Updates Page 02: Feb

(Update, 2/26/10)  There is a family of identities starting with Birck's,


(-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 was generalized by J. Zehfuss as,


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


where Birck’s was d = 0.  Generalizing this with a fifth variable e by adding just one term to the RHS will no longer work, but Alain Verghote modified this to the two (2.5.6) identities,

Form 1: 
(-a+b+c+d+e)k + (a-b+c+d+e)k + (a+b-c+d+e)k + (a+b+c-d+e)k + (a+b+c+d-e)k = (2a)k + (2b)k + (2c)k + (2d)k + (2e)k + (a+b+c+d+e)k,  for k = 1,2

Form 2:

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


Note how for e = 0, Form 1 reduces to Zehfuss', while Form 2 reverts into a tautology.  These can be further generalized as,


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


and so on. (Setting f = 0 reduces to Form 1.)  The significance of Birck's and Zehfuss' symmetric versions is that they can give rise to the multi-grade identity,


(-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


if abcd = efgh and an+bn+cn+dn = en+fn+gn+hn,  for n = 2,4.  This has an infinite number of solns and one can also let d = h = 0 for a simpler version.  However, to create an analogous multi-grade using Verghote’s Form 1 identity would entail a more complicated set of conditions on ten variables.  (End update.)

(Update, 2/23/10)  Wroblewski, Piezas

Let a2+b2 = c2 and a2+52b2 = d2, a pair of conditions which is a concordant formThen a [k,7,7] for k = 1,2,4,6,8,10 is,

(8b)k + (-5a-4b)k + (5a-4b)k +  (-a-2d)k + (a-2d)k + (-12b+4c)k + (12b+4c)k =
(-16b)k + (-4a+8b)k + (4a+8b)k + (-a+4c)k + (a+4c)k + (-3a-2d)k + (3a-2d)k 
though ratios such as a/b = {4, 12, 12/5} should be avoided as the system becomes trivial, with a non-trivial example the smallest Pythagorean triple {a,b,c} = {3, 4, 5}, with d = 29.  The two quadratic polynomials define an elliptic curve, and there is an infinite number of solns.  A pair of algebraic forms {a2+b2, a2+Nb2} that is to be made squares is called a concordant form.   (However, found throughout this book, is the more general case of  {pa2+qb2, ra2+sb2} to be made squares.)  This identity is a special case of a more general one involving 16 terms.  (End update.)

(Update, 2/22/10, 3/1/10):  Piezas


Theorem:  “The eqn a4+b4 = c6+d6 has an infinite number of unscaled, integral solns.”


Proof:  The first known example by Giovanni Resta has the form,


(m2n)4(x4+y4) = (mn)6(u6+v6)      (eq.1)


with primitive soln {m,n} = {5, 73}; and {u,v,x,y} = {3, 4, 18, 31}.  Eq.1 can be simplified as,


m2(x4+y4) = n2(u6+v6)


Hence the situation is reduced to the easier problem of finding,


(x4+y4)(u6+v6) = w2              (eq.2)


where gcd(x,y) = 1 and gcd(u,v) = 1.  However, the second factor can also be treated as a constant, so this becomes,


25(193)(x4+y4)  = w2


which is an elliptic curve. Starting with the known {x,y} = {18, 31}, an infinite number of other rational points can then be computed.  Hence, another soln to eq.1 is:


{m,n} = {5, 228420709564570748140960777}; and {u,v,x,y} = {3, 4, 56328091105014, 7382523294847}


and so on, ad infinitum.  It seems quite interesting that the Pythagorean triple {3, 4, 5} appears in this problem.  Note:  A computer search can probably find smaller co-prime solns to eq.2, though the relation u3y2-v3x2 = 0 should be avoided as it is trivial.  Using two generators of an elliptic curve, Seiji Tomita found smaller solns to x4+y4 = 193z2 as,


{x,y}= {1266, 181}; {19021, 10806}; {890858057, 810357546}; {288871421574, 44739797383}


with the next the same as the one found by this author.  (End update.)



