020: Fifth Powers (4 or 6 Terms)

III. Sum / Sums of Fifth Powers 

5.1    Four terms

5.2    Six terms

5.3    Seven terms

5.4    Eight terms

5.5    Ten terms

5.6    Twelve terms


5.1  Four terms


While it is conjectured that x1k + x2k = y1k + y2k has no non-trivial rational solns for k > 4, there are solns as roots of quadratics for k = 5.


A. Desboves


(a√2+b)5 + (-b+c√-2)5 = (a√2-b)5 + (b+c√-2)5  


where a2+b2 = c2.




(√p+√q)5 + (√p-√q)5 = (√r+√s)5 + (√r-√s)5 


{p,q,r,s} = {5vw2,  -1+uw2,  5v,  -(u+10v)+w3}, where w = u2+10uv+5v2.


One can then set v = 5t2 so that {p,r} are squares.  Whether {q,s} can then be made non-trivial squares is another matter.  More generally, the above equation for k = 5,6,7,8 has a polynomial soln and the first two can be reduced to a common form.  For k = 5 this reduces to,


√p(p2+10pq+5q2) = √r(r2+10rs+5s2)


then {p,q,r,s} = {5(bc-ad)v2,  u3-(10bc+w)v2,  5(bc-ad)u2,  -v3+(10ad+w)u2}


where {u,v,w} = {a2+10ab+5b2,  c2+10cd+5d2,  ac+5bd}


with four free variables {a,b,c,d}.  For k = 6, 

(√p+√q)6 + (√p-√q)6 = (√r+√s)6 + (√r-√s)6 


or after expanding,


(p+q)(p2+14pq+q2) = (r+s)(r2+14rs+s2)


then {p,q,r,s} = {-u2+vw,  u2-vx,  v2-ux,  -v2+uw}


where {u,v,w,x} = {a2+14ab+b2,  c2+14cd+d2,  ac+bc+13ad+bd,  ac+13bc+ad+bd}


Form: (p+cq)(p2+apq+bq2)k = (r+cs)(r2+ars+bs2)k


This generalizes the two forms above and as was already discussed in the section on Third Powers, it takes only a small transformation to assume c=0 without loss of generality. This author found the soln with c=0 as,


{p,q,r,s} = {bvwk,  -1+uwk,  bv,  -(u+av)+wk+1},  where w = u2+auv+bv2


for arbitrary u,v.  Note: This can be proven for k=1,2. Using computer algebra, it is easy to see it is also true for other small k > 2.  But I have no proof it is the case for all positive integer k.


5.2  Six terms:  x15+x25+x35 = x45+x55+x65


(For a database of solns <more than 5000> within a search radius of 17700, see Duncan Moore's results here.) 
After Euler’s work on,


ak + bk = ck + dk


for k = 3 or 4 and subsequent work by other authors, the next advance was the discovery of polynomial solns to,


ak + bk + ck = dk + ek + fk


for k = 5 or 6, with the first one for k = 5 by Sastry and Chowla in 1934 and later for k = 6 by Rao, Brudno, etc.  Whether,


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


for k = 7 or 8 will turn out to have polynomial solns as well remains to be seen since particular solns in the integers for both are now known, with more than forty for k = 7 since Ekl found the first one in 1996 and only one so far for k = 8 found by Kuosa in 2006. 

(Update: 6/20/09Special cases:


a) x15+ x25+ x35+ x45+ x55 = 0


This is a counter-example to Euler’s sum of powers conjecture and, for fifth powers, there are only three known so far:


(1967)   [27, 84, 110, 133, -144] = 0                   (by L. Lander, T. Parkin)

(1996)   [5027, 6237, 14068, -220, -14132] = 0  (by R. Scher, E. Seidl)

(2004)   [55, 3183, 28969, 85282, -85359] = 0    (by J. Frye)


Whether this is reducible to solving an elliptic curve like for the fourth power version x14+ x24+ x34 = x44 still remains to be seen.


