c+2d)k + (ad-b)k,   k = 2,6   where {a,b,c,d,e,f} = {-4x+11y,  y,  2(x-3y)(x-2y)+y2,  x-y,  2x-3y,  2x-5y},  and {x,y} satisfies x2-6y2 = 1.    It is quite interesting that this particular Pell equation appears in the context of 6th powers.  However, since the eqn is homogeneous, one can just as well solve it in the rationals.  This is also discussed several sections below.     A. Bremner, Piezas   It turns out the complete soln of S1, S2 can be given in terms of a quartic polynomial that is to be made a square, hence can be treated as an elliptic curve.  R. Guy in the same book mentions that Bremner gave a method that can find all parametric solns.  I don’t have access to this paper yet but after some experimentation found a procedure.  It suffices to use S2 and one of the eqns of S1.  Using the general form,   (p+q)k + (r+s)k + (t+u)k = (t-u)k + (r-s)k + (p-q)k   for k = 2,6 and one of the others, say c2+cf-f2 = -(a2-ad-d2) with variables changed appropriately, the complete soln is then,   {q,u} = {-rs(p+2t)/w,  rs(2p-t)/w}, where w = p2+t2  and {p,r,s,t} satisfy,   ((p2-11pt-t2)s2 + w2)r2 = (p2+pt-t2+s2)w2   (Perhaps not surprisingly, the form F5:= x2+xy-y2 appears again.)  Thus, the problem is to find {p,t,s} such that the expression,   ((p2-11pt-t2)s2 + (p2+t2)2)(p2+pt-t2+s2) = y2   which is only a quartic in the variable s is a square, though some {p,t,s} are trivial.  So, given an initial soln, this can be treated as an elliptic curve to generate more.  For ex, let,   {p,t,s} = {1, n, n-1}   This is trivial with respect to the original sextic eqn S2 but, using the same {p,t}, we can find more rational points s on the curve with the next one (using Fermat’s method) non-trivial of deg-18.  There are also small non-trivial solns,   {p,t,s} = {-n-1,  (n-1)(n-2),  5+n2} {p,t,s} = {(-3+n)(2+n),  3+3n+2n2,  3+6n+n2}   either one of which yields scaled versions of the 4th-deg Brudno-Delorme identity given below.  Again this can be used to generate more rational points with the first one yielding a rational 18-deg.  (Using the same {p,t}, there is another rational point s that is also only a quadratic, s = 3-2n+n2 for the first and s = 9+4n+n2 for the second, but these are trivial.)  Other solns are,   {p,t,s} = {2(1+n)(-1-n+n2)(-1+n+n2),  (1+n)(1-n+n2)(1-5n+n2),  (-1+n)(1+2n+7n2+2n3+n4)} {p,t,s} = {2(3+3n+2n2)(3+2n+2n2),  (3+3n+2n2)(3-2n+4n2), (1+n)(9+4n2)}   The first, after a small adjustment, yields Delorme’s 5th-deg identity given below, while the second yields a different 5th deg.  Delorme in his paper gave  identities of deg n for all 4 ≤ n ≤ 11, except n = 6,10.  Q:  Can anyone give an identity with polynomials of deg n = 6?   Note:  It can be shown that the system ak+bk+ck = dk+ek+fk,  for k = 2,6 with,   a2+nad-d2 = -(b2-nbe-e2) b2+nbe-e2 = -(c2-ncf-f2) c2+ncf-f2  = -(a2-nad-d2)   has non-trivial solns only for n = ±1.  Proof:  Using the general form and the method described above, and elimination of appropriate terms, the final resultant eqn has either trivial factors, or a quadratic factor solvable over √(-1), or if (n+1)(n-1) = 0.  Since the first two can be disregarded then only the last applies, proving the assertion.   S. Brudno (1976), J. Delorme   The 4th-deg Brudno-Delorme identity also has the side condition 3a+b+c = d+e+3f, call this S3.  Let,   ak+bk+ck = dk+ek+fk,  for k = 2,6 where,   {a,b,c} = {-n4-n3-5n2+8n+8,  (n3+7n-2)(n+2),  3(3n2+2n+4)} {d,e,f} = {(n2-n+3)(n+2)2,  -4n3-5n2-8n+8,  -n4+n2+14n+4}   (modified to reduced the size of the coefficients), then these satisfy S1, S2, S3.  