(Update, 2/22/10):  Gerardin gave the beautiful 4th power identity with small coefficients {1,2,3},


(a+3a2-2a3+a5+a7)4 + (1+a2-2a4-3a5+a6)4 =

(a-3a2-2a3+a5+a7)4 + (1+a2-2a4+3a5+a6)4


and Choudhry gave a 5th power version,


(a-a3-2a5+a9)5  +  (1+a2-2a6+2a7+a8)5 +  (2a3+2a4-2a7)5 =

(a+3a3-2a5+a9)5 + (1+a2-2a6-2a7+a8)5 + (-2a3+2a4+2a7)5  


which in fact is good for k = 1,5.  (Differences between RHS and LHS have been highlighted in blue.)  While there are identities for k = 6, none are known with such small coefficients.  (Anyone can find one?)  Other 5th deg identities can be found using the ff. two methods by Choudhry.  Given,


x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k     (eq.1)




{x1, x2, x3, x4} = {a+b+c, a-b-c, -a+b-c, -a-b+c}

{y1, y2, y3, y4} = {d+e+f, -d+e-f, -d-e+f,  d-e-f}


Expanding one side of the eqn, we get,


∑ xi1 = 0

∑ xi5 = 80abc(a2+b2+c2)


and similarly for the yi.  Eq.1 can be reduced to just 6 terms if x4 = y4.  Thus, if,


Method 1:  a+b-c = -d+e+f ; ab = de;  c(a2+b2+c2) = f(d2+e2+f2)

Method 2:  a+b-c = -d+e+f ; c = f;  ab(a2+b2+c2) = de(d2+e2+f2)


then ∑ xik = ∑ yik,  for k = 1,5.


Choudhry showed that given an initial point, even a trivial one, using appropriate substitutions and solving a system of linear equations, then an infinite number of non-trivial solns can be found.  In fact, by directly solving Method 1, one can end up with a quartic equation which has a rational root if a certain quartic polynomial is made a square, hence can be treated as an elliptic curve.  Note:  Choudhry also used Eq.1 and its substitutions to solve k = 1,3,7.


Source:  “On Equal Sums of Fifth Powers”, Indian Jour. Pure & Applied Math, Nov. 1997.


(Update, 2/22/10)It is unknown if the unbalanced multi-grade eqn,


x1k+x2k+x3k = x4k+x5k+x6k+x7k,   for k = 2,4,6
has non-trivial solns.  Q:  Anyone can provide one, preferably parametric?  In response to a post in sci.math, Giovanni Resta made a search and found that there are no solns with terms < 845.  James Waldby would later extend this search radius to < 1280 with still no soln.  However, there are a few that are good only for k = 2,6, namely,

[73, 58, 41] = [70, 65, 32, 15]
[85, 74, 61] = [87, 71, 56, 26]
[218, 167, 29]  = [224, 107, 102, 65]
[351, 265, 221] = [336, 309, 169, 73]
[493, 335, 65] = [497, 297, 155, 16]


Some exhibit interesting side relations between its terms, such as the smallest one with 73-58 = 15, 73-41 = 32, though with appropriate substitutions, they are still not enough to reduce it to a parametric form.  (End update.)



(Update, 2/18/10):  On a hunch, this author re-checked Wroblewski’s soln for (k.5.5) for k = 1,3,9 to see if there was a partition such that it would be good for k = 2 as well.  It turns out there was.  Thus,


[51, 253, 412, -621, 600]k = [187, 100, 429, 603, -624]k,  for k = 1,2,3,9


with terms obeying the constraints,


1. x1+x2+x3 = y1+y2+y3

2. x4+x5 = y4+y5

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

4. (x1-y1)(x2-y2) = (x3-y3)(x5-y5)


with the last two suggested by Wroblewski, though [4] is just a consequence of [2] and [3].  Why this “numerical curiosity” obeys such symmetric constraints is unknown.  (Note that the difference xi-yi for all five pairs is divisible by 17.)  (End update.)


(Update, 2/16/10):   As observed by Bremner and Delorme, the smallest [k.6.6] for k = 9, found by Lander et al through computer search back in 1967, is in fact good for k = 1,2,3,9,


[18, 23, 13, -10, -15, -5]k = [21, 22, 9, -13, -14, -1]k


Notice also how appropriate pairs all have the same sum,


18-10 = 23-15 = 13-5 = 21-13 = 22-14 = 9-1 = 8


This “numerical curiosity” has an explanation, which lies in the simple eqn,


142 + 192 + 92 = 172 + 182 + 52


and is just a special instance of Theorem 5 of the Prouhet-Tarry-Escott Problem.


Define, Fn = an+bn+cn+dn-(en+fn+gn+hn) and the [k.8.8],
[x+a, x+b, x+c, x+d, x-a, x-b, x-c, x-d]k = [x+e, x+f, x+g, x+h, x-e, x-f, x-g, x-h]k


1) If F2 = 0, and a variable x such that 42F4x4+28F6x2+3F8 = 0, then the system is for k = 1,2,3,9. 