b) x15+ x25+ x35+ x45+ x55 = 1


Just like three 3rd powers of signed integers can sum to 1 (a famous example of which involves the taxicab number 1729 = 13 + 123 = 93 + 103), five 5th powers can also do so.  Only eight have been found so far.  The smallest was by Lander, Parkin, and Selfridge, while three others were later given by Seiji Tomita, which were independently found by Duncan Moore who also gave four more:


[1, 89, 118] = [38, 47, 123]

[1, 127, 430] = [16, 310, 412]

[1, 328, 709] = [5, 388, 705]

[1, 588, 772] = [59, 511, 791]
[1, 561, 1151] = [401, 616, 1146]
[1, 1073, 2297] = [379, 686, 2306]
[1, 4167, 4283] = [1039, 2601, 4811]
[1, 4823, 6377] = [1089, 3501, 6611]
Note 1:  Moore searched within a radius of 17700 and these are the only solns which have a unit term.  However, this also includes one with a zero term, found by Bob Scher and Ed Seidl in 1997,
[0, 220, 14132] = [5027, 6237, 14068]
though another one of this form is unknown.  Presumably, the greater the radius, the greater the chances of finding another one.
Note 2:  The sum and differences of 5 sixth powers can sum to unity and, within a radius of 17800, Moore found five,
[1, 132, 133] = [71, 92, 147]
[1, 173, 294] = [75, 154, 295]
[1, 55, 330] = [159, 268, 311]
[1, 500, 515] = [197, 409, 556]
[1, 1123, 1143] = [573, 815, 1255]
One can make a conjecture that, if the radius is great enough, there may be a zero term as well.  (Though it is odd there are no more solns with a unit term in the range 2000-17000.) 
Note 3:  Seven 7th powers can also sum to 1, though only two are known so far, and whether they can also sum to zero like for 5th powers is probable but an example is still unknown.  See here.) 
Tomita gave a complete table of [x1, x2, x3] = [x4, x5, x6] for positive xi < 1000, and Moore made a bigger table with sum almost 18,0005.  When all terms are integers with one equal to unity, it is unknown if a parametrization can exist, unlike for third powers where there is an infinite family which recent work by this author has shown involves a Pell equation.  No soln is yet known for 1+ x25+ x35+ x45+x55= z5 with positive xi, as seen in James Waldby's database with z < 10,000.  (For seven 5th powers equal to unity, there are formulas, also dependent on Pell eqns, to be given later.) 


c) x15+ x25+ x35+ x45+ x55 = y15+ y25+ y35+ y45+ y55 = z5


The smallest number whose third power is expressible as the sum of three positive third powers in two ways is 41 as,


[2, 17, 40] = [6, 32, 33] = [41]


For fourth powers as four fourth powers in two ways, Jarek Wroblewski found it to be 31127 (see Wroblewski's database here),


[2260, 4870, 17386, 30335] = [2495, 11998, 16430, 30320] = [31127]


The first fifth power expressible as the sum of five positive fifth powers in two ways was found by Waldby as 744,


[14, 95, 545, 586, 644] = [100, 210, 414, 629, 651] = [744]


(Note:  Incidentally, the number 744 figures prominently in the j-function j(τ) since this is the constant term of its series expansion,


j(τ) = 1/q + 744 + 196884q + 21493760q2 + …


which explains why Ramanujan’s constant (and others) exceeds a cube by nearly this amount, eπ√163  6403203 + 744, though this property of 7445 is probably unrelated.)


d) a5+2b5+2c5 = d5


There are solns to a4+2b4+2c4 = d4, the smallest of which is the Pythagorean-like:  [2, 2, 3, 4, 4] = [5].  There is also one for fifth powers, the only one so far, given in Waldby’s database as:  [526, 526, 1349, 1349, 1355] = [1685]


e) x1k+x2k+x3k = x4k+x5k+x6k


The smallest soln turns out to be good for both k = 1,5:  [24, 28, 67] = [3, 54, 62]. 
For sixth powers, the smallest is good for both k = 2,6:  [3, 19, 22] = [10, 15, 23].


