022: Fifth Powers (8, 10, 12 terms)

 
 

5.4  Eight terms

 

Several results for eight terms have already been discussed in the previous section.  Before discussing the present section, it may be instructive to give an overview of some of the known “side conditions” for,

 

x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k,   k = 1,3,5, or k = 2,4,6,

 

x1k+x2k+x3k+x4k+x5k = y1k+y2k+y3k+y4k+y5k,   k = 1,3,5,7, or k = 2,4,6,8,

 

since there seems to be some general pattern.

 

Fifth powers:

1.       x1+x2 = x3+x4 = -(y1+y2)

2.       x1-x2 = x3-x4 = -(y1-y2)

3.       x1-x2 = x3-x4 = y1-y2

4.       x1+x2 = y1+y2;  x3+x4 = y3+y4

 

Sixth powers:

1.       x1+x2 = y1+y2;  x3+x4 = y3+y4

 

Seventh powers:

1.       x1+x2 = x3+x4 = y1+y2

2.       x1-x2 = x3-x4 = y1-y2

3.       x1-x2 = y1-y2;  x3-x4 = y3-y4

 

Eighth powers:

1.       x1-x2 = y1-y2;  x3-x4 = y3-y4

 

Of course these are not all of the possible side conditions but small solns tend to follow them hence are the first ones noticed.  They are useful since they give more “handles” to deal with the system and, in some cases, reduce it to merely solving a quadratic or making a low-degree polynomial a square.  (It is tempting to speculate ninth and tenth powers may obey similar rules though, since only one soln is known for each, very little can be said about their behavior.)  (Update: More solns have been found for tenth powers. See update in last section.)  Side conditions can also have interesting implications,

 

Piezas

 

Theorem 1: If ak+bk+ck+dk = ek+fk+gk+hk,  for k = 1,3,5 where a+b+c+d = e+f+g+h = 0, then the following six relations are also true,

 

[1]  5(a2+b2+c2+d2)(a3+b3+c3+d3) = 6(a5+b5+c5+d5)

 

[2]   3(a2+b2+c2+d2)(a4+b4+c4+d4-e4-f4-g4-h4) = 4(a6+b6+c6+d6-e6-f6-g6-h6

 

[3]  9(a2+b2+c2+d2)(a7+b7+c7+d7-e7-f7-g7-h7) = 7(a9+b9+c9+d9-e9-f9-g9-h9)

 

[4]  (a3+b3+c3+d3)(a6+b6+c6+d6-e6-f6-g6-h6) = (a9+b9+c9+d9-e9-f9-g9-h9

 

[5]  7(a3+b3+c3+d3)(a4+b4+c4+d4-e4-f4-g4-h4) = 12(a7+b7+c7+d7-e7-f7-g7-h7

 

Combining [2] and [3] yields the same relation as combining [4] and [5], namely,

 

[6]  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),  and,

 

One can test it with a random example, say, {a,b,c,d} = {21, -13, 9, -17} and {e,f,g,h} = {23, -3, -21, 1}, though it will work in the general case.  Note that a+b+c+d = e+f+g+h for this system implies it is also valid for k = 2.  [6] has already been given in Fourth Powers but is related to the Chernick-Lander theorem given in the previous section and is analogous to Ramanujan’s 6-10-8 Identity.  [3] seems to have analogous versions for any system k = 1,2,3,5,…n  for consecutive odd exponents n > 3, while [1] is a general identity that works on four terms a+b+c+d = 0.   

 

Proof:  Given x1k+x2k+x3k+x4k - (y1k+y2k+y3k+y4k) = 0 such that x1+x2+x3+x4 = y1+y2+y3+y4 = 0, this is completely parametricized by the Chernick-Lander form, call this L(k),

 

L(k): = (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)

 

since the three variables {a,b,c} can suffice to express the xi (and likewise {d,e,f} for the yi).  Expanding L(k) for k = 1,2,3,5, we get,

 

L(1) = 0,                      L(2):= a2+b2+c2-d2-e2-f2

L(3): = abc-def,            L(5): = abc(a2+b2+c2)-def(d2-e2-f2),

 

If we wish to set all L(k) = 0, if L(3) = 0, one can see that the eqn L(5) = 0 reduces to just L(2) = 0.  Eliminating f between L(2) = 0 and L(3) = abc-def = 0, we get a polynomial, call it P.  Substituting the terms of L(k) on any of the statements of Theorem 1 after [1], we get an expression in {a,b,c,d,e,f} and, eliminating f between this and abc-def = 0, we get P again (as a factor).  Thus, the roots of P = 0 satisfy both the condition and the theorem, which proves the latter.  (For [1], simply substract one side from the other, expand the identity, and one can see it is true if a+b+c+d = 0.)