2) If F2 = F4 = 0 and 28F6x2+3F8 = 0, then it is for k = 1,2,3,4,5,9.


Notice how appropriate pairs of terms all have the same sum 2x.  Also, it reduces to a [k.6.6] if a pair from opposite sides, say d = h = 0.  Lander's soln was simply the case, 


{a,b,c,d} = {14, 19, 9, 0}

{e,f,g,h} = {17, 18, 5, 0}


and x = 4.  Bremner and Delorme found a relationship between the terms such that the quartic in x factored into two quadratics.  Making its discriminant a square involved an elliptic curve, so there are an infinite number of solns to [1].  This is discussed more in Section 9.3 of Ninth Powers.


For non-zero {d,h}, no example is yet known for [1], but for [2], Wroblewski directly found a nonic [k.8.8] for k = 1,2,3,4,5,9, which yields,


{a,b,c,d} = {240, 63, 197, 122}

{e,f,g,h} = {167, 10, 168, 243}

and x  = 52.  Since these variables can be reduced in size with appropriate transformations given in the previous update, it is possible there are others. (End update.)


(Update, 2/15/10):  Wroblewski, Piezas


A multi-grade [k.4.4] for k = 2,4 can be used to find a nonic [k.8.8] for k = 1,2,3,4,5,9.  Given six variables {p,q,s,t,y,z},


System 1:

[10p+2q-s+t, -10p-2q-s+t, 2p-10q+y-z, -2p+10q+y-z]k =

[10p-2q-s+t, -10p+2q-s+t, 2p+10q+y-z, -2p-10q+y-z]k,  for k = 2,4


System 2:

[5p+q+s, -5p-q+s, 5p+q+t, -5p-q+t,  p-5q+y, -p+5q+y, p-5q+z, -p+5q+z]k =

[5p-q+s, -5p+q+s, 5p-q+t, -5p+q+t , p+5q+y, -p-5q+y, p+5q+z, -p-5q+z]k,  for k = 1,2,3,4,5


is identically true using the substitutions,


{p,q,s,t,y,z} = {u/2, v/2, a+b, a-b, a+c, a-c}

{b,u,c,v} = {eg-2fh, eh+fg, eg+2fh, eh-fg} 


for arbitrary {a,e,f,g,h}.  It can be shown that [2] is derived from [1] using Theorem 5 of the Prouhet-Tarry-Escott problem.


A) One can make the first system valid for k = 6 and the second for k = 6,7 as well if,


(10g2-13h2)e2-(13g2-40h2)f2 = 0


for arbitrary a.  This is reducible to an elliptic curve, with one soln as {e,f,g,h} = {171, 97, 7, 47}, and an infinite more can then be computed. 


B) However, if System 2 is to be true only for k = 1,2,3,4,5,9, then one must solve a quadratic in a with a sextic discriminant that must be made a square.  The only known soln found by Wroblewski is {a,e,f,g,h} = {104, 7, 3/2, -18, 17}, though others may exist.  Whether with the right constraints this can be reduced to a quartic polynomial to be made a square, hence an elliptic curve, is unknown. (End update)



(Update, 2/15/10):  Piezas, Wroblewski


Using a numerical search with constraints, Wroblewski found hundreds of solns to a multi-grade octic chain of form,


x1k+x2k+x3k+x4k-(x5k+x6k+x7k+x8k) = y1k+y2k+y3k+y4k-(y5k+y6k+y7k+y8k) = z1k+z2k+z3k+z4k-(z5k+z6k+z7k+z8k)


valid for k = 2,4,6,8.  An example is,


[221, 93, 73, 9]k-[219, 117, 31, 17]k = [198, 116, 96, 32]k-[192, 144, 58, 44]k = [189, 125, 105, 23]k-[175, 161, 75, 27]k


Note that it is also true x1k+x2k+x3k+x4k = x5k+x6k+x7k+x8k, for k = 2,4 and x1x2x3x4 = x5x6x7x8, relationships which also hold for the yi and zi.  It turns out the general case could be explained in the context of an old result by the first author, namely,


“If an+bn+cn+dn = en+fn+gn+hn, for n = 2,4,6 and abcd = efgh, 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.”  (If n is only up to n = 4, then k is only up to k = 8.)  Moving half of the terms of one side to the other yields,