The smallest non-trivial example is n = -3 which, after removing common factors, yields {-23, 10, 15} = {3, 19, -22}, the example given earlier.  Interestingly, it can be shown that this identity depends on the equation x2-6y2 = z2. This author found that some of the relations are enough to derive a version using: one eqn from S2 (at k = 2), two from S1, and S3.  Let,   (ad+b)k + (c-2d)k + (de+f)k = (de-f)k + (c+2d)k + (ad-b)k   and using this substitution also on the side conditions, one can derive the variables as,   {a,b,c,d,e,f} = {-4x+11y,  y,  2(x-3y)(x-2y)+y2,  x-y,  2x-3y,  2x-5y},  where x2-6y2 = 1.    One can easily solve for {x,y} in the integers as a Pell equation or, since the eqn is homogeneous, in the rationals.  Q:  Is it possible to solve S2, S3 without solving S1?   J. Delorme   Delorme gave identities of deg n for all 4 ≤ n ≤ 11 , except n = 6,10, that solves S1, S2, one of which is the 5th deg,   {x1, x2, x3} = {3n5+8n4+9n3-4n2-9n-2,  -2n5-n4+12n3+13n2+4n-1,  -n5-9n4-13n3-7n2-7n-3} {y1, y2, y3} = {2n5+9n4+4n3-9n2-8n-3,  -3n5-7n4-7n3-13n2-9n-1,  -n5+4n4+13n3+12n2-n-2}   Q: Any linear relations between the xi and yi?   Update (7/14/09): Since the xi, yi are polynomials in n, it turns out the problem is equivalent to finding six unknown rational constants pi such that p1x1+p2x2+p3x3+p4y1+p5y2+p6y3 = m for some constant m, preferably m = 0, for all n.  Expanding and collecting powers of n, set the coefficients, which are polynomials in the pi, equal to zero.  For this particular identity, we can't find pi such that m = 0, but instead find m = 40 given by the linear relation,   5x1-12x2-7x3-12y1+5y2+7y3 = 40   (End note)   Update (6/24/09): A. Choudhry   The fact that solving the systems S1, S2 is reducible to an elliptic curve implies that one soln can lead to another.  Choudhry ("On Equal Sums of Sixth Powers", 1994) has given an explicit construction for this.  Given the system,   xk+yk+zk = uk+ vk+ wk,  k = 2,6 with,   x2+xu-u2 = w2+wz-z2 y2+yv-v2 = u2+ux-x2 z2+zw-w2 = v2+vy-y2   Let {x,y,z,u,v,w} = {x1, x2, x3, y1, y2, y3} be a soln, then a new one is given by, {x,y,z,u,v,w} = {ap+x1q, bp+x2q, cp+x3q, dp+y1q, ep+y2q, y3q}, where {p,q} and {a,b,c,d,e} are,   {p,q} = {-a(2x1-y1)-b(2x2+y2)+d(x1+2y1)-e(x2-2y2),  a2+b2-d2-e2-ad+be}   {a,b,c,d,e} = {(r12+2r1r2)s2,  (r32-2r3r4)s1,  s1s2,  -(r12+r22)s2,  (r32+r42)s1}   {s1, s2} = {r22+r2r1-r12, r32+r3r4-r42},  and {r1, r2, r3, r4} = {x1-y3, x3-y1, x2-y3, x3-y2}.   Example:  The initial soln {x1, x2, x3, y1, y2, y3} = {3, 22, -19, 23, 15, 10} gives {x,y,z,u,v,w} = {4513, -104, 4693, -2273, -5099, 3352} after removing the common factor 3*392002.  (End note.)  A. Choudhry   There is a 3-parameter soln for k = 1,2,6 which entails solving an elliptic curve.  This is a class that solves S2, but not S1.   x1k+x2k+x3k = y1k+y2k+y3k   {x1, x2, x3} = { 2(α+β)m+(α-β+t)n,  -2αm+(α+β+t)n,  -2βm-(α+β-t)n} {y1, y2, y3} = {-2(α+β)m+(α-β+t)n,   2αm+(α+β+t)n,   2βm-(α+β-t)n}   for k = 1,2 is true for all variables, though is not a complete soln.  We can do the substitution {α+β, α-β, t} = {a,b,c} to get the slightly more symmetric,   {x1, x2, x3} = { 2am+(b+c)n,  -(a+b)m+(a+c)n,  -(a-b)m-(a-c)n} {y1, y2, y3} = {-2am+(b+c)n,   (a+b)m+(a+c)n,   (a-b)m-(a-c)n}   When expanded for k = 6 this has the condition,   (Poly1)m2+(Poly2)n2 = 0   where,   Poly1:= b(11a2+b2)+5(3a2+b2)c;   Poly2:= (a2+b2)(b+5c)+10(bc2+c3)   linear and cubic in c, respectively.  