015: Sums of three Fourth Powers (Part 2)

10.1. Constraint:  a+b = nc, d+e = nf


The complete soln for the case n = 1 is given by,


ak + bk + (a+b)k = ck + dk + (c+d)k,  k = 2,4


where a2+ab+b2 = c2+cd+d2.  By modifying it slightly to the form,


ak + bk + (-a-b)k = ck + dk + (-c-d)k


makes it valid for k = 1 as well.  One soln is, define {m,n} = {p-3q, p+3q}, then for k = 1,2,4,


(2p-mr)k + (2pr-n)k + (-m-nr)k = (2p-nr)k + (2pr-m)k + (-n-mr)k


for arbitrary {p,q,r}.

(Update, 1/25/10):  Wroblewski




(a2)k + (b2)k + (a2+b2)k = (c2)k + (d2)k + (c2+d2)k,  for k = 2,4    (eq.1)


or equivalently,


a4+a2b2+b4 = c4+c2d2+d4


then we get the [k.8.8] identity,


[ap+cq, -ap+cq, bp+dq, -bp+dq, dp+aq, dp-aq, cp+bq, cp-bq]k =  

[ap+dq, -ap+dq, bp+cq, -bp+cq, cp+aq, cp-aq, dp+bq, dp-bq]k,   for k = 1,2,4,8


(Notice how {c,d} of the LHS is replaced by {d,c} in the RHS.)  for arbitrary {p,q} and some constants {a,b,c,d}, hence giving an identity in terms of linear forms.  An example is {a,b,c,d} = {1, 26, 17, 22}, though Choudhry has given a 7th deg identity for,


a4+ma2b2+b4 = c4+mc2d2+d4


for arbitrary m.  See Form 3 here.  To find a [8.7.7], since {p,q} are arbitrary, one can equate appropriate xp = yq so a pair of terms cancel out.  Note:  Eq. 1, or more generally its unsquared version, is at the core of the Ramanujan 6-10-8 Identity.  (End update.)



(Update, 2/3/10):  Choudhry


Gave a soln to ak+bk+ck = dk+ek+fk,  k = 1,2,4, for integer {a,b,c,d,e,f} with the interesting property that one term is equal to 1.


(-p+v)k + (-q+v)k + (p+q-2v)k = (-p-q+2uv)k + (p+q-2uv-1)k + 1k 


where {p,q} = {2u2v+u-v,  4uv+3v+1}.


Source:  On Representing 1 as the sum or difference of kth powers of integers, The Mathematics Student, Vol. 70, 2001.


Note:  Since this also has a+b+c = d+e+f = 0, its terms obey the beautful Ramanujan-Hirschhorn Theorem given below.  (End update.)




(a+b+c)k + (b+c+d)k + (a-d)k = (c+d+a)k + (d+a+b)k + (b-c)k


if ad = bc, for k = 2,4.


This form appears in the beautiful 6-10-8 and 3-7-5 Identities,




64[(a+b+c)6+(b+c+d)6+(a-d)6-(c+d+a)6-(d+a+b)6-(b-c)6][(a+b+c)10+(b+c+d)10+(a-d)10-(c+d+a)10-(d+a+b)10-(b-c)10] = 45[(a+b+c)8+(b+c+d)8+(a-d)8-(c+d+a)8-(d+a+b)8-(b-c)8]2


M. Hirschhorn


25[(a+b+c)3-(b+c+d)3-(a-d)3+(c+d+a)3-(d+a+b)3+(b-c)3][(a+b+c)7-(b+c+d)7-(a-d)7+ (c+d+a)7-(d+a+b)7+(b-c)7] = 21[(a+b+c)5-(b+c+d)5-(a-d)5+(c+d+a)5-(d+a+b)5+(b-c)5]2


where for both, ad = bc, with the discovery of the latter inspired by the former.  These can be generalized as,