 

Lander

 

(a+b+c)5 + (a-b-c)5 + (-a-b+c)5 + (-a+b-c)5 = 80abc(a2+b2+c2)

 

(Update, 12/13/09):  Tony Rizzo showed that there are non-trivial solns to,

 

80abc(a2+b2+c2) = d5+e5      (eq.1) 

 

thus yielding a (5.3.3) and gave the example {a,b,c,d,e} = {32, 25, 50, 80, 100}.  More generally, this gives a (5.3.3) with the side condition x1+x2 = y1+y2.  The Sastry-Chowla identity has this constraint, so there is in fact an infinite number of solns to eq.1 given by {a,b,c,d,e} = {u5, 25v5, 50v5, 10u3v2, 50uv4},  for arbitrary {u,v}, with Rizzo's as {u,v} = {2,1}.  However, there are other solns as the Sastry-Chowla identity cannot give a (5.3.3) with all terms positive.  Duncan Moore has a database here and out of around 5400 solns, only 12 had the required condition that x1+x2 = y1+y2Q:  Is there any other family that solves eq.1?  (End update.)

 

Lander (k = 1,5)

 

(ax2-vxy+w1y2)k + (ax2+vxy+w2y2)k + (x2-vxy+w3y2)k + (x2+vxy+w4y2)k =

(ax2+vxy+w1y2)k + (ax2-vxy+w2y2)k + (x2+vxy+w3y2)k + (x2-vxy+w4y2)k

 

This has the side condition x1+x2 = y1+y2; x3+x4 = y3+y4 which for odd powers is equivalent to x1+x2+x3+x4 = y1+y2+y3+y4 = 0 and was discussed in the previous section.  However, this new one is only for k = 1,5 but has a surprising property.  Let v = 3a2, and {w1, w2, w3, w4} = {p+q, p-q, r+s, r-s}.  Then, the soln is,

 

{p,q,r,s} = {a7+a5-2a3+a,  3a2,  a6-2a4+a2+1,  3a5}

 

Note that {p,q,r,s} also solve,

 

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

 

To see this, if the 8-term eqn above was expanded and powers of {x,y} collected, we get a system of four auxiliary eqns in four unknowns, one of which is,

 

w14-w24- w34+w44 = 0

 

explaining why it has this property.  In fact, by solving this system one can directly derive the wi, instead of the geometric derivation which Lander used. (Note: Are there other forms of (5,4,4) that also involve the other parametrizations of (4,2,2)?)

 

Kawada, Wooley

 

(h+x)5 + (h-x)5 + (h+y)5 + (h-y)5 + (h+x+y)5 + (h-x-y)5 = 20h(x2+xy+y2+h2)2 - 14h5

 

(Update, 11/2/09):  Tony Rizzo pointed out we can let x2+xy+y2 = z2 for some z.  He gave Ramanujan's simple soln to this as {x,y} = {a-d, b+c+d} where a/b = c/d, and b = c.  For the special case when y = -x, this then becomes,

 

2(h+x)5 + 2(h-x)5 + 2h5  = 20h(h2+x2)2 - 14h5

 

where the whole eqn is now div by 2.  Note:  I realized Rizzo's observation can be extended to cover some of the identities I found given below.  For example, let z = h, and the Kawada-Wooley (KW) identity becomes,

 

(x+z)5 + (-x+z)5 + (y+z)5 + (-y+z)5 + (x+y+z)5 + (-x-y+z)5 = 66z5 

 

where x2+xy+y2 = z2, call this condition C1, the smallest soln of which is {x,y,z} = {3,5,7}.  If we choose to express 66 as a sum of powers 2(1)5+2(2)5, this results in a 10-term identity,

 

(x+z)k + (-x+z)k + (y+z)k + (-y+z)k + (x+y+z)k + (-x-y+z)k = 2(z)k + 2(2z)k

 

with the same conditon C1, though it is now valid for k = 1,2,3,4,5.  This is also discussed in Form 5.5 below.  But if we let z2 = nh2, with n = 4 or 28, then the KW identity is,

 

(h+x)5 + (h-x)5 + (h+y)5 + (h-y)5 + (h+x+y)5 + (h-x-y)5  = mh5  

 

for m = {486, 16806}, where x2+xy+y2 = nh2 for n = {4, 28}, respectively.  But if we express m as a sum of powers, then,

 

(h+x)k + (h-x)k + (h+y)k + (h-y)k + (h+x+y)k + (h-x-y)k  = 2(3h)k,   where x2+xy+y2 = 4h2 

 

(h+x)k + (h-x)k + (h+y)k + (h-y)k + (h+x+y)k + (h-x-y)k  = (-h)k + (7h)k,   where x2+xy+y2 = 28h2 

 

now valid for k = 1,3,5.  These will be discussed in Constraint I immediately after this note, though derived differently using resultants. (End update.) 

