014: Sums of three Fourth Powers (Part 1)

 
 

10. Form: ak+bk+ck = dk+ek+fk  (k = 2,4)

 

Bastien’s Theorem also given below states that the above eqn is trivial if simultaneously true for k = 1,2,3,4.  However, this does have solns when k = 2,4; k = 1,2,4, or even k = 2,3,4 and, of course, when only for single exponents.  When valid for both k = 2,4, this quadratic-quartic system will be called throughout this work as Q1, and this form is significant because it may appear as a side condition for higher powers such as for the 5th, 6th, 7th, and 8th.  First, we have these general theorems,

 

T. Sinha

 

Theorem:  If, 


a2+b2+c2 = d2+e2+f2,  

a4+b4+c4 = d4+e4+f4,


then (a2-f2)(b2-f2) = (d2-c2)(e2-c2).

 

The eqn (a2-f2)(b2-f2) = (d2-c2)(e2-c2) is quite easy to satisfy and can be reduced to solving an eqn of form x2+ny2 = z2.  However, it is not enough to imply Q1.  More generally, any two of the equations in the system implies the third, but it is not the case that one implies the other two.

 

Tarry, Escott

 

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

 

(a+t)k + (-a+t)k + (b+t)k + (-b+t)k + (c+t)k + (-c+t)k = (d+t)k + (-d+t)k + (e+t)k + (-e+t)k + (f+t)k + (-f+t)k

 

k = 1,2,3,4,5 for any t.

 

This is a special case of a theorem of the Prouhet-Tarry-Escott Problem which I’m assuming is due to Tarry or Escott since they established related theorems for this field.  Note that appropriate pairs have the common sum 2t and is known as a symmetric solution.  Any multi-grade k = 1-5 of this form is then reducible to Q1.  By Bastien's Theorem, the system,

 

x1k+x2k+…+ xmk = y1k+y2k+…+ ymk

 

for k = 1,2,…n has non-trivial solns only if m>n.  Thus, one side must have at least n+1 terms.  For the case k = 1,2,3,4,5, this implies an eqn with 6 terms on either side.  However, it is possible some terms are equal to zero.  For example, given the special case of Q1,

 

ak+bk+ck = 2dk+ek,  k = 2,4, then,

 

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

 

which has only 4 terms on the RHS.

 

Chernick

 

Theorem:  If (a+d)k + (-a+d)k + bk = (c+d)k + (-c+d)k + dk,  k = 2,4, then

 

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

 

which is one of two identities.  Note that if the terms of Q1 are expressed as xi and yi, this also satisfies x1+x2 = y1+y2.  More generally,

Piezas

 

Theorem:  If (p-q+r)k + (p+q-r)k + xk = (p+q+r)k + (p-q-r)k + yk,  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,

 

and where one of the terms of the latter eqn can be set to zero to get either of Chernick’s two solns.

 

 

Chernick, Escott

 

Theorem:  If 1k + 2k + 3k = uk + vk + (u+v)k, k = 2,4, then,

 

(u-7)k + (u-2v+1)k + (3u+1)k + (3u+2v+1)k = (u+7)k + (u-2v-1)k + (3u-1)k + (3u+2v-1)k,  k = 2,4,6.

 

Sinha

 

Theorem:  If (a+3c)k + (b+3c)k + (a+b-2c)k = (c+d)k + (c+e)k + (-2c+d+e)k,  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,  k = 1,3,5,7

 

excluding the trivial case c = 0.  This was derived using the next theorem.

 

Birck, Sinha

 

Theorem:  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 + (2d)k + (2e)k + (2f)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.

 

Piezas

 

Theorem:  If ak+bk+ck = dk+ek+fk for k = 2,4.  Define {x,y} = {a2+b2+c2,  a4+b4+c4}.  Then,

 

a6+b6+c6-d6-e6-f6 = 3(a2-d2)(b2-d2)(c2-d2) = 3(a2b2c2-d2e2f2)

3(a8+b8+c8-d8-e8-f8) = 4(a6+b6+c6-d6-e6-f6)(x)

6(a10+b10+c10-d10-e10-f10) = 5(a6+b6+c6-d6-e6-f6)(x2+y)

 

If we define Fk:= ak+bk+ck-dk-ek-fk, then for even k > 4, Fk is of the form F6Poly1 and also a highly composite number.  More significantly, the identities can be combined and for a,b,c,d,e,f with appropriate properties, these can be woven together into a very neat form.  For example, define n as,

 

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

 

and Q1 implies the general identity,

 

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

 