Theorem 1a:  If ak+bk+ck = dk+ek+fk, k = 2,4  (eq.1)  where a+b+c = d+e+f = 0, then,


64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2   (eq.2)


25(a3+b3+c3-d3-e3-f3)(a7+b7+c7-d7-e7-f7) = 21(a5+b5+c5-d5-e5-f5)2           (eq.3)


Proof:  By doing the substitution {a,b,c,d,e,f} = {p, q, -p-q, r, s, -r-s}, all three equations have the common factor p2+pq+q2-(r2+rs+s2) = 0 so the problem is reduced to finding expressions {p,q,r,s} that satisfy this.  (End proof)  It has a complete soln.  Let {p,q,r,s} = {-u+v, u+v, -x+y, x+y}, and the condition becomes,


u2+3v2 = x2+3y2


with the complete soln to u2+nv2 = x2+ny2 as, (ac+nbd)2 + n(bc-ad)2 = (ac-nbd)2 + n(bc+ad)2There are also interesting relations using odd powers k = 1,3,

Theorem 1b:  If ak+bk+ck = dk+ek+fk for k = 1,3  (eq.1) where a+b+c = d+e+f = 0,  then,


9abc(a6+b6+c6-d6-e6-f6) = 2(a9+b9+c9-d9-e9-f9)     (eq.2)


Proof:  Doing the same substitution {a,b,c,d,e,f} = {p, q, -p-q, r, s, -r-s}, the two equations have the common factor pq(p+q)-rs(r+s) = 0.  (End proof)  This is also discussed in Form 13 of Third Powers.   Another uses k = 1,3,5,




Theorem 2:  If ak+bk+ck+dk = ek+fk+gk+hk,  k = 1,3,5 where a+b+c+d = e+f+g+h = 0, then,


7(a4+b4+c4+d4-e4-f4-g4-h4)(a9+b9+c9+d9-e9-f9-g9-h9) = 12(a6+b6+c6+d6-e6-f6-g6-h6)(a7+b7+c7+d7-e7-f7-g7-h7)


which again has a complete soln.  For higher powers,


Theorem 3:  If ak+bk+ck+dk = ek+fk+gk+hk,  k = 2,4,6, define n as, a4+b4+c4+d4 = n(a2+b2+c2+d2)2 then,


25(a8+b8+c8+d8-e8-f8-g8-h8)(a12+b12+c12+d12-e12-f12-g12-h12) = 12(n+1)(a10+b10+c10+d10-e10-f10-g10-h10)2


Theorem 4:  If ak+bk+ck+dk+ek = fk+gk+hk+ik+jk,  k = 2,4,6,8 define n as a4+b4+c4+d4+e4 = n(a2+b2+c2+d2+e2)2 then,


72(a10+b10+c10+d10+e10-f10-g10-h10-i10-j10)(a14+b14+c14+d14+e14-f14-g14-h14-i14-j14) = 35(n+1)(a12+b12+c12+d12+e12-f12-g12-h12-i12-j12)2


and so on based on higher systems k = 2,4,…2m.  Also, buried in the musty pages of an old journal, there is a similarly beautiful identity, Sinha's 2-4-6 Identity,




5[a2+b2+c2]*[(a+b+c)4 + (a+b-c)4 + (a-b+c)4 + (-a+b+c)4 - (2a)4 - (2b)4 - (2c)4] = [(a+b+c)6 + (a+b-c)6 + (a-b+c)6 + (-a+b+c)6 - (2a)6 - (2b)6 - (2c)6]


which this author realized could be extended (and no further) as,



[(a+b+c-d)4 + (a+b-c+d)4 + (a-b+c+d)4 + (-a+b+c+d)4 - (2a)4 - (2b)4 - (2c)4 - (2d)4] =

[(a+b+c-d)6 + (a+b-c+d)6 + (a-b+c+d)6 + (-a+b+c+d)6 - (2a)6 - (2b)6 - (2c)6 - (2d)6]


One can see that Sinha’s is just the case d=0.  How this was derived will be discussed in the section on Eighth Powers.


