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It is quite well-known that,

$$1^2+2^2+\dots+24^2 = 70^2$$

Not so well-known is,

$$15^3+16^3+\dots+34^3 = 70^3$$

The formula for the sum of $m$ consecutive squares starting with $a^2$ is,

$$F(a,m) = (m/6)(6a^2-6a+6am+1-3m+2m^2)$$

while the sum of $n$ consecutive cubes starting with $b^3$ is,

$$C(b,n) = (n/4)(2b+n-1)(2b^2-2b+2bn-n+n^2)$$

Question: Is the only solution in positive integers to the simultaneous equations,

$$F(a,m) = x^2$$

$$C(b,n) = x^3$$

given by $(a,b;m,n;x) = (1,\,15;\,24,\,20;\,70)$? (I have searched within a relatively small range, but didn't find any new solution.)

Update (Jan 2025). For those curious, there are an infinite number of solutions to,

$$C(b,n) = x^3$$

when $n$ is a cube not divisible by 3. Hence,

$$n = v^3,\quad b =\frac{(v^3 - 2v^2 - 4v - 4)(v - 1)}{6}$$

mentioned by Dave Rusin in 1997 (though in Dickson's book it seems A. Martin found a version in 1871) and explaining the cubes found by Oleg567 below.

P.S. The largest non-cube $n$ so far is $n = 11859210 = 10\times1089^2$ found in 2015 by Vladimir Pletser,

$$\sum_{k=0}^{11859210-1}(21709458+k)^3=6398174475^3$$

though there is no explanation how it was found.

Rosie F
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Tito Piezas III
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    Maybe check out http://www.math.niu.edu/~rusin/known-math/97/cube.sum – Ian Coley Oct 10 '13 at 04:05
  • @IanColey: Yes, I am familiar with that site. – Tito Piezas III Oct 10 '13 at 04:11
  • That site no longer exists. – Akiva Weinberger Jan 26 '25 at 02:30
  • @AkivaWeinberger Fortunately, the Internet Archive saved a copy about 20 years ago, https://web.archive.org/web/20050405011153/http://www.math.niu.edu/~rusin/known-math/97/cube.sum I wonder where's Dave Rusin now. – Tito Piezas III Jan 26 '25 at 02:37
  • @AkivaWeinberger Rusin's analysis though needs an update. He states, "...there is some evidence here to suggest that Brown's 1997 list (as amended) could be complete." Unfortunately, Brown's list stops at just $k =512=8^3$ cubes. Oleg567's 2013 list below goes up to $k= 63001 = 251^2$ cubes. Rusin did mention, "Other generators of rank >1 curves (and linear combinations thereof) as for k=3, 20, 25, 49, 99, 153, 288". The highest k now known is $k = 11859210=10\times1089^2$. How that was found, I have no idea, but there is probably an infinite number of *non-cube* $k$. – Tito Piezas III Jan 26 '25 at 02:59
  • It might or might not be relevant that $k=11859210=2\cdot3^4\cdot5\cdot11^4$ and $k+1=7\cdot13\cdot19^4$ are the largest consecutive integers which are both 19-smooth. – Rosie F Jan 26 '25 at 12:49
  • @RosieF Oh, I totally forgot that Pletser's $k=11859210=10\cdot 33^4$ as I was inspecting their factorizations earlier. Their being 19-smooth is certainly interesting, but I also don't know how to utilize that fact, nor how Pletser found that huge $k$. – Tito Piezas III Jan 26 '25 at 12:56

1 Answers1

10

Just as a wide comment.

My search limitation was $x<2 \times 10^6$.

There is only $x=70$ in this range.

The set of such $x$, which can be written as sum of consequent cubes, is rather pure: here is list:

