017: Sums of four or more Fourth Powers


16. Form: v4+x4+y4+z4 = ntk


There is yet no polynomial identity for a4+b4+c4+d4 = t4 so this form is the closest so far.


E. Fauquembergue


(2a2bc3)4 + (2ab2c3)4 + (2ab(a4+b4))4 + ((a2-b2)c4)4  = (4a2b2(a4+b4)2-c12)2,  if a2+b2 = c2


Any other polynomial soln to a4+b4+c4+d4 = ntk?


(Note: Yes, there is for n = 18, 63, etc. Add the ones by Haldemann.)

(Update, 6/23/13):  Jean-Joël Delorme solved,

x4 y4 z4 t4  = (x2+y2+z2-t2)2 

If we let a2+b2 = c2 and,

{x, y, z, t} = {(a2-b2)c4,   2a2bc3,   2ab2c3,   2ab(a4+b4)}

then this is equivalent to Fauquembergue's. Delorme also gave a new solution as,

{x, y, z, t} = {2a2b2c(a4+b4),   a(a8-b8),   b(a8-b8),   abc5(a2-b2)}

and pointed out that if {x, y, z, t} is a solution, then so is {1/x, 1/y, 1/z, 1/t}.


(Update, 2/9/10):  Lee Jacobi, Daniel Madden


The complete soln to a4+b4 = c4+d4 and a4+b4+c4 = d4 can be reduced to solving an elliptic curve.  It turns out that for a special case of four 4th powers equal to a 4th power namely,


a4+b4+c4+d4 = (a+b+c+d)4       (eq.0)


with variables as ±, then one can do so as well, a small example of which is,


9554+17704+(-2634)4+54004  = (955+1770-2634+5400)4 = 54914


Jacobi's and Madden’s clever soln depended on the identity,


a4+b4+(a+b)4 = 2(a2+ab+b2)2


an identity also useful for solving a4+b4+c4+d4+e4 = f4 as quadratic forms.  (See Form 21 below.)  Their method starts by adding (a+b)4 + (c+d)4 to both sides of eq.0,


a4+b4+(a+b)4+c4+d4+(c+d)4 = (a+b)4+(c+d)4+(a+b+c+d)4


and, using the identity, it is seen that eq.0 is equivalent to the special Pythagorean triple,


(a2+ab+b2)2 + (c2+cd+d2)2 = ((a+b)2 + (a+b)(c+d) + (c+d)2)2


Sparing the reader the intermediate algebraic manipulations, using the transformation, {a,b,c,d} = {p+r, p-r, q+s, q-s}, the eqn,


(p+r)4+(p-r)4+(q+s)4+(q-s)4 = (2p+2q)4      


can be solved if {p,q,r,s} satisfy two quadratics to be made squares,


(m2-7)p2+4(m2-1)pq+4(m2-1)q2 = (m2+1)r2     (eq.1)

8mp2+8mpq-(3m2-8m+3)q2 = (m2+1)s2           (eq.2)


for some constant m.  This is easily proven by solving {r,s} in radicals and substituting it into the eqn to see that it holds.  These two quadratic conditions define an elliptic curve.  One can then try to find a suitable m and find rational {p,q,r,s}.  Or, given a known {p,q,r,s}, a value m can be derived as,


m = (3q2+s2) / ((p+2q)2-r2)


Example:  Given {a,b,c,d} = {5400, 1770, -2634, 955}, then {p,q,r,s} = {3585, -1679/2, 1815, -3589/2}, and m = 961/61.  This initial soln set can be used to find an infinite more.  Since eq.1 and eq.2 are homogeneous, we can assume q = 1 without loss of generality.  Starting with a soln to eq.1 as p1 = -2(3585)/1679, a complete parameterization is,


p = (30/1679) (4291863+112196282u+110805419u2)/(448737-463621u2)


which makes eq.1 a square for any u.  But substituting this to eq.2, and still with q = 1, produces a rational 4th deg polynomial in u that still has to be made a square.  To find an initial rational point u, equating p-p1 = 0 yields the factor 55770003+56098141u = 0, hence this is one suitable value.  Treating the quartic polynomial as an elliptic curve, an infinite more rational u can be computed, thus proving that the eqn,


a4+b4+c4+d4 = (a+b+c+d)4


has an infinite number of distinct, rational solns.  (End update.)



17. Form: vk+xk+yk+zk = ak+bk+ck+dk,  k = 2,4




For the special form,


[0]   (p+u)k + (-p+u)k + (q+v)k + (-q+v)k = (r+u)k + (-r+u)k + (s+v)k + (-s+v)k,  k = 1, 2, 4


one must satisfy two conditions,


p2+q2 = r2+s2


p2+3u2 = s2+3v2


The system is also equivalent to the two simultaneous equations,


[1]  p2+3u2 = s2+3v2


[2]  r2+3u2 = q2+3v2


which is quite easily solved, though the trivial case p2-3v2 = q2-3u2 and others that merely permute the xi, yi should be avoided.  If [0] is to be valid for k = 6 as well, then a third condition must be met,


[3]  p2+q2 = 5(u2+v2)


See also the section on Crussol's approach.