In fact, out of the 198 solns in Tomita's database for [5.3.3] cited above, which is complete for terms < 1000, almost 2/3 (66%) are good for k = 1,5.  It should be interesting to check the percentage of [6.3.3] with terms below a bound that is good for k = 2,6.  Why they suddenly become good for two powers is food for thought, but may simply be due to the increasing number of terms and the symmetric nature of the equation.  For higher powers also with a minimal number of terms, there are solns valid for three powers, like,


[344, 902, 1112, 1555] = [479, 662, 1237, 1535],  for k = 1,3,7


and others found by A. J. Choudhry whenever a certain multi-variable cubic has a rational root.  And also,


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


found by Wroblewski and the only one so far.  There is only one known for [8.4.4] found by Kuosa, but I’m betting one will eventually be found good for k = 2,8 or, if signed, even good for k = 1,2,8, like some k = 1,2,6.  (End note)
(Update, 11/3/09):  f) x1k+x2k+x3k = x4k+x5k+x6k,   with x1+x2+x3 = x4+x5+x6 = 0
At the bottom of this webpage, I asked if this system for k = 1,5 has non-trivial solns since the analogous version for 7th powers with eight terms and side-condition x1+x2+x3+x4 = x5+x6+x7+x8 = 0 is solvable, good for k = 1,3,7 and reduced by Choudhry to finding a rational root of a multi-variable cubic.  Wroblewski checked his database from 2002 and found four, namely:
[105, 1153, -1258] = [455, 582, -1037]
[31, 47242, -47273] = [5681, 9717, -15398]
[3914, 51858, - 55772] = [19264, 25403, -44667]
[12292, 59070, -71362] = [27745, 37983, -65728] 
The system, in fact, is also reducible to finding a non-trivial rational root of a cubic.  Use the form L1,

(a+bp+q)k + (b-bp+q)k + (c+ap+q)k = (a+cp+q)k + (b+ap+q)k + (c-cp+q)k


which is already true for k = 1, a general form to be proven later.  Let p = q/a+1, and this will equate x1+x2 = x5.  Expanding for k = 5, and after removing trivial factors, will result in a cubic in q with coefficients in {a,b,c}.  (End update)
(Update 6/26/09): In response to an email, Tomita was kind enough to create a complete [6.3.3] table with terms < 1000.  True enough, of the 59 solns, almost 3/4 (75%) were good for k = 2,6, a greater percentage than for k = 1,5.  For the [7.4.4]  in Wroblewski's database (which is not a complete list below a certain bound), and excluding 18 solns to Choudhry's cubic with is tailored to yield k = 1,3,7, there are 40 remaining and almost 1/4 are for k = 1,7.  If a complete list with terms < 10,000 can be given, I believe the percentage will be greater.  But why there is this tendency to be multigrade is unknown.  (End note)
(Update 7/17/09): While browsing the Net, I came across (again) Tom Womack's page. I had read his dissertation years ago and, re-reading it, realized he observed the same point about k = 2,6 being abundant (and Peter Montgomery as well as others before him).  In fact, Womack had made a search and found that, out of 207 solns with common sum < 57176, almost 90% were for k = 2,6 (only 22 were just for k = 6), a much higher percentage than the one given by Tomita's smaller soln set.  How high will this percentage get for a complete list with a much higher common sum?  And what does this imply for eight positive terms valid for k = 1,7 or k = 2,8?  (Considering the only one known so far for k = 8 is not multigrade.)  (End note.)
(Update 10/7/09):  Duncan Moore, who has done extensive work on taxicab and cabtaxi numbers, has a bigger databasewhich includes [5.3.3] and [6.3.3].  Below a certain bound, for 5th powers, he found almost 5400 solns, about 59% of which are multigrade for k = 1,5.  For 6th powers, there were 405, with 92% multigrade for k = 2,6, an increase from Womack's database.  Whether these percentages asymptotically approach a definite value is unknown.  (End update.)
Various identities for 5th powers have been found, after transposition to symmetric form where {xi, yi} are not necessarily positive, that belong to four classes:
1. x15+x25+x35 = y15+y25+y35,  x1+x2 = y1+y2,
2. x15+x25+x35 = y15+y25+y35,  (Moessner's, with unknown linear relations between the xi, yi)
3. x1k+x2k+x3k = y1k+y2k+y3k,  k = 1,5
4. x1k+x2k+x3k = y1k+y2k+y3k,  x1-x2 = y1-y2,  k = 1,5
If you know of another, pls send it so it can be added to the collection.  The first two classes have beautiful identities.