For the special case when a+b = ±c, since,

 

a4+b4+(a+b)4 = (1/2)(a2+b2+(a+b)2)2

 

then n = 1/2, and the above becomes,

 

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

 

which is a concise version of Ramanujan’s 6-10-8 Identity discussed in the next section.  (In the general case, this turns out to be just the first in a family of identities.)  (Update, 7/31/09):  This can be extended to the "solution chain" a1k+a2k+a3k = a4k+a5k+a6k = a7k+a8k+a9k = a10k+a11k+a12k = ..., k = 2,4, for a chain divisible into pairs.  For the special case when {a1+a2, a4+a5, a7+a8, a10+a11} = {a3, a6, a9, a12}, then,  

 

64(a16+a26+a36+a46+a56+a66-a76-a86-a96-a106-a116-a126)(a110+a210+a310+a410+a510+a610-a710-a810-a910-a1010-a1110-a1210)

= 45(a18+a28+a38+a48+a58+a68-a78-a88-a98-a108-a118-a128)2,

 

One can test this by the soln chain [28, 175, 203] = [77, 140, 217] = [107, 113, 220] = [5, 188, 193] given here, found by Gloden in the 1940's.  (End update.)  It seems then the system Q1 is quite important and worth a second look.  To solve,

 

ak+bk+ck = dk+ek+fk

 

for k = 2,4, one trick is to reduce the degree of the two eqns by using the substitution {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s, p-q} to get,

 

pq+rs+tu = 0

p3q+pq3+r3s+rs3+t3u+tu3 = 0

 

By eliminating u between them, one ends up with a quartic in t,

 

(pq+rs)t4-(p3q+pq3+r3s+rs3)t2+(pq+rs)3 = 0

 

or a cubic in either p,q,r,s.  However, it is possible to use another substitution that can reduce the degree even further.

 

Theorem: "The system ak+bk+ck = dk+ek+fk for k = 2,4, call it Q1, can be reduced to solving a quadratic. Its discriminant D is a quartic polynomial in one variable and the problem of making D a square can be treated as an elliptic curve."

 

(In fact, as was already mentioned, similar results can be given for the three systems k = 1,4, or k = 1,5, or k = 1,2,6.  Another substitution will be used for these which is slightly more efficient since the first system ends up with a discriminant D that is only a quadratic polynomial.)

 

Proof: We simply use a different substitution, call this form F1,

 

{a,b,c,d,e,f} = {p+qu, r+su, t+u, t-u, r-su, p-qu} to get,

 

pq+rs+t = 0

p3q+r3s+t3+(pq3+rs3+t)u2 = 0

 

where the degree of the variable u has been reduced to merely a quadratic.  Eliminating t between them yields,

 

(Poly1)u2-(Poly2) = 0

 

where,

 

Poly1:= p(q-q3)+r(s-s3)

Poly2:= p3(q-q3)-3pqrs(pq+rs)+r3(s-s3)

 

and the problem is reduced to making its discriminant a square,

 

y2 = (Poly1)(Poly2)

 

which is just a quartic in p (or r), thus proving the theorem.  This quartic in fact has a square leading and constant term so is easily made a square and from that initial point, other rational ones may be computed.  To prove that this is the complete characterization of non-trivial solns, solve for p,q… in terms of the original variables to get,

 

{p,q,r,s,t,u} = {(a+f)/2,  (a-f)/(c-d),  (b+e)/2,  (b-e)/(c-d),  (c+d)/2,  (c-d)/2}

 

and even if there is division by a variable, when c = d the system ak+bk = ek+fk for k = 2,4 has only trivial solns hence this exceptional case is of no interest and does not affect the generality of the theorem.  While it has been proven that Q1 can be reduced to solving a quadratic, a third constraint may be useful to explore various solns.  This constraint is an appropriate linear or quadratic relation between the terms such that the final eqn remains a quadratic and involve a simpler elliptic curve.  This author found nine relations, six linear and three quadratic,

 

10.1  a+b = nc; d+e = nf

10.2  a+b ≠ c; d+e ≠ f

10.3  a+b±c = n(d+e±f)

10.4  na+b+c = d+e+nf

10.5  na+b = e+nf

10.6  a+d = n(c+f)

10.7  (a2-f2)c2 = -(b2-e2)d2

10.8  a2+nab+b2 = d2+nde+e2

10.9  a2+nad+d2 = b2+nbe+e2

 

for any n.  All of these have a complete soln in terms of an elliptic curve though may simplify into quadratic forms for particular values of n or be trivial when n = 0.  The linear relations will be discussed in the next section.

 

 

Previous Page        Next Page  

Comments