Solving Dudeney's "A Question of Cubes": sum of consecutive cubes is a square

Question

The following is a puzzle of Dudeney:

Professor Rackbrane pointed out one morning that the cubes of successive numbers, starting from $1,$ would sum to a square number... He stated that if you are forbidden to use the $1,$ the lowest answer is the cubes of $23,24,25,$ which together equal $204^2.$ He proposed to seek the next lowest number using more than three consecutibe cubes and as many more as you like excluding $1.$

There is an answer provided but a proof is not included:

The cubes of $14,15,$ up to $25$ inclusive (twelve in all) add up to... the square of $312.$ The next lowest answer is the five cubes of $25,26,27,28,$ and $29,$ which together equal $315^2.$

My query is more general: Classify all finite sets of consecutive positive integers, the sum of whose cubes is a square. Any idea how we can do this? If no one manages to answer this question, I will accept an answer that shows how Dudeney came to the minimal solutions.

Here are my thoughts on the matter so far: Clearly the sum of first $n$ positive cubes works. The sum of the cubes of $m+1,m+2,\ldots,n$ for positive $n$ and non-negative $m$ is $$(1^3+2^3+\cdots+n^3)-(1^3+2^3+\cdots+m^3) = \left[\frac{n(n+1)}{2}\right]^2-\left[\frac{m(m+1)}{2}\right]^2.$$ If this is equal to a square, it is equivalent to seeking all Pythagorean triples such that at least two of the elements of the triple are triangular numbers. At that point, I tried to use the classification of all (primitive) Pythagorean triples, but that went nowhere.

Answer 1

R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica 97: 295-307, 1995, is available at http://www.numdam.org/article/CM_1995__97_1-2_295_0.pdf The abstract reads,

In this paper estimates of linear forms in elliptic logarithms are applied to solve the problem of determining, for given $n\ge2$, all sets of $n$ consecutive cubes adding up to a perfect square. Use is made of a lower bound of linear forms in elliptic logarithms recently obtained by Sinnou David. Complete sets of solutions are provided for all $n$ between $2$ and $50$, and for $n = 98$.

The author proves that nontrivial solutions exist for all odd values of $n$, which at least shows that there are infinitely many solutions.

The largest numbers that show up in the Stroeker paper are the $49$ cubes starting with $117576^3$, which add up to $282298800^2$.

There is more up-to-date information at https://oeis.org/A253679

Answer 2

Euclid's formula $\quad A=x^2-y^2\quad B=2xy\quad C=x^2+y^2\quad$ is the one most commonly used for generating Pythagorean triples. Your expression

$$\left[\frac{n(n+1)}{2}\right]^2-\left[\frac{m(m+1)}{2}\right]^2$$ is the same as that for generating the A-component above. To find triples for a given side-A, we can solve the A-function for y and test x-values to see which, if any yield integers.

\begin{equation} A=x^2-y^2\implies y=\sqrt{x^2-A}\qquad\text{for}\qquad \lfloor\sqrt{A+1}\rfloor \le x \le \frac{A+1}{2} \end{equation} The lower limit ensures $y\in\mathbb{N}$ and the upper limit ensures $x>y$. $$A=81\implies \lfloor\sqrt{81+1}\rfloor=9\le x \le \frac{81+1}{2} =41\\\quad\land\quad x\in\{15,41\}\implies y \in\{12,40\} $$ $$f(15,12)=(81,360,369)\qquad \qquad f(41,40)=(81,3280,3281) $$

You can substitute any square number into the formula but the difficulty is finding x and y where both are the sum of integers $\big(\frac{n(n+1)}{2}\big)$ as shown inside your square brackets. E.g. \begin{equation} f(5,4)=(9,40,41)\quad f(5,3)=(16,30,34)\\ f(13,12)=(25,312,313)\quad f(10,8)=(36,160,164)\\ f(25,24)=(49,1200,1201)\quad \mathbf{f(10,6)=(64,120,136)}\\ f(17,15)=(64,510,514)\quad f(15,12)=(81,360,369)\\ f(41,40)=(81,3280,3281)\quad f(26,24)=(100,1248,1252)\\ f(61,60)=(121,7320,7321)\quad f(13,5)=(144,130,194)\\ f(15,9)=(144,270,306)\quad f(20,16)=(144,640,656)\\ \mathbf{f(45,36)=(729,3240,3321)}\quad f(123,120)=(729,29520,29529)\\ f(365,364)=(729,265720,265721)\quad f(221,85)=(41616,37570,56066)\\ f(255,153)=(41616,78030,88434)\quad \mathbf{f(325,253)=(41616,164450,169634)}\\ \end{equation}