 

Piezas

 

So far, most of the identities discussed have the condition x1+x2+x3+x4 = y1+y2+y3+y4 = 0.  One can avoid this and explore other constraints such as:  

 

a) x1+x2 = x3+x4 = x5+x6,  as (a+c)k + (-a+c)k + (b+c)k + (-b+c)k + (a+b+c)k + (-a-b+c)k + uk + vk = 0

b) x1-x2 = x3-x4 = x5-x6,     as (a+c)k + (a-c)k + (b+c)k + (b-c)k + (a+b+c)k + (a+b-c)k + uk + vk = 0

c) x1-x2 = x3-x4 = -(x5-x6),  as (3p+q)k + (p+q)k + (p+r)k + (-p+r)k + (-3p+q)k + (-p+q)k + uk + vk = 0  

 

I. Constraint: x1+x2 = x3+x4 = x5+x6

 

We can focus on a special case of this constraint by defining the polynomial, P(k): = (a+c)k + (-a+c)k + (b+c)k +(-b+c)k + (a+b+c)k + (-a-b+c)k which is one side of the Kawada-Wooley equation.  This has the common sum of two terms as 2c.  Note this also obeys x1-x2+x3-x4 = x5-x6. If we set the equation,

 

E(k): = P(k) + (r)k + (s)k = 0

 

to be simultaneously true for k = 1,3,5, then this has a complete parametrization.  By eliminating {r,s} from the three equations E(1), E(3), E(5) using resultants, easily done using Mathematicas Resultant[] function, one gets a final eqn,

 

(a2+ab+b2-28c2) (a2+ab+b2-4c2) = 0

 

so there are two identities involving P(k), namely,

 

(a+c)k + (-a+c)k + (b+c)k +(-b+c)k + (a+b+c)k + (-a-b+c)k = 2(3c)k,   where a2+ab+b2 = 4c2, and,

 

(a+c)k + (-a+c)k + (b+c)k +(-b+c)k + (a+b+c)k + (-a-b+c)k  = (-c)k + (7c)k,   where a2+ab+b2 = 28c2,

 

both for k = 1,3,5.  Also, the polynomial Pk as a sum is quite peculiar since a common factor turns up for k = 1,2,3,4,5.  To illustrate, for the first, set {a,b,c} = {2(u2-v2),  2(2uv+v2),  (u2+uv+v2)}, then for k = 1,2,3,4,5 we have the sums,

 

Pk = {2(3)(u2+uv+v2),   2(11)(u2+uv+v2)2,   2(33)(u2+uv+v2)3,   2(83)(u2+uv+v2)4,   2(35)(u2+uv+v2)5}

 

while for the second, let, {a,b,c} = {2(u2+6uv+2v2),  2(2u2-2uv-3v2),  (u2+uv+v2)}, then,

 

Pk + ck = {7(u2+uv+v2),   7(17)(u2+uv+v2)2,   73(u2+uv+v2)3,   7(545)(u2+uv+v2)4,   75(u2+uv+v2)5}

 

Thus, in general, for these k and no higher, they have the common factor u2+uv+v2.   These identities also have other properties.  The first, by solving a2+ab+b2 = t2k, gives a soln to,

 

x15 + x25 + x35 + x45 + x55 + x65 = 2(35y5k)

 

the quintic analogue of Ramanujan’s family of solns to x14 + x24 + x34 = 2(y2k).  See Form 11 of Fourth Powers.  Also, since the a,b,c are binary quadratic forms, these can provide solns to,

 

x15 + x25 + x35 + x45 + x55 + x65 + x75 = 1

 

by setting one term as equal to ±1 and solving the resulting Pell eqn.  For ex, we have,

 

(a+c)k + (b+c)k + (a+b+c)k + (-a-b+c)k  + (-a+c)k + 2(-3c)k = ±1

 

where {a,b,c} = {2(p+q)(p+5q),  8(p+4q)q,  p2+8pq+19q2},  for k = 1,3,5,  if p2-13q2 = ±1. 