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

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


and since either sign of {±d, ±h} can be used and still satisfy abcd = efgh, then the RHS has two distinct octuplets for a given {a,b,c,d,e,f,g,h}.  Thus, given,

p1k+p2k+p3k+p4k = q1k+q2k+q3k+q4k, k  = 2,4


where p1p2p3p4 = q1q2q3q4, then this automatically leads to an octic chain of length 3.  Wroblewski’s solns satisfies the constraints,


p1p2 = q3q4

p3p4 = q1q2

p12+p22 = q12+q22

p32+p42 = q32+q42

p14+p24+p34+p44 = q14+q24+q34+q44


The first author found the complete soln to this system as an elliptic “curve”, namely,


{p1, p2, p3, p4} = {ar+bc, ac-br, ds+ef, df-es}

{q1, q2, q3, q4} = {ar-bc, ac+br, ds-ef, df+es}


r = (abc2+def2)(d2-e2)(f/y),  s = (abc2+def2)(a2-b2)(c/y)


and {a,b,c,d,e,f} must satisfy,


(abc2+def2)(c2de)(a2-b2)2 + (abc2+def2)(abf2)(d2-e2)2  = y2


which is only a quartic polynomial in {c,f} to be made a square.  Given one soln, then an infinite more can be found.




Another set of conditions is,


p1p4 = q1q4

p2p3 = q2q3

p1+p2 = q1+q2

p1k+p2k+p3k+p4k = q1k+q2k+q3k+q4k,  k = 2,4


The soln can be given as,


{p1, p2, p3, p4} = {a(p+q), b(r-s), c(r+s), d(p-q)}

{q1, q2, q3, q4} = {a(p-q), b(r+s), c(r-s), d(p+q)}


{p,q,r,s} = {a(b2-c2),  be,  (a2-d2)b,  ae}


and where {a,b,c,d,e} satisfy,


a2(3b2-c2)+e2 = (b2+c2)d2


This can be easily solved parametrically as quadratic forms.  (End update)



(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 here.)  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, which 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((p+2q)2-r2) = 3q2+s2


For 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.  But 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.  But substituting this to eq.2, and still with q = 1, produces a rational quartic 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 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.)



(Update, 2/8/10):  Piezas


If p4+q4+r4 = 1, then (p+q)4 - r4 + 1 = 2(p2+pq+q2)2  also holds.



(Update, 2/8/10):  Demjanenko, Elkies


The complete soln to


p4+q4+r4 = 1                          (eq.0)


can be given in the form,


(x+y)4 + (x-y)4 + z4 = 1




ay2 = (8mn-3a)x2-2bx-2mn       (eq.1)

±az2 = 4bx2+8mnx-b                 (eq.2)


and {a,b} = {2m2+n2, 2m2-n2}, for some constants {m,n} hence is a problem of making two quadratics in x as squares and, for appropriate {m,n}, reduces to solving an elliptic curve.  One can solve {y,z} in the radicals to see that it holds and it is easily proven that this is the complete soln.  Proof:  Let {x+y, x-y, z} = {p,q,r}.  As eq.1 and eq.2 are also quadratics in {m,n}, solve for m, assume eq.0 as true to rationalize the discriminant, and take the positive sign of the square root to get,


m1 = n((p+q)2+r2-1) / (2(p2+pq+q2+p+q))

m2 = n(p2+pq+q2-p-q) / ((p+q)2-r2-1)


One can use the identity given by the author to rationalize the disciminant of m2.  It will then be seen that m1-m2 = 0 if eq.0 is true, hence one can always find rational {m,n} in terms of {p,q,r}.  For example, using the smallest {p,q,r} = {217519/s, 414560/s, 95800/s} and s = 422481, we get {m,n} = {6007, 30080}.  (End proof)  However, one can also start with {m,n} as it is possible they are relatively small.  Elkies’ first soln used only {m,n} = {8,-5}, and (eq.1) becomes,


153y2 = -779x2-206x+80


which has initial soln {x1,y1} = {3/14, 1/42}.  This yields a parametrization for x as,


x = (51p2-34p-5221)/(238p2+10906)


which, when substituted into (eq.2), becomes the problem of finding a value such that the quartic polynomial in p is a square.  This can be treated as an elliptic curve and since x1 = x = 3/14 has p = -3779/17, from this initial point one can calculate more, proving that eq.0 has an infinite number of distinct solns.  Clever transformations may then reduce the size of the coefficients.  Explicitly, a particular case can be given as the identity, 


(85v2+484v-313)4 + (68v2-586v+10)4 + (2u)4 = (357v2-204v+363)4


if u2 = 22030+28849v-56158v2+36941v3-31790v4,


a soln of which is v = -31/467.  This gives, after removing common factors, Elkies' soln.  One can then use this initial point to generate an infinite more. 