Theorem: If ak+bk+ck+dk = ek+fk+gk+hk,  for k = 2,4, and abcd = efgh.  Define n as a4+b4+c4+d4 = n(a2+b2+c2+d2)2 then,


32(a6+b6+c6+d6-e6-f6-g6-h6)(a10+b10+c10+d10-e10-f10-g10-h10) = 15(n+1)(a8+b8+c8+d8-e8-f8-g8-h8)2


Note if d = h = 0, the condition abcd = efgh disappears and this reduces to the theorem given earlier of which Ramanujan’s 6-10-8 Identity is a special case.  Proof:  Given,


a2+b2+c2+d2 = e2+f2+g2+h2       (eq.1)

a4+b4+c4+d4 = e4+f4+g4+h4       (eq.2)


Eliminate h between the two to get a quadratic in {a,b,c,d} and solve for any, say d.  Solve for h in eq.1 and we get the complete radical soln of eq.1 and eq.2.  Substitute these two values into the 6-10-8 identity and it has a factor, call this P, which is the same polynomial that results when the values are substituted into abcd-efgh = 0.  Thus if the latter is satisfied so is the former. (End proof.)   A solution will be given below.


Theorem. If ak+bk+ck+dk = ek+fk+gk+hk,  for k = 2,4,6, where abcd = efgh, and a+b ≠ ±(c+d); e+f ≠ ±(g+h), (call the entire system V1) then,


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

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


for k = 1,2,4,6,8,10.


Again, for d = h = 0, the condition abcd = efgh disappears so reduces to Birck’s Theorem (given later). This is then an extension to tenth powers and the proof will be discussed in that section.  Even for non-zero variables it is possible one or two of the addends will vanish, as we shall see later.  This author  hasn’t been able to find a parametric soln to V1 but was able to do so for the simpler case when valid only for k = 2,4.  Given,


x1k + x2k + x3k + x4k = y1k + y2k + y3k + y4k,  k = 2,4


assume it to have the form, call it F1,


ak(p+q)k + bk(r-s)k + ck(r+s)k + dk(p-q)k = ak(p-q)k + bk(r+s)k + ck(r-s)k + dk(p+q)k


which has x1x4 = y1y4 and x2x3 = y2y3.  This sufficient but not necessary constraint is completely parameterized by F1.  To see this, use the Mathematica command,


Solve[{a(p+q), d(p-q), a(p-q)} = {x1, x4, y1}, {a,p,d}]


to express {a,d,p} in terms of {x1, x4, y1}. Substituting these into d(p+q) = y4, we get,


x1x4/y1 = y4


which, if true, yields the appropriate {a,d,p}, with q a free variable, and similarly for {b,c,r}.  For example, given,


16k + 48k + 58k + 99k = 22k + 29k + 96k + 72k,  k = 2,4


which is the smallest in distinct integers with NO sign changes such that x1+x2+x3+x4 = y1+y2+y3+y4 = 0.  This has 16(99) = 22(72);  48(58) = 29(96), then,


{a,d,p,q} = {1, 9/2, 19, -3}

{b,c,r,s}  = {1/2, 1, 77, -19}


with {q,s} arbitrary.  F1 takes care of the necessary requirement that x1x2x3x4 = y1y2y3y4.  To simplify matters, assume further that x1+x2 = y1+y2 (not obeyed by the example above).  After some algebra, we get the soln,


{p,q,r,s} = {ab2-ac2,  be,  a2b-bd2,  ae},  where a2(3b2-c2) + e2 = (b2+c2)d2.


This conditional eqn is easily solved parametrically.  First, let n = 3b2-c2 and assume LHS as a2n+e2 = y2. One can easily find {a,e}. Then assume RHS as b2+c2 = z2 and find {b,c}, to get the form y2 = z2d2 which then gives d.  For example, let {b,c} = {3,4} entails solving the conditional equation,


11a2+e2 = (5d)2


which is easily done.


(Note:  If the particular form F1 is extended to k = 6, it yields only trivial solns so another form has to be used.)