 

 

II. Constraint: x1-x2 = x3-x4 = x5-x6

 

We can focus first on a special case.  Define the eqn,

 

(a+c)k + (a-c)k + (b+c)k + (b-c)k + (a+b+c)k + (a+b-c)k + (u)k + (v)k = 0,  for k = 1,3,5,   (eq.1)

 

This has the common difference of the first three pairs of terms as 2c and also has a complete parametrization, but now involves an elliptic curve.  Note that this also has the sum of its first four terms as x1+x2+x3+x4 = x5+x6.  Let {a,b,u,v} = {p+q, -p+q, d-4q, -d-4q}, then eq.1 is true if,

 

c2+11d2 = 4p2,  and 3c2+d2 = 12q2

 

where the ratio c/d = {1, 1/3} should be avoided.  A small soln is {c,d} = {37, 15} and it is easy to convert the two quadratic conditions into an elliptic curve.  A special case solvable as quadratic forms is,

 

(a+b+c)k + (a+b-c)k + (-a+b+c)k + (-a+b-c)k + (8b+c)k + (8b-c)k = (7b)k + (13b)k,  for k = 1,3,5,   (eq.2)

 

where -a2+126b2 = 5c2.

 

The common difference of the first three pairs is also 2c but the sum of its first four terms is now x1+x2+x3+x4 = (x5+x6)/4.  From the initial soln {a,b,c} = {1,1,5}, one can get a parametrization.  However, for the complete soln to the general case, let, 

 

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

 

Expanding for k = 1,3,5, we can eliminate d,n and find the resultant in terms of m which is only a quadratic.  Explicitly, the complete soln is, d = a+b+c,  and {m,n} = {u/(2d),  v/(2d)} where {u,v} are,

 

(a+b)4 + (a+b)(4a2+9ab+4b2)c + (3a+2b)(2a+3b)c2 + 4(a+b)c3 + c4 = u2

 

(a2-b2)2 – 3ab(a+b)c – (2a2+3ab+2b2)c2 + c4 = v2

 

One must find {a,b,c} such that these two quartic polynomials in c are squares which in general is not easily done.  (Surprisingly, for a k = 1,3,5,7 with a similar constraint, it involves only two quadratics to be made squares which is easier.)  Without loss of generality, we can set a = 1 and one soln for the first eqn is,

 

c = -(2+3b)/(3+4b)

 

It reduces the second eqn, after removing square factors, to solving,

 

25+24b-48b2+64b4 = y2

 

with initial b = 3/16 from which other rational points can then be computed.  (Update, 8/13/09):  One can use a variant,

 

(p+aq)k + (p+bq)k + (p+cq)k + (d+q)k = (p-aq)k + (p-bq)k + (p-cq)k + (d-q)k,  for k = 1,3,5

 

already satisfying the constraint (after appropriate transposition of terms).  Solving this system involves only making a quartic and quadratic polynomial into squares, hence is a slight improvement over the previous method.  (End update.)

 

 

III. Constraint: x1-x2 = x3-x4 = -(x5-x6)

 

This is a generalization of a single numerical result by Tarry.  It can be the case that a k = 1,3,5 can give rise to a k = 2,4,6.  Given,

 

(3p+q)k + (p+q)k + (p+r)k + (-p+r)k + (-3p+q)k + (-p+q)k + uk + vk = 0,  for k = 1,3,5, then,

 

(4p+q)k + (2p+r)k + (p+u)k + (p+v)k = (4p-q)k + (2p-r)k + (p-u)k + (p-v)k,  for k = 2,4,6

 

which has a complete soln in terms of an elliptic curve as,

 

{p,q,r,u,v} = {2x2+x+1,  x2+6x+1,  -8(x2+2x),  6x2+4x-2+y,  6x2+4x-2-y}

 

and where x,y must satisfy,

 

y2 = 15x4+104x3+223x2+54x+4

 

though some of the small rational solns are trivial.  (Update, 8/4/09):  Alternatively, the system can be expressed as,

 

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

 

(a+c-d)k + (a+c+d)k + (-a+2b+2c)k + (b-4c)k  = (a-c-d)k + (a-c+d)k + (-a+2b-2c)k + (b+4c)k,  for k = 1,2,4,6, 
 

where {a,b,c,d} satisfy a2+ab+2b2 = 8c2 and a2-7ab+42b2 = 8d2

 

This is easily proven by solving for {c,d} in terms of {a,b} and subsituting into the two systems, though the ratio a/b = {-1/3, -1, 2, 5} should be avoided as they are trivial, with a small non-trivial soln as {a,b} = {15,1}.  (End update)

 

Note 1:  These are the only known classes for k = 1,3,5.  Any other class not discussed here?