Sastry, Chowla  (1934)


(u5+25v5)5 + (u5-25v5)5 + (10u3v2)5  = (u5+75v5)5 + (u5-75v5)5 + (-50uv4)5


If {u,v} are chosen such that Abs[u/v] = x is within the range 251/5 < x < 751/5, or roughly 1.904 < x < 2.371, then it yields a (5.1.5) with all terms positive. Outside this range, it becomes a (5.2.4), but there is no {u,v} such that it gives a (5.3.3) with all positive terms.  This has the basic form,

(u5+av5)5 + (u5-av5)5 - (u5+bv5)5 - (u5-bv5)5 = 20(a2-b2)(u3v2)5 + 10(a4-b4)(uv4)5


If rational {a,b} can be found such that,


20(a2-b2) = mp5       (eq.1)

10(a4-b4) = nq5        (eq.2)


then it yields an identity for (5,4,m+n).  For m = n = 1, Sastry and Chowla found {a,b} = {75, 25} and {p,q} ={10, 50}.  Q:  Are there others not simply fifth power multiples of these two, or other small m,n?


(Update, 11/18/09):   Dave Rusin showed that the system can reduced to a certain hyperelliptic curve, thus has only finitely many rational solns, and probably the known soln is the only non-trivial one.  He gave the transformation,


{a,b,p,q} = {m2y/(5n),  m3n,  -2mx,  20m2x},  where m = (4x5-5)/(5n2)


and eq.1, eq.2 are true if,


-42x10+ 52 = y2 


the rational soln of which apparently is only {x,y} = {1,3}.  It is hard not to notice this is essentially the smallest Pythagorean triple, -42+52 = 32, and it is interesting that it appears in the context of 5th powers.  (End update.)


(Update, 12/15/09):   Tony Rizzo pointed out that if {a,b} = {75r3, 25r3}, then {m,n} ={r, r2}.  For r = 2, this gives the (5,4,6) identity,


(u5+600v5)5 + (u5-600v5)5 - (u5+200v5)5 - (u5-200v5)5 = 2(2u3v2)5 + 4(4uv4)5


If {u,v} are chosen such that Abs[u/v] = x is within the range 2001/5 < x < 6001/5, then it yields a (5.1.9) with all terms positive.  (End update)



Update, 11/23/09):   


Moessner (1951)


 t6(a+20t)5 + t3(b-20t2-20t4)5 + (c+20t5)5 = t6(a-20t)5 + t3(b+20t2+20t4)5 + (c-20t5)5


where {a,b,c} = {t6+8t4+12t2-1,  t6+8t4-8t2-1,  t6-124-8t2-1}.


Since t is arbitrary, one can set t = v5 to have equal sums of 5th powers.  Note the nice symmetry of terms and how the coefficients, after negation, are palindromic.  Surely this is not an isolated result.  Anyone else knows of a similarly simple identity, as well as linear relations among the xi, yi, if any, for this?  (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 LHS and RHS are 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, 11/24/09):   Choudhry (1999) has considered the general eqn,


ax5+by5+cz5 = au5+bv5+cw5     (eq.1)


for the case a+b+c = 0.  Excluding the situation when abc(a-b)(a-c)(b-c) = 0, he gave the soln to,


ax5+by5+cz5 = t2 (au5+bv5+cw5)