(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 




{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, by the Ramanujan-Hirschhorn Theorem, its terms obey the beautiful relations,


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


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


(End update)



(Update, 2/3/10):  Choudhry


Sophie Germain’s Identity is given by,


a4+4b4 = (x2+2xy+2y2)(x2-2xy+2y2)


Choudhry solved the eqn,


a4+4b4 = c4+4d4    (eq.1)


using 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 had 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.)



(Update, 2/3/10):  Choudhry


(-8m6+2mn5)k + (8m5n+n6)k + 2(8m6)k =  (8m6+2mn5)k + (8m5n-n6)k + 2(n6)k,  for k = 1,5


A soln in positive terms can be acquired if n > m and within the range 22/5 < n/m < 23/5, or approx. 1.32 < n/m < 1.51.  With terms transposed to one side,


(-8m6+2mn5)k + (8m5n+n6)k - (8m6+2mn5)k - (8m5n-n6)k + 2(8m6)k =  2(n6)k


For n = 1, the identity proves that the integer 2 is the sum/difference of six integral 5th powers in infinitely many ways.  Source:  The Diophantine Equation x15+x25+2x35 = y15+y25+2y35, Ganita, Vol. 48, No. 2, 1997, 115-116.


Note:  In one sense, this is a 5th power version of,


(-6x3+y3)3 + (6x3+y3)3 - (6x2y)3 = 2(y3)3


where, for y = 1, proves that 2 infinitely is the sum is the sum/difference of three integral 3rd powers. 


Q:  Anyone can give a similar identity expressing 2 as integral 7th powers?  (End update.)

(Update, 2/2/10):  Piezas


More generally, for the equivalent form (disregarding signs),


[a+b, -a+b,  b+u, -b+u, c+v, -c+v, 2d]k = [c+d, -c+d, a+u, -a+u, d+v, -d+v, 2b]k,  for k = 1,2,4,6,8,


the {a,b,c,d,u,v} must satisfy the simultaneous eqns,


a2+2b2+u2 = c2+2d2+v2

(a2-b2)(b2-u2) = (c2-d2)(d2-v2)


excluding signed combinations of trivial cases (a+b)(c+d)(b+d)(a+b+c+d)(a2+3b2-c2-3d2) = 0.  For the special case u = v, the first condition reduces to,


a2+2b2 = c2+2d2


the complete soln of which {a,b,c,d} = {pr+2qs, ps-qr, pr-2qs, ps+qr} reduces the second condition to the easy problem of making a quadratic polynomial into a square.  (End update.)



(Update, 2/1/10):  This update covers a family that includes 6th, 8th, and 10th power multi-grades.  Whether it can be extended for 12th powers with a minimal number of terms remains to be seen.


Piezas  (6th powers)


Given the general form,


[xy+ax+3by-c,  xy-ax-3by-c,  xy-3bx+ay+c,  xy+3bx-ay+c]k =

[xy+ay+3bx-c,  xy-ay-3bx-c,  xy-3by+ax+c,  xy+3by-ax+c]k


Notice how {x,y} just swap places or alternatively, “a” in one side is replaced by “3b” in the other side.  With terms expressed as {pi, qi}, this form has the constraints,


p1+p2 = q1+q2,

p3+p4 = q3+q4,

p1k+p2k+p3k+p4k = q1k+q2k+q3k+q4k,  k = 1,2


We can make this valid for k = 1,2,4,6.  For k = 4, let c = ab.  For k = 6 and some constants {a,b}, then {x,y} must satisfy the simple eqn,


10a2b2-(a2+9b2)(x2+y2)+10x2y2 = 0


This is an elliptic “curve” in disguise.  Let {x,y} = {u, v/(10u2-a2-9b2)} and one is to solve,


(a2+9b2-10u2)(10a2b2-a2u2-9b2u2) = v2


which is a quartic polynomial in u to be made a square.  Trivial ratios are a/b = {0,1,3,9}.  It can be proven that for any rational {a,b} that do not have a trivial ratio, one can always find an infinite number of rational {x,y}.  Proof:  A polynomial soln can be given starting from the trivial point u = 3b.   This yields the non-trivial,


u = 9b(a2+15b2)/(5a2-117b2)


from which one can then successively compute an infinite sequence of rational polynomials. 



