## Sum of Consecutive Nth Powers Equals an Nth Power

```The famous "cannonball stacking" problem of Lucas (1875) requires
a sum of consecutive squares, beginning with 1, equal to a square.
The only non-trivial solution is

1^2 + 2^2 + 3^2 + ... + 24^2  =  70^2

However, as discussed in Laurent Beeckmans' article in the May 94
Mathematical Monthly, if we allow ANY set of k consecutive squares
(not necessarily beginning with 1) there are solutions for k = 1,
2, 11, 23, 24, 33, 47,etc.  For each of these we have infinitely
many sequences of k consecutive squares whose sum is a square.
For example, with k=24 we not only have the above sequence, we
also have

9^2 + 10^2 + 11^2 + ... + 32^2  =  106^2

20^2 + 21^2 + 22^2 + ... + 43^2  =  158^2

etc.

In general, the sum of the 24 squares beginning with m^2 is a
square for m = 1, 9, 20, 25, 44, 76, ...  These values can be
determined by solving the corresponding Pell equation, since the
sum of k consecutive squares beginning with m^2 is

(k/6)(6m^2 - 6m + 6mk + 1 - 3k + 2k^2)

Setting this equal to an arbitrary square N^2 gives, for any fixed
k, a Pell equation in m and N.  For example, with k=24 we have

24m^2 + 552m + 4324  =  N^2

Solving this for m gives
____________
-138 + / 6N^2 - 6900
m  =  --------------------
12

so there must be an integer q such that

6N^2 - q^2  =  6900

There are six infinite families of solutions [N,q], whose smallest
members are

[70,150]  [106,246]  [158,378]  [182,438]  [274,666]  [430,1050]

For any of these six fundamental solutions [N0,q0] the rest of the
corresponding infinite family is given by
_    _     _       _    _    _
|  Nj  |   |  5   2  |j |  N0  |
|      | = |         |  |      |
|_ qj _|   |_ 12  5 _|  |_ q0 _|

Equivalently, both the sequences Nj and qj satisfy the second order
recurrence s[j] = 10 s[j-1] - s[j-2].  For any q the corresponding
value of m is (q-138)/12.

If we go on to consider sums of higher powers, it appears that there
are no sums of two or more consecutive 4th powers equal to a 4th
power, or in general sums of two or more consecutive nth powers
equal to an nth power for any n>3.  Can anyone supply a proof,
reference, or counter-example?

This leaves the case of cubes, for which there are certainly some
solutions, but they don't reduce to simple Pell equations so they
aren't as easy to analyze as in the case of squares.  Dickson's
"History of the Theory of Numbers" discusses the more general problem
of finding a cube that is equal to the sum of the cubes of consecutive
elements of any arithmetical progression.  The special case of
consecutive cubes is discussed on page 582.  C. Pagliani (1830) treated
the problem of finding 1000 consecutive integers the sum of whose cubes
is a cube.  Notice that the sum of the cubes of  x+1, x+2,...,x+m  is
s = m(y+1)(y^2 + 2y + m^2)/8  where  y = 2x+m.  If we set m = 8n^3
then s is the cube of n(y+4n^2)  IF  y=0  OR  y satisfies the
equation
3 (4n^2 - 1) y  =  2 (32n^6 - 24n^4 + 1)

Letting v denote 2n, the above is equivalent to the statement that

(x+1)^3 + (x+2)^3 + ... + (x+v^3)^3   =   [ vx + v^3 (v+1)/2 ]^3

if 6x = (v^2 - 1)^2 - 3(v^3 + 1).  This implies that x is an integer
if v is not divisible by 3.  For example, the cases v = 2, 4, and 10
give
3^3 + 4^3 + 5^3 = 6^3

6^3 + 7^3 + ... + 69^3 = 180^3

1134^3 + ... + 2133^3 = 16830^3

So this gives an infinite family of solutions.  However, not all sums
of consecutive cubes equal to a cube are members of this family.  Note
that the sum of k consecutive cubes beginning with m^3 is

(k/4)(2m+k-1)(2m^2 - 2m + 2km - k + k^2)

Setting this quantity equal to a cube N^3, the first several
solutions in integers are

k    m     N
---  ---  ----
3    3     6
4   11    20
20    3    40
20   15    70
25    6    60
49  291  1155
64    6   180
99   11   330
99  556  2805
125   34   540
153  213  1581
153  646  3876
288  273  2856
343  213  2856
512  406  5544

It's easy to see that the solutions with k equal to a cube (such as
k=64, 125, 343, and 512 in the above list) are members of the infinite
family found by Pagliani.  But the remaining solutions are more
interesting.  Note that the values of k include the squares 4, 25,
49, and 64.  Also there are *pairs* of solutions for k=20, 99, 153,
and also with m=3, 11, 6, and 213.  It's also interesting that the
cube 2856^3 is expressible as a sum of consecutive cubes in two
distinct ways.