{x,y,z} = { (p1t2-p1t+q1)r1,  (p2t2-p2t+q2)r2,  (p3t2-p3t+q3)r3 }  

{u,v,w}= { (q1t2-p1t+p1)r1,  (q2t2-p2t+p2)r2,  (q3t2-p3t+p3)r3 }




{p1, p2, p3} = {17a2+14ab+14b2,  14a2+14ab+17b2,  17a2+20ab+17b2}

{q1, q2, q3} = {2a2-ab-b2,  -a2-ab+2b2,  2a2+5ab+2b2}

{r1,  r2, r3} = {a+2b,  -2a-b,  a-b}


given constant {a,b}, for arbitrary t.  One can then set t = v5 to satisfy eq.1.  As an example, for {a,b,c} = {1,2,-3}, after transposing terms, we get a soln to,


x5+2y5+3z5 = u5+2v5+3w5


{x,y,z} = {101t10-101t5-4,  4t12-88t7+88t2,  25t10-25t5+4}

{u,v,w}= {-4t12-101t7+101,  88t10-88t5+4,  4t12-25t7+25t2}


(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.)


Lander would later find a three-parameter family which in general would involve 9th degree polynomials.  This author found that for special cases, it can be reduced to 7th or 8th deg.  Whether there is a 6th deg is unknown.  We have already seen that given an initial soln to a1k+a2k = b1k+b2k for either k = 2, 3, or 4, one can use this to find more solns.  It turns out there is a quintic version of this, albeit with a small condition,


Theorem (Lander):  "Given a rational solution to a1k+a2k+a3k = b1k+b2k+b3k for both k = 1,5, call this system S5, then this can generate subsequent ones."


In “Geometric Aspects of Diophantine equations involving Equal Sums of Like Powers”, Lander started out with the form,


(u+x)k + zk + (v+y)k = (v+x)k + yk + (u+z)k,   (call this L0)


a general soln for k = 1.  This can also be true for k = 5 using appropriate values.  (Note:  Incidentally, it should be pointed out that the eqn, x1k + x2k + …+ xk+1k = 0 simultaneously valid for three odd powers k = 1,3,n only has trivial solns when n = 5.  Using the form above and Mathematica’s Resultant[] function to eliminate one variable at k = 3 and 5, it will be seen that there are only trivial solns.  In contrast, there are non-trivial solns when k = 1,3,7 as found by Choudhry and k = 1,3,9 by Wroblewski.  Whether there is a k = 1,3,11 remains to be seen.)  

Anyway, L0 when expanded out for k = 5 has the nice symmetrical form,


(x-z)u4+2(x2-z2)u3+2(x3-z3)u2+(x4-z4)u = (x-y)v4+2(x2-y2)v3+2(x3-y3)v2+(x4-y4)v


where all fifth powers vanish and as such is a quartic surface.  After a clever geometric analysis, Lander gave a method to derive polynomial expressions for all the variables.  Given an initial soln {ai, bi} to L0, this leads to a new one, typically much larger, in two ways.  First,


1. Using point P1(x1,y1,z1)


Define {u,v,x,y,z} = {-a3+b3,  a2-b2,   x0+(x1-x0)t ,  y0+(y1-y0)t,  z0+(z1-z0)t}, 




x1 = y1 = z1 = (x0d1+y0d2+z0d3)/(d1+d2+d3)

{d1, d2, d3} = {a14-b14,  a24-b24,  a34-b34}


{x0, y0, z0} = {a1+a3-b3,  b2,  a3}


However, it still remains to find t.  When all these expressions are substituted into L0, if {ai, bi} is a soln, then the eqn is a quartic with four roots {0,0,1,t} so is just a linear eqn in t which is then easily solved.  One can also use a second approach,


2. Using point P2(x2,y2,z2)


Define {u,v,x,y,z} = {-a3+b3,  a2-b2,  x0+(x2-x0)t ,  y0+(y2-y0)t,  z0+(z2-z0)t}, 




{x2, y2, z2} = {x1-d0,  x1-d0+u,  x1-d0+v}

x1 = (x0d1+y0d2+z0d3)/(d1+d2+d3)


d0 = (d2u+d3v)/(d1+d2+d3)


and {d1, d2, d3}, {x0, y0, z0} are defined the same way as in the first point.  Using these, eq.0 again reduces to a linear eqn in t.  It is also possible to find a polynomial soln using a trivial initial one.  Let,


{a1, a2, a3, b1, b2, b3} = {p, -p, q, q, r, -r}

{a1, a2, a3, b1, b2, b3} = {p, -p, q, r, q, -r}


for the first and second points respectively and these will yield 9th degree polynomials. 