Note 2: Also, in Third Powers, it was seen that [a,b,c] = [d,e,f] for k = 1,3 where abc = def has solns. Is the corresponding [a,b,c,d] = [e,f,g,h] for k = 1,3,5 where abcd = efgh also solvable?

 

 

5.5  Ten terms

 

The system of eqns, x1k+ x2k+…+ xnk = y1k+ y2k+…+ ynk,  for k = 1,2,3,4,5 is non-trivial only when n > 5, but two terms on one side can be equal to zero.

 

Piezas

 

Use the form,

 

(a+c)k + (-a+c)k + (b+c)k + (-b+c)k + dk + ek = fk + gk  + 0 + (2c)k + 0 + (2c)k

 

where {d,e,f,g} = {(cw+u)/w,  (cw-u)/w,  (cw+v)/w,  (cw-v)/w}.  Note that this is a symmetric ideal solution with pairs of terms having the common sum 2c.  The system can be solved if {a,b,c} satisfy the two conditions,

 

(a2+ab+b2-c2)(a2-ab+b2-c2)w-w3 = u2

(a2+ab+b2-c2)(a2-ab+b2-c2)w = v2

 

where w = a2+b2-2c2.  The cases (a2-c2)(b2-c2)(a2+b2-2c2) = 0 must be avoided since it gives either trivial results or there is division by zero.  An example of an appropriate soln is {a,b,c} = {-11, 13, -7} which gives {u,v} = {384, 2688}.  For the special case a2+ab+b2-c2 = 0, this implies v = 0 and f = g.  After some manipulation, we then get a nice soln to the form,

 

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

 

where a2+ab+b2 = c2

 

 

5.6  Twelve terms

 

Piezas

 

This is the fifth power counterpart to the third and fourth power identities found by this author.  For k = 1,2,3,4,5,

 

(a1x+v1y)k + (a2x-v2y)k + (a3x+v3y)k + (a4x-v3y)k + (a5x+v2y)k + (a6x-v1y)k =

(a1x-v1y)k + (a2x+v2y)k + (a3x-v3y)k + (a4x+v3y)k + (a5x-v2y)k + (a6x+v1y)k

 

and the sum-product identity,

 

(a1x2+2v1xy+3a6y2)k + (a2x2-2v2xy+3a5y2)k + (a3x2+2v3xy+3a4y2)k + (a4x2-2v3xy+3a3y2)k + (a5x2+2v2xy+3a2y2)k + (a6x2-2v1xy+3a1y2)k = (a1k+a2k+a3k+a4k+a5k+a6k)(x2+3y2)k

 

where {a1, a2, a3, a4, a5, a6} = {a+c, b+c, -a-b+c, a+b+c, -b+c, -a+c} and {v1, v2, v3} = {a+2b, 2a+b, a-b}, for five arbitrary variables a,b,c,x,y.  (Note that the vi are the same as the ones for fourth powers for Form 21.)  These are symmetric ideal solns, with the common sum 2cx for the linear forms and 2c(x2+3y2) for the quadratic forms.  They have the form,

 

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

 

where appropriate pairs of terms have the common sum 2n and can be solved by any soln to am+bm+cm = dm+em+fm, m = 2,4, for any n.  For the special case when a+b = d+e, then it is also true that,

 

(a+n)k + (a-n)k + (b+n)k + (b-n)k + (c+n)k + (-c+n)k = (d+n)k + (d-n)k + (e+n)k + (e-n)k + (f+n)k + (-f+n)k,  k = 1,2,3,4,5   (Eq.M)

 

when a+b = 2n.  Expressed as {xi, yi}, in addition to having the common sum x1+x4 = x2+x3 = x5+x6 = a+b, (likewise for the yi since a+b = d+e), it also has the near-common difference of x1-x4 = x2-x3 = x5+x6 = 2n, and same for the yi.  One simple soln to Eq.M is {a,b,c,d,e,f,n} = {5y,  x-2y,  x+2y,  x, 3y, z, (x+3y)/2} based on the simple identity,

 

(5y)k + (x-2y)k + (x+2y)k = xk + (3y)k + zkfor k = 2,4, 

 

if x2+24y2 = z2.

 

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