Dave Rusin approached the problem from the standpoint of elliptic
curves and classified several types of (possible) solutions:

1. The trivial solutions --  m=(1-k)/2 (so N=0) for odd  k, and
m=-k/2 or -k/2+1 for even  k
2. The torsion solutions -- m=(u^3-2*u^2-4*u-4)*(u-1)/6  when k=u^3
3. Other elements in the subgroup generated by 1. and 2. (Conjecturally
only k=4, m=11)
4. Other elements of rank-1 curves (None such? see Sect. VI)
5. Other generators of rank >1 curves (and linear combinations thereof)
as for k=3, 20, 25, 49, 99, 153, 288

He speculated that Types 3 and 4 may provide no more solutions, but
that Type 5 might be more likely to yield additional solutions.

The "torsion solutions" correspond to the infinite family of solutions
found by Pagliani.  Beyond these, it seems that even today we have
no way of finding solutions other than by searching, possibly with
guidance from the algebraic factorization of the sum.  Jim Buddenhagen
searched for solutions with k equal to a square, and found the cases
k = 119^2 and k = 251^2.

It was also pointed out by Dave Rusin that if we let A denote the
number of consecutive cubes and B denote the sum of the first and
last cubes, then we have A=k and B=2m+k-1, which can be substituted
into the basic equation

(k/4)(2m+k-1)(2m^2 - 2m + 2km - k + k^2) = N^2

and, multiplying both sides by 8, we have

A B (A^2 + B^2 - 1)  =  K^3                (1)

where K = 2N.  Note that this equation appears symmetrical in A and
B, but since A+B = 2(m+k)-1 is odd, it's clear that A and B have
opposite parity.  Rusin notes that the "torsion points" are given
by  A= u^3, B = 3( (u^2-1)/3 )^2, and goes on to classify the other
solutions in terms of the greatest common divisor of A and B.

We could expand on this a little by characterizing the solutions in
terms of the mutually coprime components of the three factors on the
left side of (1), where I'll stipulate that A is odd and B is even.
If we define

x = gcd( A, A^2+B^2-1 )
y = gcd( A, B )
z = gcd( B, A^2+B^2-1 )
g = gcd( A, B, A^2+B^2-1 )

then x,y,z are pairwise coprime, and we have pairwise coprime integers
a,b,c such that

A = axyg        B = byzg       (A^2 + B^2 - 1) = cxzg

If we substitute for A and B into the right hand relation, it's clear
that g must equal 1, and we have

(axy)^2 + (byz)^2 - 1  =  cxz                  (2)

Also, substituting into (1) gives

(abc)(xyz)^2 = K^3                      (3)

Letting G denote the greatest common divisor of abc and xyz, there must
be coprime integers u,v such that

abc = Gu^3         xyz = Gv^3                    (4)

and so K = Guv^2.  Following is a list of these parameters for all
the solutions with A,B less than 30000.  This includes two solutions
that haven't been mentioned before.

A     B      a    b      c       x    y    z       G    u    v
----  ----    ---  ---   -----    ---  ---  ---    ----  ---  ---
*    3     8      1   1        3      3   1    8        3    1   2
25     4      5   1       32      5   1    4       20    2   1
25    20      5   1      256      1   5    4       20    4   1
25    36      5   3       32      5   1   12       60    2   1
49    20      7   1       20      7   1   20      140    1   1
49   630      7   3    13310      1   7   30      210   11   1
*   75    64      5   8       81     15   1    8      120    3   1
99   120      3   1       55     11   3   40      165    1   2
99  1210      3  11    49130      3  11   10      330   17   1
*  125   192    125   8     2187      1   1   24       24   45   1
153   578      3  17    59582      3  17    2      102   31   1
153  1444      3  19      544     51   1   76     3876    2   1
*  343   768    343  16    14739      1   1   48       48  119   1
833   288      7   3       68    119   1   96     1428    1   2
* 1323   512      7  64     1331    189   1    8       56   22   3
* 1331  4800   1331  40   206763      1   1  120      120  451   1
* 2197  9408   2197  56   555579      1   1  168      168  741   1
* 3267  1000     11 125     4913    297   1    8       11   85   6
* 4913 27648   4913  32   912673      1   1  864       32 1649   3
6591  7200   2197  15   595508      1   3  160       60  689   2
*12675  2744     65 343   107811    195   1    8      195  231   2
13923 19942      3  13    14042    357  13  118   547638    1   1
14161 17408    119  32  2248091      7  17   32     3808  131   1
*21675  4096     85 512   238521    255   1    8     2040  172   1
?     ?
