021: Fifth Powers (7 or 8 Terms)


5.3  Seven terms


As was pointed out by Lander, quite a lot of the small solns found by computer search to,


x1k+x2k+x3k+x4k = x5k+x6k+x7k+x8k        (eq.0)


for k = 1,5 have a transposition such that x1+x2+x3+x4 = x5+x6+x7+x8 = 0, call this condition J1.  Furthermore, this transposed form is valid for k = 1,2,3,5.  This has a complete soln in terms of binary quadratic forms since it can be reduced to solving the beautifully simple equation ax2+by2 = cz2 using Chernick’s approach discussed below.  In fact, it can be showed that any soln to eq.0 for k = 1,2,3,5 must satisfy J1.  To prove this, use the form F2,


(a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k


where it can be set a = 1 without loss of generality.  This can be satisfied for k = 1,2 using f = -(1+b+d), g = -(ab+cd+ef).  Expanding for k = 3,5 gives,


(Poly01)h2 + (Poly02) = 0

(Poly11)h4 + (Poly12)h2 + (Poly13) = 0


where the Polyi are in {b,c,d,e}.  Eliminating h between the two gives a resultant which completely factors into linear eqns, some trivial but the non-trivial ones all imply J1, thus proving the assertion. 


A large class involving eq.0 can be reduced to solving x1k+x2k+x3k = y1k+y2k+y3k for k = 2,4 with the constraint x1+x2 = y1+y2.


Theorem (Piezas):  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


This, in turn, can be completely solved in terms of binary quadratic forms as was already discussed in Fourth Powers (Part 3).  Note that the one for k = 1,2,3,5 have sums of terms on either side equal to zero so, by Theorems 1 and 6 of the Prouhet-Tarry-Escott Problem, is essentially an ideal soln of degree 3 in standard form.  The special case of eq.0,


x1k+x2k+x3k+x4k = x5k+x6k+x7k+x8k      


when one term is equal to zero entails making a certain quartic polynomial a square (to be discussed later) and has has an infinite family of polynomial solns the smallest of which found by this author is of fifth degree, one by Chernick as ninth degree, though there are others also due to Chernick which need solving an elliptic curve but are notable due to their aesthetic form.




One can derive a soln such that the same variables solve two sets of systems, for k = 2,4 and k = 1,2,3,5.  There are two different pairs, special cases of the more general identity given above with the first being,


A.  If (2a+c)k + (-2a+c)k + (2b)k = (3c)k + (-c)k + (2d)k,  for k = 2,4, then,


(2a-2c)k + (-2a-2c)k + (4c)k = (2b+c)k + (-2b+c)k + (-c-2d)k + (-c+2d)k,   for k = 1,2,3,5.


This can be solved as,


2a2+4b2 = 9c2              (eq.1)

2a2+b2 = 2c2+d2          (eq.2)




2a2+4b2 = 9c2 

14a2+b2 = 9d2 


with the second condition easily found by substituting c from eq.1 into eq.2, though trivial {a,b} should be avoided.  Similarly for the second pair,


B.  If (a+d)k + (-a+d)k + bk = (c+d)k + (-c+d)k + dk,  for k = 2,4, then,


(b-d)k + (-b-d)k + (2d)k = (a+2d)k + (-a+2d)k + (-c-2d)k + (c-2d)k,   for k = 1,2,3,5.




11a2+4b2 = 9c2            (eq.1)

2a2+b2 = 2c2+d2          (eq.2)


or alternatively,


11a2+4b2 = 9c2

-4a2+b2 = 9d2 


which can be similarly treated as above.  These are special cases of the general theorem given above.  Recall that, for certain {p,q,r,u,v},


(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


Since, after minor sign changes of p, the variables {q, r} are interchangeable (as well as the {u,v}), then there are two ways to set one term equal to zero, either:


1) RHS:  2p-v = 0

2) LHS:  -p+2r = 0


for example, though one can choose any term on one side.  Chernick's two identities have,


1) {p,q,r,u,v} = {c, b, d, 2a, 2c}

2) {p,q,r,u,v} = {d, b/2, d/2, a, c}


which, when substituted into the theorem, explains why one term is equal to zero.



J. Chernick, L. Lander


Theorem:  "If a2+b2+c2 = d2+e2+f2, and abc = def, then,


(a+b+c)k + (a-b-c)k + (-a-b+c)k + (-a+b-c)k = (d+e+f)k + (d-e-f)k + (-d-e+f)k + (-d+e-f)k,  for k = 1,2,3,5."