Here, among these examples are $$f(10,6)=f\bigg(\frac{4(5)}{2},\frac{3(4)}{2}\bigg) \implies A=(10^2-6^2)=64=4^3=8^2 $$ and $$f(45,36)=f\bigg(\frac{9(10)}{2},\frac{8(9)}{2}\bigg) \implies A=(45^2-36^2)=729=9^3=27^2 $$ and $$f(325,253)=f\bigg(\frac{25(26)}{2},\frac{22(23)}{2}\bigg) \implies A=(325^2-254^2)=23^3+24^3+25^3=204^2 $$

Note that in these examples, the 2nd number of the y-product and the first number of x-product suggest, perhaps by coincidence, the range of cubes to be summed. E.g. $$4(5),3(4)\rightarrow 4^3\quad 9(10),8(9)\rightarrow 9^3\quad 25(26),22(23)\rightarrow 23^3+24^3+25^3$$

A search is needed if all squares are used but the square of any of the numbers in sequence A126200 will yield x,y values that meet the criteria above.

Answer 3

We show the solutions of $y^2 = x^3 + (x+1)^3 +...+ (x+n-1)^3$ using Stroeker's method.
We consider below diophantine equation.

$$y^2 = nx^3 + \frac{3}{2}n(n-1)x^2 + \frac{1}{2}n(n-1)(2n-1)x + \frac{1}{4}n^2(n-1)^2\tag{1}$$

$$\sum_{k=0}^{n-1} k = \frac{1}{2}n(n-1)$$ $$\sum_{k=0}^{n-1} k^{2} = \frac{1}{6}n(n-1)(2n-1)$$ $$\sum_{k=0}^{n-1} k^{3} = (\frac{1}{2}n(n-1))^2$$

Hence we get equation $(1)$.

Equation $(1)$ is birationally equivalent to an elliptic curve below.

$$Y^2 = X^3 + \frac{1}{4}n^2(n^2-1)X\tag{2}$$

$X=nx+\frac{1}{2}n(n-1), Y=ny$

Multiply both sides of the equation $(1)$ by $n^2$ and substitute $\frac{1}{2n}(2X-n^2+n)$ for $x$, then we get equation $(2)$.

$1$. Equation $(1)$ always has the integer solutions for odd values of $n$

According to Stroeker, we know equation $(2)$ has a solution $(X,Y) = ( n^2(n-1)(n+1), \frac{1}{2}n^2(n-1)(n+1)(2n^2-1) ).$

Hence we obtain $(x,y)=( \frac{1}{2}(n-1)(2n^2+2n-1), \frac{1}{2}n(n-1)(n+1)(2n^2-1) ).$

Thus $(x,y)$ is always the integer for odd values of $n$.

We show how this solution is derived from group law.

First, equation $(2)$ has a solution $(X,Y)=( 0,0 )$.
Substitute $X=\frac{1}{2}n(n+1)$ to equation $(2)$, then we get $Y=\frac{1}{2}n^2(n+1)$.
Hence we know equation $(2)$ has a solution $(X,Y)=( \frac{1}{2}n(n+1), \frac{1}{2}n^2(n+1) )$.
Let $P1(X,Y)=( 0,0 )$ and $P2(X,Y)=( \frac{1}{2}n(n+1), \frac{1}{2}n^2(n+1) )$ and computing $P1-2P2$ using group law.
We get a solution $P1-2P2=( n^2(n-1)(n+1), \frac{1}{2}n^2(n-1)(n+1)(2n^2-1) )$.

$2$. Equation $(1)$ has infinitely many integer solutions for even values of $n=2m^2$

Vladimir Pletser had already got this result.
We rediscovered it by brute force search as follows.
As a candidate of solution, $(X,Y^2)=(\frac{1}{2}n^3-\frac{1}{2}n, \frac{1}{8}n^5(n-1)^2(n+1)^2)$ was found.
Hence $\frac{1}{8}n^5(n-1)^2(n+1)^2)$ needs to be a square number.

Let $n=2m^2$, then equation $(2)$ has a solution $(X,Y) = (4m^6-m^2, 2m^5(2m^2-1)(2m^2+1))$.