Theorem (Piezas):  "The complete soln to S5, or x1k+x2k+x3k = y1k+y2k+y3k, for k = 1,5 involves solving only a quadratic equation with a discriminant D that is quartic polynomial in one variable.  The problem of making D a square can then be treated as an elliptic curve."


Proof:  It can be shown that Lander’s result can be concisely encoded in the form L1,


(a+bp+q)k + (b-bp+q)k + (c+ap+q)k = (a+cp+q)k + (b+ap+q)k + (c-cp+q)k


which we’ve come across before. This is already true for k = 1.  It is a subset of Lander's L0,


(u+x)k + zk + (v+y)k = (v+x)k + yk + (u+z)k


excluding the case uv(u-v) = 0.  To see this, simply equate the first five terms of L1 with that of L0, and one can always find rational {a,b,c,p,q} using,


{a,b,c,p,q} = {(1+t)(st+y-z)/(2t),  -(1+t)(st-y+z)/(2t),  (1+t)(y+tu-tx+ty-z)/(2t),  t/(1+t),  (st-y+z+2tz)/(2t)},  where {s,t} = {u+x-y,  -(u-v)/u}
hence the cases with denominator (1+t)t = uv(u-v) = 0 should be excluded, otherwise there is division by zero.  Using these expressions, the sixth terms of L1 and L0 are also seen to be equal.  The fact that L1 is less general than L0 does not matter since for the exceptional case, L0 reduces to the four-term form,


n1k + (n2+n3)k = (n1+n2)k + n3k


which, after expanding, is easily seen to have trivial solns for k < 6 other than k = 1, hence is of no interest.  With that out of the way, by expanding L1 at k = 5, this reduces to a polynomial in p that is just a quadratic of form,


(Poly1)p2+(Poly2)p+(Poly3) = 0   (eq.1)


disregarding the trivial factor p(p-1).  The Polyi are in {a,b,c,q} and as polynomials in q are of degree 1,1,3, respectively.  (Unfortunately, they are tedious to explicitly write down.)  The discriminant D is,


D:= (Poly2)2-4(Poly1)(Poly3)


of deg 1+3 and so is a quartic in q, thus proving the theorem.  To find a polynomial soln, there are two ways:
1st method:  Since Poly1 is linear in q, all we have to do is to set Poly1 = 0 and solve for q, with the eqn given by,


2(b+c)(b2+c2)q = (a4-c4)-(a+b)(b+c)(b2+c2)


and (eq.1) reduces to the linear eqn (Poly2)p+(Poly3) = 0 where p is easily solved for.  Consistent with Lander’s result, this yields ninth deg polynomials in one of the variables {a,b,c}.  Incidentally, the sums of the terms as x1+x2, y1+y3 have the simple forms, 


(a+bp+q) + (b-bp+q) = (a4-c4)/((b+c)(b2+c2))

(a+cp+q) + (c-cp+q) = (a4-b4)/((b+c)(b2+c2))


for {p,q} as defined above.  This author found out that the degree of the polynomials can be reduced if we assume certain linear relationships between the terms.  It’s known that the system for k = 1,5 may have the side condition x1-x2 = y1-y2.  Let,


(a+bp+q) - (b-bp+q) = (a+cp+q) - (b+ap+q)  


which is true if c = a+2b.  Using this constraint, the polynomials reduce to seventh deg.  Since the eqn is homogeneous, it can be set b = 1 without loss of generality and we have the identity,