V. Tariste


(23x-57y)k + (40x+6y)k + (17x+63y)k = (23x+57y)k + (40x-6y)k + (17x-63y)k


The identity is in fact valid for k =2,4.  There is nothing really special about these particular values as these can be generalized as,




Let {u1, u2, u3, c} = {a+2b, 2a+b, a-b, a+b},  then for k = 2,4,


(ax+u1y)k + (bx-u2y)k + (cx-u3y)k = (ax-u1y)k + (bx+u2y)k + (cx+u3y)k


(ax+u2y)k + (bx-u3y)k + (cx+u1y)k = (ax+u3y)k + (bx+u1y)k + (cx+u2y)k


And as sums of one side,


(ax+u1y)k + (bx-u2y)k + (cx-u3y)k = (ak+bk+ck)(x2+3y2)k/2


(ax+u2y)k + (bx-u3y)k + (cx+u1y)k = (ak+bk+ck)(x2+3xy+3y2)k/2




(ax2+2u1xy-3ay2)k + (bx2-2u2xy-3by2)k + (cx2-2u3xy-3cy2)k = (ak+bk+ck)(x2+3y2)k


where the last is a template identity that can generate quadratic parametrizations using an initial soln, as long as it satisfies the condition a+b = c.  For example, the smallest soln, 24+24+44+34+44 = 54, was used by Ramanujan.  Another, found by this author, is 504+504+1004+44+154 = 1034 which yields, 


(50x2+300xy-150y2)4 + (50x2-300xy-150y2)4 + (100x2-300y2)4 + (4x2+12y2)4 + (15x2+45y2)4 = 1034(x2+3y2)4

and so on for others involving a+b = c.  Note that the expressions {a+2b, 2a+b, a-b, a+b} are intimately connected to the form x2+3y2 since,


(a+2b)2 + 3a2 = (2a+b)2 + 3b2 = (a-b)2 + 3(a+b)2 = 4(a2+ab+b2)


However, for general n, given the system,
a+b = nc;  d+e = nf

a2+b2+c2 = d2+e2+f2

a4+b4+c4 = d4+e4+f4


and the substitution {a,b,c,d,e,f} = {p+q,r+s, t+u, t-u, r-s, p-q}, where set t = 1 without loss of generality, solve for {p,s,u} and substitute these into the last eqn.  This will have a quadratic factor in r with discriminant D,


D:= (1+n2)2q4 + 8n(2+n2)q3 - 2(-17-6n2+n4)q2 - 8n(2+n2)q + (1+n2)2


which must be made a square.  Since this quartic polynomial has a square leading and constant term, this is easily done.  Two small solns are,


q1 = -2/n,   q2 = (n-1)/(n+1)


Though these lead to a trivial soln, they can be used to compute other rational points on the curve.  A non-trivial point found using Fermat’s method is,


q3 = (n6+2n4+n2-4)/(4n(n2+1)2)


10.2.  Constraint:  a+b ≠ c; d+e ≠ f


Birck, Sinha


Theorem 1: If ak+bk+ck = dk+ek+fk,  k = 2,4


where a+b ≠ ±c; d+e ≠ ±f, then,


(a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k + (2e)k + (2f)k + (2g)k =

(d+e+f)k + (-d+e+f)k + (d-e+f)k + (d+e-f)k + (2a)k + (2b)k + (2c)k


for k = 1,2,4,6,8. 


This author found an extension of this to tenth powers and will be discussed later.  The proof for this particular theorem of Birck will be discussed on the section on Eighth Powers.  From this, Sinha derived a version applicable to odd exponents,


T. Sinha


Theorem 2: If ak+bk+ck = dk+ek+fk,  k = 2,4  (or Q1)


where a+b-c = 2(d+e-f) and a+b ≠ c, d+e ≠ f, then,


(2a-3h)k + (2a-h)k + (2b-3h)k + (2b-h)k + (2f+h)k = (2c+h)k + (2c+3h)k + (2d-h)k + (2e-h)k + (3h)k


for k = 1,3,5,7, where 2h = d+e-f. 