For indefinite k, call this system C1.  Labeled as,


x1k + x2k + x3k + x4k = x5k + x6k + x7k + x8k,      (eq.0)


note that this also satisfies x1+x2+x3+x4 = x5+x6+x7+x8 = 0.  The history of C1 is quite interesting.  In 1913, Crussol considered a transposed form of C1 to solve k = 1,2,4,6.  In 1937, Chernick studied this system in the context of an ideal solution of degree 3 in standard form and gave the complete soln, though he knew by Theorem 6 it could be extended to fifth powers.  In the late 1960s, Lander would revisit this system, with a small rearrangement of terms, focusing only on odd powers and solving the cases k = 1,5 and k = 1,3,5.  Surprisingly, this system can be extended to solve seventh powers as well as was done by A. Choudhry in the 1990s to find the first numerical solns to k = 1,3,7 though the variables must fulfill a more complicated condition.  Note that either side is a special case of Boutin’s Theorem,


(a+b+c)3 + (a-b-c)3 - (a+b-c)3 - (a-b+c)3 = 24abc


a theorem which generalizes the difference of two squares to other kth powers.  Also, in the guise,


(-a-b-c)k + (a-b-c)k + (-a-b+c)k + (-a+b-c)k = (-2a)k + (-2b)k + (-2c)k,  for k = 1,2,


this appears in the Birck-Sinha Theorem which involves eighth powers.  Whether C1 has a soln for k = 1,8 or even k = 1,2,8 still remains to be seen, but perhaps there is.  For k = 1,2,3,5, the problem then is to solve the two conditions,


a2+b2+c2 = d2+e2+f2     (eq.1)

abc = def                       (eq.2)


By adding the condition a±b±c = 0 (for any choice of signs), this implies one of the xi of eq.0 will vanish and hence will involve only seven terms.


A. Partial Solutions


The section on Sums of n Squares gives a lot of identities for the first eqn such as the nice symmetrical one by Goldbach,


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


and it satisfies eq.2 as well if  p2+pq+q2 = r2. Lander also gave the partial soln,


{a,b,c} = {q(pr+s),  (p-r)(pr-s),  pq(p+r)}

{d,e,f} = {q(pr-s),  (p+r)(pr+s),  pq(p-r)}


where s = q2-r2.  For the special case when one of the terms xi vanishes, let one of the four (d±e±f) be equal to zero.  It suffices to consider d+e±f = 0, hence,


q(pr-s) + (p+r)(pr+s) ± pq(p-r) = 0


Solving this quadratic in p, its discriminant must be made a square and involves an elliptic curve.  Using the positive, then negative case yields,


5q4-8q2r2+4r4 = t12

-3q4+12q3r+4q2r2-8qr3+4r4 = t22


respectively, both of which have small solns, one being {q,r} = {4,1}, etc.



B. Complete Solution


Completely solving the two eqns,


a2+b2+c2 = d2+e2+f2     (eq.1)

abc = def                       (eq.2)


just involves binary quadratic forms.  Proof:  Let,


{a,b,c} = {p+qu, r+su, t+u}

{d,e,f} = {p-qu,  r-su,  t-u}




t = -(pq+rs)

p = ((1-q2-s2)r + w) /(2qs)


and w satisfies,


r2(-1+q-s) (1+q-s) (-1+q+s) (1+q+s) + 4q2s2u2 = w2


Since this is of the form,


u2+dv2 = w2


hence is easily completely solved in binary quadratic forms. (End proof.)  Chernick used another approach by setting {c,d,e} = {mnf, ma, nb} which solves eq.2 and transforms eq.1 to,


c1a2 + c2b2 = (m2n2-1)f2


where {c1, c2} = {m2-1, n2-1} with its complete soln as,


a = c1nu2+2c2uv-c2nv2

b = c1u2-2c1nuv-c2v2

f = c1u2+c2v2


for arbitrary {m,n,u,v}.  We can give our own proof that this is complete.  Proof:  To remove scaling factors, divide the eqn by f2 to get the simpler form,


c1p2 + c2q2 = (m2n2-1)             (eq.3)


where {p,q} = {a/f,  b/f}.  Given any rational soln {p,q} to eq.3, one can always find rational {u,v} using the formulas,


u = (n2-1)(q2-m2)(p+n)

v = (m2-1)(p2-1)(q-1)


Verifying via Mathematica, one can see that the eqns {p-a/f = 0, q-b/f = 0} using this u,v have eq.3 as a factor, hence will be true if {p,q} make this zero.  Thus, Chernick’s soln completely solves the two conditions as binary quadratic forms in four parameters {m,n,u,v}. 


For the special case when one of the terms xi vanishes, again let one of the four (d±e±f) be equal to zero.  They can be easily constructed using {d,e} = {ma, nb} with {a,b,f} as defined above.  As was seen, they are binary quadratic forms in {u,v} and interestingly this quartet have the same discriminant D which if made a square y2,


D:= (2m2-1)m2n4-(m4+3m2-2)n2+(2m2-1) = y2


implies there are rational u,v such that one of the terms xi can be chosen to vanish. (The other quartet also have just one discriminant.)  This author found two small solns,


n = 1/(m+1),  or  n = (-2m2+5)/(2m2+1)


After solving one appropriate xi = 0 for {u,v}, both solns, after removal of common factors, yield xi as just 5th degree polynomials in contrast to the 9th degree ones found by Chernick who used a different method.  (Which is ironic since he could have used this one.)   Making D a square can then be treated as an elliptic curve and with these initial solns, subsequent ones can then be found.



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