The problem then is:  Given constants a,b, find c such that,   y2 = -(Poly1)(Poly2)   avoiding the cases c = -b/3 and y = Poly1 = 0 since these give trivial results.  For a non-trivial ex., let {α,β} =  {1, 7}, hence {a,b} = {8,-6} giving,   y2 = -6(-74+19c)(-60+50c-6c2+c3)   one soln being c = 38/23 with this elliptic curve having an infinite number of rational points. (Choudhry gave more than twenty {a,b} with another as {11,-9}.)   Piezas   While Choudhry did not give the complete soln of k = 1,2,6, it can be shown that the general soln entails making a certain quartic polynomial into a square (just like for the case k = 2,6 plus S2).  We can show this in two ways.    First method:  Using a small variant of the general form,   (p+q)k + (r+s)k + (t+u)k = (-t+u)k + (-r+s)k + (-p+q)k   and letting r = np for some rational constant n, one can completely solve k = 1,2 with,   {t, u} = {-p(n+1),  (q+ns)/(n+1)}   With these values, k = 6 becomes a polynomial of form (Poly1)p2+(Poly2) = 0, where Poly1 is linear and Poly2 is cubic in {q,s}, so their product is a quartic and the problem is reduced to finding,   y2 = -(Poly1)(Poly2)   a situation similar to Choudhry’s, though the tricky part is finding appropriate rational n which may be the ratio of relatively large integers.    Second method:  We simply use the old form L1 again.  Expanding for k = 2, the non-trivial condition is b+c = 0, hence we modify it to,   (a+bp+q)k + (b-bp+q)k + (-b+ap+q)k = (a-bp+q)k + (b+ap+q)k + (-b+bp+q)k   which solves k = 1,2.  Expanding for k = 6 results in a quadratic in b of form (Poly1)b2+(Poly2) = 0, where Poly1 is linear and Poly2 is cubic in {p,q}.  The objective then is to make its discriminant D a square,   y2 = -(Poly1)(Poly2)   similar to the first method, but now the expressions are simple enough to write down,   Poly1:= a(1+p)(1+p2) + 5(1-p+p2)q Poly2:= a3(1+p)(1+p2) + 5a2(1+p+p2)q + 10a(1+p)q2 + 10q3   Thus any solution to k = 1,2,6 must satisfy D = y2.  For example, using a result of Choudhry’s,   43k + (-372)k + 371k = 307k + (-405)k + 140k   for k = 1,2,6, we equate its first four terms with that of the modified L1 to get,   {a,b,p,q} = {291, -388, 33/97, -116}   Using the values for a,p gives the elliptic curve in q,   y2 = -(818844+7297q)(7369596+123291q+780q2+2q3)   where a square numerical factor has been removed.  A rational point, of course, is Choudhry's q = -116 (with another one as q = -130 though this gives a trivial result).  From these, more points q can then be computed though typically they are rational numbers with many digits.  Note:  Incidentally, if we are to add more constraints to this system, then there are no non-trivial rational solns to k = 1,2,6 plus any of the eqns of S1, for ex. c2+cf-f2 = -(a2-ad-d2).  These four eqns are enough to linearly derive a soln.  Using the general form,   (p+q)k + (r+s)k + (t+u)k = (t-u)k + (r-s)k + (p-q)k   then,   {q,r,u} = {(p+2t)s/v,  -(p2+t2)/v,  -(2p-t)s/v},  where v = p-3t   and {p,s,t} satisfies p3-2p2t+pt2+8t3-(p-8t)s2 = 0.  One must then solve the elliptic curve,   (p3-2p2t+pt2+8t3)(p-8t) = y2   and as far as I checked this has only trivial solns.  However, the curve,   (p3-2p2t+pt2+8t3)(p-8t) = -y2   does have non-trivial ones, like {p,t,y} = {24, 13, 1280} and many more thus giving an imaginary value to s.  So this system is solvable in terms of Gaussian rationals.   ◄
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