To solve Q1 and the side conditions, Sinha gave a system of linear relations (modified by this author to be consistent with the others), call it S,


a+b-c = 2(d+e-f)

(a+b)-(d-e) = 0; 

13a+13b+11c+4f = 0


These can then define three of the variables.  For example, solving for {a,c,f}, and substituting into Q1 for either k = 2 or 4, this forces the remaining three variables, say they are {x,y,z}, into a quadratic eqn of form,


a1x2 + a2y2 + a3z2 + a4xy + a5xz + a6yz = 0


One can solve this by using an identity of Desboves’ (found in the section on Second Powers) that needs an initial soln {x,y,z}.  Another approach, producing more aesthetic results, is that since this is a quadratic in any of the variables, say z, for rational solns its discriminant must be made a square,


b1x2+b2xy+b3y2 = t2


where the bi are in terms of the ai.  Thus, if one can find a system of useful linear relations between the {a,b,c,d,e,f}, the problem will be reduced to solving this eqn.  As this author found out, there are at least two systems S,


  1. a+b-c = 2(d+e-f);  (a+b) + (d-e) = 0;  13a+13b+11c+4f = 0
  2. a+b-c = 2(d+e-f);  (a-b) + (d+e) = 0;  4c+7d+7e-f = 0

These then yield polynomial solns to Sinha's problem as, 


1. (-5x+2y+z)k + (-5x+2y-z)k + (6x-4y)k = (9x-y)k + (-x+3y)k + (16x-2y)k


where 126x2-5y2 = z2,  D = 9(70).  (Sinha’s)


2. (6x+3y)k + (4x+9y)k + (2x-12y)k = (-x+3y+3z)k + (-x+3y-3z)k + (-6x-6y)k


where x2+10y2 = z2,  D = -10.


where D is the discriminant of the conditional eqn and some manipulation was done to attain these simple forms.  Since all have a square leading coefficient, these can be transformed to the form u2+dv2 = w2 which has a complete soln.


Note 1: Any other linear relations or S that is not merely a permutation of variables and change of signs?  For ex, the first pair (a+b)±(d-e) = 0 is equivalent to either of the ff: (a+b)±(d+f), (a-c)±(d-e), (a-c)±(d+f), while the second pair (a-b) ± (d+e) = 0 is the same as (a-b)±(d-f), (a+c)±(d+e), (a+c)±(d-f).  I do not know if the above four solns are the only ones in terms of binary quadratic forms.


Note 2:  Sinha’s theorem can be expressed more concisely as given,


(a+3c)k + (b+3c)k + (a+b-2c)k = (c+d)k + (c+e)k + (-2c+d+e)k, 


for k = 2,4, then,


ak + bk + (a+2c)k + (b+2c)k + (-c+d+e)k = (a+b-c)k + (a+b+c)k + dk + ek + (3c)k


for k = 1,3,5,7, (excluding the trivial case c = 0). 




There can be more solns to Q1 with a+b-c = 2(d+e-f) though now an elliptic curve has to be solved.  The linear relation used is either case of (a+b) = ±(d+e).  The general form is,


(-y+t)k + (y+t)k + pk = (-z±t)k + (z±t)k + qk, 


where t is chosen so the eqn satisfies a+b-c = 2(d+e-f).  For the positive case,


(-y+t)k + (y+t)k + (48-2t)k = (-12x+t)k + (12x+t)k + (24)k,  where t = 13-x2,  and y2 = -2x4+100x2+46


This elliptic curve has trivial solns x = {1,3,5,7} but from these we can find non-trivial ones such as 4307x = 2367, and so on.  For the negative case,


(-y+t)k + (y+t)k + (2u+6v)k = (-z-t)k + (z-t)k + (-2u)k,  where t = u+v,


and u,v must satisfy the simultaneous eqns u2+6v2 = y2 and u2+12uv+24v2 = z2.  Solving the first as {u,v} = {p2-6, 2p}, the second becomes,


p4+24p3+84p2-144p4+36 = z2


with integral solns p = {-6,-4,-2,0,1,3,5} from which rational ones can be derived though this has to be checked for triviality. 




