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

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.

• How are you defining lowest answer? The sum of the cubes of 9 through to 25 sum to 323 squared. Dec 13, 2020 at 16:18
• @OldPeter I believe it is about finding the smallest numbers whose squares are equal to the sum of the consecutive cubes. Dec 13, 2020 at 16:22
• So it’s 9 to 25? Dec 13, 2020 at 16:31
• @OldPeter no, I want the smallest numbers that are equal to the sum of the cubes. Dec 13, 2020 at 18:45
• At OEIS (A163393, A126200, A238099) there doesn't seem to be much after Dudeney excluding what you have already discovered independently. Dec 14, 2020 at 16:11

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.

$$$$A=x^2-y^2\implies y=\sqrt{x^2-A}\qquad\text{for}\qquad \lfloor\sqrt{A+1}\rfloor \le x \le \frac{A+1}{2}$$$$ The lower limit ensures $$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. $$$$f(5,4)=(9,40,41)\quad f(5,3)=(16,30,34)\\ f(13,12)=(25,312,313)\quad f(10,8)=(36,160,164)\\ f(25,24)=(49,1200,1201)\quad \mathbf{f(10,6)=(64,120,136)}\\ f(17,15)=(64,510,514)\quad f(15,12)=(81,360,369)\\ f(41,40)=(81,3280,3281)\quad f(26,24)=(100,1248,1252)\\ f(61,60)=(121,7320,7321)\quad f(13,5)=(144,130,194)\\ f(15,9)=(144,270,306)\quad f(20,16)=(144,640,656)\\ \mathbf{f(45,36)=(729,3240,3321)}\quad f(123,120)=(729,29520,29529)\\ f(365,364)=(729,265720,265721)\quad f(221,85)=(41616,37570,56066)\\ f(255,153)=(41616,78030,88434)\quad \mathbf{f(325,253)=(41616,164450,169634)}\\$$$$

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.

• Euclid's formula does not capture all Pythagorean triples unless we are allowed to scale them up by some factor. So it doesn't look like your answer proves that Dudeney's solutions are the minimal ones or how we can find all solutions. Could you clarify the main point of your response please? Perhaps I am missing something. Dec 20, 2020 at 1:02
• @Favst Euclid's formula generates all primitives, doubles, and square multiples of primatives without a multiplier. We are addressing the square-difference formula in the OP and the square or double-squared of some target side-A which can be any odd number $>3$. Since we are looking for cubes that sum to squares, finding these squares should be sufficient to find all candidates that match the OP square-difference criterion. Dec 20, 2020 at 1:25

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

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 $$n=2m^2$$

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]

• The result for $n=2m^2$ is already in arxiv.org/abs/1501.06098 Dec 20, 2020 at 2:29
• @Gerry Myerson Thank you for your information. Dec 20, 2020 at 3:19