(at+p+q)k + (t-p+q)k + (ct+ap+q)k = (at+cp+q)k + (t+ap+q)k + (ct-cp+q)k,  k = 1,5,


c = a+2

t = 4a(a-2)(a+3)(a2+4a+5)2

p = (a2+1)(a2+8a+11)(a3+15a2+33a+31)

q = -2a(a+1)(a-2)(a2+4a+5)(a3+15a2+33a+31)


where the common factor (a+1) of the terms has to be removed.  (A. Moessner also found a seventh deg soln in 1951 but I do not know if this is the same since Lander gave the source as an Italian journal.)  Alternatively, if c = a-2b is used, they reduce to eighth deg though I haven’t been able to find a linear relationship between the terms.  Q: So there are 5th, 7th, 8th, and 9th deg solns. Any for 6th deg?  (Note:  As was already discussed, the form L1 can also be used for appropriate higher powers k.  Later, it will be used to prove that the system k = 1,2,6 can also be reduced, again, to merely solving a quadratic.)
(Update 7/13/092nd method:  We directly find a q such that the discriminant D is a square. To find a trivial soln, one begins by expanding L1 for k = 5,7 and eliminate p between them. Factoring the resultant, one finds a linear relationship between the {a,b,c,q} given by,
2(b+c)q = a2-b2-c2-ab-ac-bc
Using this q, D becomes the simple square,
D: = (a2-b2)2(a2-c2)2(b-c)2/(b+c)2
This is a trivial soln but, as was shown in the first method that D is a quartic in q, from this initial point, one can then find an infinite number of non-trivial ones. For simplicity, one can set c = a+2b and the next point yields terms for the xi, yi as 13th-deg polynomials after removing common factors.



The complete criterion to solve the system x1k+x2k+x3k = y1k+y2k+y3k for k = 1,5 has already been given above, with a subset also satisfying x1-x2 = y1-y2.   We can also explore other methods for the sake for the exercise.  This subset can be completely determined by either of two nice forms.  Let,


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


already true for k = 1 and if expanded for k = 5 reduces to the simple eqn,


(a4+b4-2c4) + 2(a2+b2-8c2)d2 - 6d4 = 0


One way to solve this is to treat it as a quadratic in d2 and make its discriminant (a quartic polynomial) a square, though is not easily done for this particular quartic. However, we can use the alternative form,


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


which, for convenience, after scaling can be assumed s = 1 without loss of generality.  By expanding this at k = 5, this is only a quadratic in r2.  Solving for r2,


r2 = -p2-pq-q2 ± y,


where y is the square root of the discriminant,


y2 = 4q4+(8+5p)q3+2(4+4p+3p2)q2+(4+8p+8p2+5p3)q+(1+2p+2p2)2        (eq.1)


This is more easily made a square being a polynomial with a square leading and constant term, and illustrates how using the appropriate form can simplify a problem.  This quartic is also rather special since its solns are plentiful enough that they can be parametricized linearly in three ways, namely {q1, q2, q3} = {-(80p+40)/(57p+80),  -(p+2),  -(3p+2)/3}, though the last two are trivial with respect to the original eqn.  However, we can always derive non-trivial points.  Three parametric solns will then be given, each of which involves a distinct elliptic curve.  Thus,


(p+r)k + (p-r)k + (2q+1)k = (q+r)k + (q-r)k + (2p+1)k,  k = 1,5


has solns,


q = – (80p+40)/(57p+80),  r = z/(57p+80),  402+10160p+22999p2+14478p3+3249p4 = z2


q = 3(13p2+10p+8)/(9p-16),  r = z/(9p-16),  82+656p+1999p2+2922p3+1629p4 = z2


q = (13p2+14p+8)/(3p+16),  r = z/(3p+16),   242+1232p+1183p2+542p3+181p4 = z2


where one must solve the elliptic curve in p.  The first q was found was found using Fermat’s method.  There are trivial p like {-1/2, -80/57, 0}, but these can lead to non-trivial ones, like p = -15/38, etc.  The next two q’s were found using Fermat’s method on the two trivial linear solns. These do not completely give all values that make the discriminant a square but it can be shown that the three q's as linear polynomials are the only ones of form q = (ap+b) or q = (ap+b)/(cp+d) for some rational {a,b,c,d}.  To show this, given eq.1, substitute q = (ap+b) into the eqn and assume the resulting polynomial to be a square,