Chris Smyth  (8th powers)


This is a variation of the Letac-Sinha 8th power multi-grade.  For k = 1,2,4,6,8,


[xy+ax+ay-d, -(xy-ax-ay-d), xy-bx+by+d, -(xy+bx-by+d), cx+cy]k =

[xy+bx+by-d, -(xy-bx-by-d), xy-ax+ay+d, -(xy+ax-ay+d), cx-cy]k


where {a,b,c,d} = {1, 3, 4, 11} and {x,y} satisfies the eqn,


x2y2-13(x2+y2)+112 = 0


One can note its affinity with the 6th power multi-grade and see also in the general form how “a” in one side has just been replaced with “b” in the other side.  With terms expressed as {pi,qi}, this obeys the constraints,


p1-p2 = q1-q2

p3-p4 = q3-q4


If we set that,


p1k+p2k+p3k+p4k+p5k = q1k+q2k+q3k+q4k+q5k,  k = 1,2


then one must have 2ak-2bk+ck = 0,  for k = 1,2, and the consequence is that,


p1k+p2k+p5k = q1k+q2k, 

p3k+p4k = q3k+q4k+q5k


for k = 1,2 which, naturally enough, is the same set of constraints obeyed by the Letac-Sinha identity.  Expanding the system for k = 4,6,8, it will be seen that the only non-trivial soln in the rationals is {a,b,c,d} = {m, 3m, 4m, 11m2} where one can set m = 1 without loss of generality and {x,y} satisfies the eqn,


x2y2-13(x2+y2)+112 = 0


Let {x,y} = {u, v/(u2-13)} and this is the elliptic curve,


(u2-13)(13u2-112) = v2


Smyth’s form for (k.5.5) can be generalized by adding two terms on each side to create a (k.7.7).  A possible addition, given in italics, may be,


[xy+ax+ay-c, -(xy-ax-ay-c), xy-bx+by+c, -(xy+bx-by+c), dx+dy, ex+fy, fx-ey]k =

[xy+by+bx-c, -(xy-by-bx-c), xy-ay+ax+c, -(xy+ay-ax+c), dx-dy, ey+fx, ex-fy]k


which is the same pair successfully added to the (k.4.4) to find the first (k.6.6) identity for 10th powers.  For non-zero {e,f}, whether this has a non-trivial soln up to k = 8, 10, or even 12 is unknown.



Choudhry, Wroblewski   (10th powers)


The 6th power (k.4.4) identity can be generalized to the form (k.6.6) by adding the italized pair of terms,


[-(axy+bx+cy-d), axy-bx-cy-d, axy-cx+by+d, -(axy+cx-by+d), ex+fy, fx-ey]k =

[-(axy+by+cx-d), axy-by-cx-d, axy-cy+bx+d, -(axy+cy-bx+d), ey+fx, ex-fy]k


where again, {x,y} merely swap places.  (Since the eqn is homogeneous, one may assume the leading coefficient of the terms as a = 1 without loss of generality.)  This obeys the same basic constraints (up to sign changes),


p1-p2 = q1-q2,

p3-p4 = q3-q4,

(-p1)k+p2k+p3k+(-p4)k = (-q1)k +q2k+q3k+(-q4)k,  k = 1,2

p52+p62 = q52+q62


If we assume,


p1+p2+p5 = q1+q2+q5, 

p3+p4+p6 = q3+q4+q6, 


then it must be the case that 2(b-c) = e-f   (eq.1).  Choudhry and Wroblewski found for k = 2,4,6,8,10, the equivalent non-trivial solns,


{a,b,c,d,e,f} = {1, 1, 2, 14, 3, 5},  where 2x2y2-17(x2+y2)+392 = 0

{a,b,c,d,e,f} = {2, 1, 2, 7, 3, 5},  where 8x2y2-17(x2+y2)+98 = 0


Disregarding eq.1, two other distinct non-trivial solns found (by Piezas), though only up to k = 2,4,6,8, are,


{a,b,c,d,e,f} = {5, 3, 7, 3, 2, 10},  where 125x2y2-53(x2+y2)+45 = 0

{a,b,c,d,e,f} = {5, 9, 11, 11, 10, 12}, where 125x2y2-221(x2+y2)+605 = 0


though this author is not certain if these are the same 8th power multi-grades found by Wroblewski.  (End update.)



You can email author at tpiezas@gmail.com.