\begin{array}{rlr} 6^3 = & 3^3+4^3+5^3 & (3 ~terms) \\ 20^3 = & 11^3 + ... + 14^3 & (4= \color{darkviolet}{2^2}~terms) \\ 40^3 = & 3^3 + ... + 22^3 & (20 ~terms) \\ 60^3 = & 6^3 + ... + 30^3 & (25 = \color{darkviolet}{5^2}~terms) \\ 70^3 = & 15^3 + ... + 34^3 & (20 ~terms) \\ 180^3 = & 6^3 + ... + 69^3 & (64 = \color{darkviolet}{8^2}= \color{red}{4^3} ~terms) \\ 330^3 = & 11^3 + ... + 109^3 & (99 ~terms) \\ 540^3 = & 34^3 + ... + 158^3 & (125 = \color{red}{5^3} ~terms) \\ 1155^3 = & 291^3 + ... + 339^3 & (49 = \color{darkviolet}{7^2} ~terms) \\ 1581^3 = & 213^3 + ... + 365^3 & (153 ~terms) \\ 2805^3 = & 556^3 + ... + 654^3 & (99 ~terms) \\ 2856^3 = & 213^3 + ... + 555^3 & (343 = \color{red}{7^3} ~terms) \\ 2856^3 = & 273^3 + ... + 560^3 & (288 ~terms) \\ 3876^3 = & 646^3 + ... + 798^3 & (153 ~terms) \\ 5544^3 = & 406^3 + ... + 917^3 & (512 = \color{red}{8^3} ~terms) \\ 16830^3 = & 1134^3 + ... + 2133^3 & (1000 = \color{red}{10^3}~terms) \\ 27060^3 = & 1735^3 + ... + 3065^3 & (1331 = \color{red}{11^3}~terms) \\ 62244^3 = & 3606^3 + ... + 5802^3 & (2197 = \color{red}{13^3}~terms) \\ 82680^3 = & 305^3 + ... + 6895^3 & (6591 ~terms) \\ 90090^3 = & 4966^3 + ... + 7709^3 & (2744 = \color{red}{14^3}~terms) \\ 175440^3 = & 8790^3 + ... + 12885^3 & (4096 = \color{darkviolet}{64^2} = \color{red}{16^3}~terms) \\ 237456^3 = & 11368^3 + ... + 16280^3 & (4913 = \color{red}{17^3}~terms) \\ 249424^3 = & 1624^3 + ... + 15784^3 & (14161 = \color{darkviolet}{119^2}~terms) \\ 273819^3 = & 3010^3 + ... + 16932^3 & (13923 ~terms) \\ 413820^3 = & 18171^3 + ... + 25029^3 & (6859 = \color{red}{19^3}~terms) \\ 431548^3 = & 34228^3 + ... + 36076^3 & (1849 = \color{darkviolet}{43^2}~terms) \\ 534660^3 = & 22534^3 + ... + 30533^3 & (8000 = \color{red}{20^3}~terms) \\ 860706^3 = & 33558^3 + ... + 44205^3 & (10648 = \color{red}{22^3}~terms) \\ 1074744^3 = & 40381^3 + ... + 52547^3 & (12167 = \color{red}{23^3} ~terms) \\ 1205750^3 = & 18551^3 + ... + 51674^3 & (33124 = \color{darkviolet}{182^2}~terms) \\ 1306620^3 = & 4880^3 + ... + 54655^3 & (49776 ~terms) \\ 1630200^3 = & 57084^3 + ... + 72708^3 & (15625 = \color{darkviolet}{125^2} = \color{red}{25^3} ~terms) \\ 1764070^3 = & 46690^3 + ... + 71890^3 & (25201 ~terms) \\ 1962820^3 = & 11170^3 + ... + 74170^3 &(63001 = \color{darkviolet}{251^2}~terms) \\ 1983150^3 = & 67150^3 + ... + 84725^3 & (17576 = \color{red}{26^3} ~terms) \\ \end{array}

And a few curious equalities of this search:

Power $3$:

$$\sum_{k=0}^{343-1}(213+k)^3=\sum_{k=0}^{288-1}(273+k)^3 = \large{2856^3}$$

Power $2$:

$$\sum_{k=0}^{32}(7+k)^2=\sum_{k=0}^{10}(38+k)^2 = {\large{143^2}}\quad$$

$$\sum_{k=0}^{73}(294+k)^2=\sum_{k=0}^{10}(854+k)^2 = {\large{2849^2}}\quad$$

$$\sum_{k=0}^{55^2-1}(2175+k)^2=\sum_{k=0}^{49}(29447+k)^2 = {\large{208395^2}}$$

$$\sum_{k=0}^{15872}(9401+k)^2=\sum_{k=0}^{5977}(26181+k)^2 = {\large{2259257^2}}$$

Tito Piezas III
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Oleg567
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