34 + (2v4-1)4 + (4v5+v)4 = (4v4+1)4 + (6v4-3)4 + (4v5-5v)4


This has the nice form of involving a constant term and can be generalized as,




For any constant n,


(2n)k + (n+p)k + (n-p)k = (2r)k + (q+r)k + (q-r)k,  k = 2,4


{p,q,r} = {mx2+nx+3m,  mx2+nx-3m,  2mx+n},  


and three free variables {m,n,x}.  Example, let n = 1, x = 2, so,


2k + (7m+3)k + (7m+1)k = (3m-1)k + (5m+3)k + (8m+2)k,  k = 2,4


though for any {m,n,x} this family belongs to the case a+b = c, d+e = f so can’t be used for Sinha’s problem.


R. Norrie


(a4-2)4b4 + a4(b4+2)4  + (2a3b)4 = (a4+2)4b4 + a4(b4-2)4 + (2ab3)4 



10.3. Constraint:  a+b±c = n(d+e±f)




It turns out the system,


a+b±c = n(d+e±f)          (eq.1)

a2+b2+c2 = d2+e2+f2     (eq.2)

a4+b4+c4 = d4+e4+f4     (eq.3)


also has a soln for any n of which Sinha’s problem was just the special case n = 2.  The general problem is equivalent to finding rational points on a certain elliptic curve (though this curve may be degenerate for some n).  To show this, first express the variables in the form,


{a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s,  p-q}


reducing the degrees of eq.2 and 3, a trick which will be also useful when dealing with sixth powers.  As the explicit expressions are a bit messy, only the steps will be given:


  1. Solve for p,u using eqs.1,2. 
  2. Substituting these into eq.3, it has a quadratic factor in r with discriminant D.
  3. D is a quartic in s.  Hence this must be made a square with two small solns for any n as s = q±t.  By treating D = y2 as an elliptic curve, from an initial soln subsequent ones can then be found.

Example: Using Sinha’s problem n = 2, and the negative case a+b-c = 2(d+e-f), doing the substitutions, we get,


3p-q-r+3q-3t+u = 0                  (eq.1)

pq+rs+tu = 0                             (eq.2)

p3q+pq3+r3s+rs3+t3u+tu3 = 0    (eq.3)


Solve for {p,u} and substitute into eq.3.  One factor will be a quadratic in r.  The square-free discriminant D of this is of form,


D: = c1s4+c2s3+c3s3+c4s+c52 = y2


with a square constant term and where the ci are polynomials in {q,t}.  Four small solns are:  s = {q, q-t, q+t, (q-t)/7}. 


These may then be used to generate subsequent ones.  For general n, using s = q+t with t = 1, a quadratic form soln can be found for the + case of eq.1 as,


{a,b,c} = {1+2nq+r,  n(1+2q),  -1+n-r}

{d,e,f} = {n+q+nq+r,  1-n+q-nq-r,  1+2q}


where r = q+(n+1)q2, for arbitrary {n,q}.  This also satisfies {a-b+c, d+e-f} = {0,0} though one can also find parametric ones such that this is not the case.