u1p4+u2p3+u3p2+u4p+u5 = (v1p2+v2p+v3)2

Since the ui are in terms of {a,b}, then the unknowns are {a,b,v1,v2,v3}.  Expanding the above and collecting powers of p, one then has a system of five eqns in five unknowns, and the final eqn in this case has two rational solns.  A similar approach for q = (ap+b)/(cp+d) will give a single rational soln. 
By adding yet another constraint, one can also find a simple form for S5.  An example is,
(v+8)k + (u-8)k + (-u+2v+7)k = (-u+2v+8)k + (v-8)k + (u+7)k
This is already true for k = 1. Expanding for k = 5 yields only a quadratic in v and is easily solved.  For rational v, one must make its discriminant, a quartic polynomial in u, as a square,
y2 = -43281+40110u-4597u2+390u3-31u4
with one small soln as u = 3/2, giving [389, -208, 442] = [474, -123, 272].  From this initial point, one can then compute an infinite number of rational points. This, in fact, is the complete soln to the system, call this M1,
x1-x2 = y1-y2

x2-nx3 = -ny1+y3

x1k+x2k+x3k = y1k+y2k+y3k


for k = 1,5 and n = 15.  Other than the case n = 1 which is trivial, I haven’t checked yet for what other n this system is solvable. 
(Update 7/12/09): Another one obeying the system M1 but with n = 2 is,
(1-2q+2r)k + (-3-2q)k + (1+2q)k = (2+q+r)k + (-2+q-r)k + (-1-4q+2r)k
Again, it is already true for k = 1 and, when expanded for k = 5, yields a quadratic in r.  Its discriminant is a quartic in q and, if made a square is,
y2 = -(384+1112q+1297q2+738q3+171q4)
with a small rational point at q = -16/15.  Q:  Any other simple soln to system M1 for some integer n?  (End note.)

Note 1:  Incidentally, given the monic quartic of form,


y2 = x4+(Poly1)x3+(Poly2)x2+(Poly3)x+(Poly4)


to be made a square and where the Poly_i are polynomials in a variable p of degree 1,2,3,4 respectively (like eq.1 above), what is the maximum number possible of linear solns of form x = (ap+b) or x = (ap+b)/(cp+d)?


Note 2:  The complete criterion of x1k+x2k+x3k = y1k+y2k+x3k,  k = 1,5, where x1-x2 = y1-y2 has also been given by Bremner though I haven’t seen that work yet.  I will update this section to compare methods when I’ve read it. 


Note 3:  It can be proven that k = 1,2,5 yields only trivial rational results.  The complete soln to k = 1,2 is by L. Dickson, (Form 4 in Sums of Three Squares)


(ad+e)k + (bc+e)k + (ac+bd+e)k = (ac+e)k + (bd+e)k + (ad+bc+e)k


and expanding this at k = 5 gives merely a quadratic polynomial in e with discriminant D,


D = -2(a2-ab+b2)(c2-cd+d2)


which must be made a square.  But let {a,b,c,d} = {u+v, 2v, x+y, 2y} and it becomes,


D = -2(u2+3v2)(x2+3y2)


and it is obvious there are no real values such that D becomes a positive real square.  The same conclusion was arrived at by Choudhry in his paper, The system of simultaneous equations x1k+x2k+x3k = y1k+y2k+y3k, k = 1,2,5 has no non-trivial solns in integers, The Mathematics Student, Vol 70, 2001.  (It is not known, however, if the case k = 2,5 has a soln or not, though it is easily proven using Moore's database that there are none within a radius of 17000.)


Note 4:  Is a5+b5+c5 = d5+e5+f5 where a+b+c = d+e+f = 0 solvable?  (The analogous eqn for seventh powers does have solns.)  This can be completely parametricized as,


p5+q5+(-p-q)5 = r5+s5+(-r-s)5


which when expanded is,


pq(p+q)(p2+pq+q2) = rs(r+s)(r2+rs+s2)


though it seems to be unknown if this is non-trivially solvable.  (Update: Wroblewski has proven this has non-trivial solns, the smallest of which is {p,q,r,s} = {105, 1153, 455, 582}, yielding,

1055 + 11535 + (-105-1153)5 = 4555 + 5825 + (-455-582)5

though they are sparse in the range 10^5.  See update 11/03/09 above.)
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