18. Form: 2(v4+x4+y4+z4) = (v2+x2+y2+z2)2


The equation,


2(a4+b4+c4) = (a2+b2+c2)2


has the simple soln a+b = c, which is central to Ramanujan's 6-10-8 Identity.  Turns out when there are four terms,


2(v4+x4+y4+z4) = (v2+x2+y2+z2)2    (eq.1)


it has a parametric soln as well.  Note that this is a special case of Descartes' Circle Theorem,


2(x12+x22+x32+x42) = (x1+x2+x3+x4)2


where the xi are squares.  Euler’s clever soln uses the fact that eq.1 is equivalent to either,


(2vx)2 + (2yz)2 = (v2+x2-y2-z2)2

(2vy)2 + (2xz)2 = (v2-x2+y2-z2)2

(2vz)2 + (2xy)2 = (v2-x2-y2+z2)2


so is a special case of three simultaneous Pythagorean triples.  More generally, for any d, we are to solve, 


2(d2v4+x4+y4+z4) = (dv2+x2+y2+z2)2   (eq.2)


and Euler's soln can be given as {v,x,y,z} = {2pqrs(a2-t2),  t(a2-t2),  t(abrs-cpqt),  a(abrs-cpqt)}, where {a,b,c} = {pq(r2-ds2),  p2+dq2,  r2+ds2}, and {p,q,r,s,t} satisfies the conditional eqn,  


pqrs(p2-dq2)(r2-ds2) = t2


For d = 1, one soln is {p,q,r,s} = {2m2-n2,  m2+n2,  m2+n2,  m2-2n2}.  Note:  Strictly speaking, Euler’s analysis was focused on the case d = ±1 and this author inserted the d to generalize it.  Any soln for other d?


(Update, 9/13/09):  I noticed we can have a better form if we assume t = rs(r2-ds2) and the conditional eqn becomes the symmetric,


pq(p2-dq2) = rs(r2-ds2)


This has appeared already in Section 013 and has many solns and is connected to the two distinct eqns,


a4+b4 = c4+d4

ak+bk+ck = dk+ek+fk,  k = 2,4


Thus, we can also solve 2(n2v4+x4+y4+z4) = (nv2+x2+y2+z2)2 as:


I.  When n = -1:


{v,x,y,z} = {2(ab-cd)(ab+cd),  (a2+b2+c2+d2)(a2-b2+c2-d2),  2(ac-bd)(a2+c2),  2(ac-bd)(b2+d2)},  where a4+b4 = c4+d4.


II.  When n = 1:


Let ak+bk+ck = dk+ek+fk,  for k = 2,4, where a2+b2 = f2,  d2+e2 = c2,  and ab = de, then,


{v,x,y,z} = {ae-bd,  (a+d)(a-d),  a(c+f),  d(c+f)}


A soln given by Gloden is {a,b,c,d,e,f} = {p2-q2, p2+r2, 2pq, p2+q2, p2-r2, 2pr},  where p2 = q2+qr+r2.  This can be solved by {p,q,r} = {u2+uv+v2,  u2-v2, 2uv+v2}.  (End update.)



19. Form: x1k+x2k+x3k+x4k+x5k = y1k+y2k+y3k,  k = 1,2,3,4


The system of eqns,


x1k+x1k+…+ xnk = y1k+y1k+…+ ynk,  k = 1,2,3,4


has non-trivial solns only for n > 4.  However, it is possible two terms on one side are equal to zero.




Given ak+bk+ck+dk+ek = fk+gk+hk,


for the special case when h = a+b+c, then,


{d,e,f,g} = {(p+u)/(2q),  (p-u)/(2q),  (p+v)/(2q),  (p-v)/(2q)}


where p = (a+b)(a+c)(b+c),  q = (ab+ac+bc), and {a,b,c} satisfies the two quadratic conditions,

p(p+4abc) = u2

p(p+4abc) - 4q3 = v2


An example of a particular soln is {a,b,c} = {-21, -4, 12} which yields the eqn,


(-63)k + (-12)k + 36k + (-35)k + 10k = (-62)k + 37k + (-39)k


Q: Any general polynomial identity for this?



20. Form: x1k+x2k+x3k+x4k+x5k = y1k+y2k+y3k+y4k+y5k,  k = 1,2,3,4


The complete soln to this system, call this E4, is unknown. Use the form,


(a+bj)k+(c+dj)k+(e+fj)k+(g+hj)k+(i+j)k = (a-bj)k+(c-dj)k+(e-fj)k+(g-hj)k+(i-j)k


and let i = -(ab+cd+ef+gh), h = -(1+b+d+f) to make it identically true for k = 1,2.  

I. Case x2+x3 = y2+y3:

Let f = -d to satisfy the constraint.  It can be shown that the end result involves making a quartic polynomial a square.  Expand further the form above at k = 3,4, to get the auxiliary resultants,


(Poly10)j2+(Poly20) = 0                                   (eq.1)

(Poly11)j4+(Poly21)j2+(Poly31) = 0                 (eq.2)


where the Polyi are in {a,b,c,d,e,g}.  Eliminate j between eq.1 and 2 (set j = √v for convenience) and one gets a final resultant R which is only a quadratic in a,c,e,g.  Solving for g, this has a discriminant D which is only a quadratic in e so to solve D = y2, one uses a quadratic form e = Q(x) for some variable x.  After a rational g is found, to find j, solve,


-(Poly10)(Poly20) = z2


which is a quadratic in e. Substituting the quadratic form e = Q(x) into this, it becomes,


Poly(x) = z2          (eq.3)


which is a quartic in x, hence any soln to E4 with the constraint x2+x3 = y2+y3 must satisfy eq.3, a quartic polynomial to be made a square.