Hence we obtain $(x,y)=( m^2(2m^2-1), m^3(2m^2-1)(2m^2+1) )$.
$m$ is arbitrary.

           m<20
           [  n       x         y ]

           [  8      28        504]
           [ 18     153       8721]
           [ 32     496      65472]
           [ 50    1225     312375]
           [ 72    2556    1119528]
           [ 98    4753    3293829]
           [128    8128    8388096]
           [162   13041   19131147]
           [200   19900   39999000]
           [242   29161   77947353]
           [288   41328  143325504]
           [338   56953  250991871]
           [392   76636  421651272]
           [450  101025  683434125]
           [512  130816 1073737728]
           [578  166753 1641349779]
           [648  209628 2448874296]
           [722  260281 3575480097]

$3$. Numerical solutions

We extened the search range n to 99.
Search results of the integer points for equation $(1)$ using Online Magma Calculator.

           [  n         x          y ]

           [  3         23        204]
           [  5         25        315]
           [  5         96       2170]
           [  5        118       2940]
           [  7        333      16296]
           [  8         28        504]
           [  9        716      57960]
           [ 11       1315     159060]
           [ 12         14        312]
           [ 13        144       6630]
           [ 13       2178     368004]
           [ 15         25        720]
           [ 15       3353     754320]
           [ 15      57960   54052635]
           [ 17          9        323]
           [ 17        120       5984]
           [ 17       4888    1412496]
           [ 18        153       8721]
           [ 18        680      76653]
           [ 19       6831    2465820]
           [ 21         14        588]
           [ 21        144       8778]
           [ 21       9230    4070220]
           [ 23      12133    6418104]
           [ 25      15588    9742200]
           [ 27      19643   14319396]
           [ 28         81       4914]
           [ 29      24346   20474580]
           [ 31      29745   28584480]
           [ 32         69       4472]
           [ 32        133      10296]
           [ 32        496      65472]
           [ 33         33       2079]
           [ 33      35888   39081504]
           [ 35        225      22330]
           [ 35      42823   52457580]
           [ 37      50598   69267996]
           [ 39        111       9360]
           [ 39      59261   90135240]
           [ 40       3276    1196520]
           [ 41      68860  115752840]
           [ 42         64       5187]
           [ 43      79443  146889204]
           [ 45        176      18810]
           [ 45      91058  184391460]
           [ 47     103753  229189296]
           [ 48         64       5880]
           [ 48        410      62628]
           [ 48      19881   19455744]
           [ 48      60040  101985072]
           [ 49     117576  282298800]
           [ 50       1225     312375]
           [ 51     132575  344826300]
           [ 53     148798  417972204]
           [ 54        265      36729]
           [ 54       1272     343917]
           [ 55     166293  503034840]
           [ 57       1625     507471]
           [ 57     185108  601414296]
           [ 59     205291  714616260]
           [ 60        118      14160]
           [ 61     226890  844255860]
           [ 63        217      31248]
           [ 63        837     203112]
           [ 63       1121     310464]
           [ 63     249953  992061504]
           [ 64        105      13104]
           [ 64      34272   50827392]
           [ 65     274528 1159878720]
           [ 67     300663 1349673996]
           [ 69         81      10695]
           [ 69     328406 1563538620]
           [ 71     357805 1803692520]
           [ 72       2556    1119528]
           [ 73      16864   18771220]
           [ 73      26937   37849113]
           [ 73     388908 2072488104]
           [ 75     421763 2372414100]
           [ 76        295      53200]
           [ 77        144      22022]
           [ 77     456418 2706099396]
           [ 79     492921 3076316880]
           [ 81     531320 3485987280]
           [ 82        144      23247]
           [ 83     571663 3938183004]
           [ 85     613998 4436131980]
           [ 87        232      43065]
           [ 87     658373 4983221496]
           [ 89     704836 5583002040]
           [ 91       4785    3202290]
           [ 91     753435 6239191140]
           [ 92       4992    3429530]
           [ 93     804218 6955677204]
           [ 94        400      91979]
           [ 95     857233 7736523360]
           [ 97      23668   35970704]
           [ 97      33660   60951696]
           [ 97     912528 8585971296]
           [ 98         25       7497]
           [ 98         97      18333]
           [ 98        216      43309]
           [ 98        745     221697]
           [ 98        760     227997]
           [ 98       3961    2513511]
           [ 98       4753    3293829]
           [ 99     970151 9508445100]