33124 70225     91  53      100    364   1 1325   482300    5   1
63001 85340    251   1 33094240      1 251  340    85340    5   1

The two new [A,B] solutions are [6591,7200] and [13923,19942].  The
second of these is interesting because it shows that [49,20] is not
the only solution with u=v=1.

The solutions with asterisks are those where either A or B is the
cube of some integer m, so they are members of the infinite family
of "torsion" solutions.  Over the range covered in this table these
torsion solutions all share the property that two of the three
parameters x,y,z are cubes, and the third is either (m^2 - 1) or
3(m^2 - 1), depending on whether or not v is divisible by 3.

It's interesting that for the entire list of solutions, including
the two larger solutions with A,B greater than 30000 found by Rusin
and Buddenhagen, the parameter v is always 1, 2, 3, or (in one case)
6.  What is the smallest solution with v not equal to one of these
four values?  Also, notice that the only solutions in this table with
v = 3 or 6 are members of the torsion set, so if we eliminate those,
all the remaining solutions have v = 1 or 2.  What is the smallest
non-torsion solution with v greater than 2?

Not surprisingly, the value of G is always greater than 1 in the above
table, because if we had G=1 then equations (4) would imply that each
of the numbers a,b,c and x,y,z is a cube, and so equation (2) would
have the form

[(a'x'y')^2]^3  +  [(b'y'z')^2]^3   =  [(c'a'z')]^3   +  1       (5)

where the primed variables are cube roots of the corresponding unprimed
variables.  This is a restricted case of the "near-miss" solution of
Fermat's equation for cubes, which is interesting because it's another
case where there is a known infinite family of solutions, as well as
a semingly plentiful supply of "sporadic" solutions.  In other words,
the equation

X^3 + Y^3  =  Z^3 + 1                           (6)

has infinitely many solutions because of identities such as

(1 + 9m^3)^3  +  (9m^4)^3  =  (9m^4 + 3m)^3  +  1            (7)

but there are other solutions as well, and it doesn't seem to be known
if all the solutions are members of some infinite family.

In any case we can certainly show that (5) has no solution of the form
(7), because that would require 1 + 9m^3 to be a square, which implies
an integer r such that

9m^3  =  r^2 - 1  =  (r+1)(r-1)

Note that r+1 and r-1 cannot both be divisible by 3, so one of them
is divisible by 9 and the other is coprime to 3.  Thus, if r is even
we can choose its sign to give integers s,t such that

r+1 = 9s^3          r-1 = t^3

Subtracting one from the other gives  2 = 9s^3 - t^3  which is
impossible (mod 9) where 0,+1,-1 are the only cubes.  On the other
hand, if r is odd then we know m is even, and so 2 divides one of
(r+1),(r-1) and 4 divides the other.  This gives two possibilities

r+1 = 18s^3  ,  r-1 = 4t^3
or
r+1 = 36s^3  ,  r-1 = 2t^3

In the first case we can substract the two equations to give
2=18s^3-4t^3, which is the same as 1 = 9s^3 - 2t^3, clearly
impossible (mod 9).  In the second case we have 2 = 36s^3 - 2t^3,
which is the same as 1 + t^3 = 18s^3.  This factors as

(1+t) (1-t+t^2)  =  18s^3

and each factor is divisible by 3 if and only if t=2 (mod 3), so we
have an integer q such that t=3q+2 and the above equation becomes

(q+1) (1 + 3q + 3q^2)  =  2 s^3

From this it's clear that q must be odd, so we have integer k such
that q = 2k-1, which gives

k(12k^2 - 6k + 1) = s^3

The factors are coprime so we have integers m,n such that

k = m^3           12k^2 - 6k + 1 = n^3

Solving the second of these for k gives
__________________            ___________
6 +- / 36 - 48(1 - n^3)       3 +- / 12n^3 - 3
k = -------------------------  =  ------------------
24                          12

To yield a rational result there must be an integer h such that

h^2 = 12n^3 - 3  =  3(4n^3 - 1)

which implies that 4n^3 - 1 must be of the form 3d^2 for some
integer d.  Thus we have

3d^2 + 1 = 4n^3

But notice that if we define the rational number f = (3d-1)/2 we
have
f^2 + f + 1 = (9/4)d^2 + (3/4) = 3n^3

Furthermore, we have the convenient identity

(2+f)^3 + (1-f)^3  =  9(1 + f + f^2)

so we can combine this with the previous relation to give

(2+f)^3  +  (1-f)^3  =  (3n)^3

This is Fermat's equation for cubes, which of course has no rational
solutions.  So we've shown that if G=1 in our original problem then
there must be an integer solutions of equation (6), but it can't be
of the form (7).  Therefore, it would have to be one of the other
(much less dense) parametric solutions, or a sporadic solution.
```