II. Case x1+x2 = y1+y2x2+x3 = y2+y3:

This is just a special case of the one above which is much simpler. Let b = f = -d to satisfy the two constraints. After eliminating j between eq.1 and 2, the final resultant is only linear in g.  Substituting this into eq.1 and eq.2, one must make a quadratic polynomial in e as a square.  Since its leading coefficient is already a square, then this is easily done, giving a multi-variable polynomial parametrization to E4  


Note:  For Case 1, since the final resultant involves six variables {a,b,c,d,e,g}, the explicit soln is a mess.  There might be a way to clean this up though using a more suitable form than the one given without losing its generality.  But using the simpler form,


(a+b)k+(c+d)k+(e+f)k+(g+h)k+(i+j)k = (a-b)k+(c-d)k+(e-f)k+(g-h)k+(i-j)k


for example, is messier.



21. Form: x14+x24+…xn4, n > 4



E. Fauquembergue


(4x4-y4)4 + 2(4x3y)4 + 2(2xy3)4 = (4x4+y4)4




(2x2+12xy-6y2)4 + (2x2-12xy-6y2)4 + (4x2-12y2)4 + (3x2+9y2)4 + (4x2+12y2)4 = 54(x2+3y2)4


This gives as a first instance 24+24+44+34+44 = 54.  Similar identities were found by A. Martin and Ramanujan.  A generalization has been found by this author.




(ax2+2v1xy-3ay2)k + (bx2-2v2xy-3by2)k + (cx2-2v3xy-3cy2)k = (ak+bk+ck)(x2+3y2)k


where {v1, v2, v3, c} = {a+2b, 2a+b, a-b, a+b}, for k = 2,4


Thus, it suffices to decompose a sum which starts as a4+b4+(a+b)4+… into a sum and difference of any number of biquadrates and apply it to the identity.  For example, using the eqn 504+504+1004+44+154 = 1034, this yields, 


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


R. Norrie


(x+y)4 + (-x+y)4 + (2y)4 + (u2-v2)4 + (2uv)4 = (u2+v2)4,  if  2uv(u2-v2) = x2+3y2


Note that {u2-v2, 2uv} are the legs of a Pythagorean triple.  Thus, if the product of the legs can be expressed as x2+3y2, it then gives a soln to the above equation, the smallest of which is the particular equation given earlier, {2,2,4,3,4; 5}.  For the sum of five or six biquadrates,


Sophie Germain


Sophie Germain’s Identity is given by


x4+4y4 = ((x+y)2+y2) ((x-y)2+y2) = (x2+2xy+2y2)(x2-2xy+2y2)


The quartic form can be generalized as,


x4+(-a2+2b)x2y2+b2y4 = (x2+axy+by2)(x2-axy+by2)


where Germain's was the case {a,b} = {2,2}.  Another interesting case is {a,b} = {2,-1},


x4-6x2y2+y4 = (x2+2xy-y2)(x2-2xy-y2) = (x2-y2)2 - (2xy)2  


which sometimes appear when dealing with Pythagorean triples.



(Update, 2/3/10):  Choudhry


Given the eqn,


a4+4b4 = c4+4d4    (eq.1)


use the transformation,


{a,b,c,d} = {(x2+2x-2)z+xy, 2xz+y, (x2+2x+2)z+xy, y}


so that, substituted into eq.1, it has a linear factor in y.  This first soln can be used to compute an infinite sequence of polynomial solns, as eq.1 is an elliptic curve in disguise. The transformations, {a,b,c,d} = {p+q, r-s, p-q, r+s}, then {p,q,r,s} = {u, mv, nu, v} reduce eq.1 to the form,


(m-4n3)u2+(m3-4n)v2 = 0


so one is to solve,


-(m-4n3)(m3-4n) = z2


One soln is,


{p,q,r,s} = {8x+2x3,  -4+2x2,  4+4x2,  -2x+x3}


from which others can then be computed.  Source:  The Diophantine Equation A4+4B4 = C4+4D4, Indian Journal of Pure and Applied Mathematics, Vol. 29, 1998.  (End update.)




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


Note that the form (a4+2b4)4 can also be expressed as a sum of both squares and biquadrates,


E. Fauquembergue


(a4-2b4)4 + (2a3b)4 + (8a2b6)2 = (a4+2b4)4


(2a2b2)4 + (2a3b)4 + (a8-4a4b4-4b8)2 = (a4+2b4)4




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