10.4. Constraint:  na+b+c = d+e+nf




Just like the previous, this has a soln for any n.  The inspiration for this form is the smallest polynomial soln to the sixth power multi-grade ak+bk+ck = dk+ek+fk,  k = 2,6 which, among others, surprisingly obeys the side condition 3a+b+c = d+e+3f.  (Other n for sixth powers obeying na+b+c = d+e+nf are now known.)  For the quartic version, for general n again this can be reduced to finding rational points on an elliptic curve.  Let,


na+b+c = d+e+nf          (eq.1)

a2+b2+c2 = d2+e2+f2     (eq.2)

a4+b4+c4 = d4+e4+f4     (eq.3)


and likewise {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s,  p-q},


  1. Solve for r,u in eq.1,2. When substituted into eq.3, this becomes a quadratic in q.
  2. Discriminant D of eq.3 is a quartic polynomial in s (but a sextic in p,t).
  3. For general n, four small solns are s = {p-nt,  p-t,  p+t,  p}.  Using these and others, subsequent ones can then be found.  Thus, any soln to eq.1,2,3 must solve D = y2.

 Example: Using s = p+t with t = 1 yields,


{a,b,c} = {2n-3p+np,  -n-n2+3p+np,  -n(-1+n-2p)}

{d,e,f} = {n(1+n-2p),  -n+n2+3p-np,  -2n+3p+np}


for arbitrary {n,p}, with this particular soln also satisfying {a+b-c, d-e+f} = {0,0}.



10.5. Constraint:  na+b = e+nf




For n = 1, this can be seen as the case n = 0 of a+b+nc = nd+e+f of the system above, and can be completely solved as polynomials thus providing a four-parameter soln to ak+bk+ck = dk+ek+fk, for k = 2,4.  Interestingly, this ultimately involves solving the simple equation x2+my2 = z2.  Furthermore, any soln implies ak+bk+ck+dk = ek+fk+gk+hk for k = 1,2,3,5.


Theorem:  If (p+u)k + (p-u)k + (2q)k = (p+v)k + (p-v)k + (2r)k,  for k = 2,4, then,


(p+2q)k + (p-2q)k + (-p+2r)k + (-p-2r)k = (-2p+u)k + (-2p-u)k + (2p+v)k + (2p-v)k,  for k = 1,2,3,5


The complete soln is given by the simultaneous eqns,


-3p2+q2+3r2 = u2

-3p2+3q2+r2 = v2


Proof:  Just solve for {u,v} in terms of {p,q,r}.  This is simpler than it looks since the complete parametrization to both is in linear forms.  Alternatively, one can just directly solve the form,


ak+bk+ck = dk+ek+fk for k = 2,4 (or Q1)


with a+b = d+e, eliminate e,f using resultants, giving the final eqn,


2a2+5ab+2b2-2c2-3ad-3bd+3d2 = 0


Solving this for a, its discriminant must be made a square and after suitable transformations, turns out to involve the eqn,


x12+nx22 =  x32+nx42


which has a complete soln.  Sparing the reader some algebra, the final eqn, with e = a+b-d,  is solved as,


{a,b,c,d,e} = {-3pr-4qr+2ps+qs,  5qr-ps,  3pr+qs,  3qr+ps, -3pr-2qr+qs}


It remains to find f.  Substituting these values into Q1, one ends up with a quadratic function to be made a square.  To simplify matters, let,


{f, r,s} = {(4p+2q)z,  (2p+q)2y,  2x+(6p+q)(p+2q)y}


and this final condition becomes,


x2-6q(3p+q)(p2-q2)y2 = z2


This is an eqn of form x2+ny2 = z2 which can be completely solved as {x,y,z} = {u2-nv2,  2uv,  u2+nv2} so we now have a four-parameter soln in the variables {p,q,u,v}.  For a specific example, let {p,q} = {0,1} and we get the nice identity,


(5y)k + (x-2y)k + (x+2y)k = xk + (3y)k + zk, 


if x2+24y2 = z2.  (And so on for all p,q.)   An alternative formulation can be given as,


Theorem:  If (p-q+r)k + (p+q-r)k + xk = (p+q+r)k + (p-q-r)k + yk,  for k = 2,4, then,


(2p+q+r)k + (2p-q-r)k + (-2p-q+r)k + (-2p+q-r)k = (p+x)k + (p-x)k + (-p+y)k + (-p-y)k,  k = 1,2,3,5


