Solving Dudeney's "A Question of Cubes": sum of consecutive cubes is a square
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.
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 $n=2m^2$
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]