Note that the k = 2,4 system still has x1+x2 = y1+y2.  The complete soln is given by,


{r,p} = {(x2-y2)/(8q),  s/(24q)}, where s2 = -192q4 +96(x2+y2)q2-3(x2-y2)2


so the problem is to find {q,x,y} which makes the quartic a square.  However, we already know that this variant of Q1 has a polynomial soln so after a little reverse algebra, we find,


{q,s,x,y} = {vw/2, 6vw(2t+vw), uw+v, uw-v},  with w = (3tv+z)/(2u2-2v2),  if 4(u2-v2)2 + 3(4u2-v2)t2 = z2


Just like before, the last condition is of the form x2+dy2 = z2 and is easily solved completely as,


t = 4mn(u2-v2)/(m2-3(4u2-v2)n2)


for free variables {m,n,u,v}.  Given the constraint na+b = e+nf on Q1, we have discussed the case n = 1.  For general n,


na+b = e+nf                   (eq.1)

a2+b2+c2 = d2+e2+f2     (eq.2)

a4+b4+c4 = d4+e4+f4     (eq.3)

nb+c = -(nd+f)               (eq.4)


where again {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s,  p-q} and the fourth eqn is optional.  Solve for {u,s} in eq.1,2.  Substituted into eq.3 this becomes the quadratic in q, 


v1q2 = v2t2where v1 = (p-nr)3-(p-n3r)t2,  v2 = (p3-nr3)-(p-nr)t2 


Thus the objective is to find v1v2 = y2 which is only a quartic in t.  Four small solns are:  t = {p+r,  p-r,  (p-nr)/(n+1),  (p-nr)/(n-1)}.


Non-trivial soln can be computed from these trivial ones.  If the fourth eqn is also to be used, solve for q and eq.3 becomes a quadratic in t with a discriminant D that is only a quartic in p,r.  Without loss of generality, let r =1 and this is,


D:= (4+n2)2p4 - 4n(4+5n2)p3 + 2(4+3n2+4n4)p2 - 4n(5+4n2)p + (1+4n2)2


Since this has a square leading and constant term, this is easily solved with three small solns:  p = {n,  1/n,  -1/n}.



10.6. Constraint:  a+d = n(c+f)


Lastly, we have the system,


a+d = n(c+f)                  (eq.1)

a2+b2+c2 = d2+e2+f2     (eq.2)

a4+b4+c4 = d4+e4+f4     (eq.3)


Let {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s,  p-q}.  Solve for u,p.  Substituted into eq.3, this has a quadratic factor in r.  The square-free discriminant D of this is a quartic in s.  Small solns to finding rational points on the curve D = y2 are:  s = {q+t,  q-t,  (1+n)q+(1-n)t}.


Note 1:  Any other system consisting of Q1 and one linear relation between the terms (or two, as long as both are new) which would suffice to reduce the problem to solving a quadratic or an elliptic curve?

Note 2:  There is another system but it involves a quadratic relation between the terms,


J. Chernick


If pq(p2-q2) = rs(r2-s2), then (p2+q2)k + (r2-s2)k + (2rs)k = (2pq)k + (r2+s2)k + (p2-q2)k,  for k = 2,4.


This can be re-phrased as,


If a2+b2 = f2, and c2 = d2+e2, and ab = de, then ak+bk+ck = dk+ek+fk,  for k = 2,4.  A soln was given by Gloden as,


{a,b,c,d,e,f} = {p2-q2, p2+r2, 2pq, p2+q2, p2-r2, 2pr},  where p2 = q2+qr+r2, and can be solved by {p,q,r} = {u2+uv+v2,  u2-v2, 2uv+v2}.


This form is useful for solving a special case of Descartes' Circle Theorem when all variables are squares,


2(w4+x4+y4+z4) = (w2+x2+y2+z2)2


discussed in the next page, Form 18.


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