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The identities

$1+2=3$
$4+5+6=7+8$
$3^2+4^2=5^2$
$10^2+11^2+12^2=13^2+14^2$
$21^2+22^2+23^2+24^2=25^2+26^2+27^2$

are examples of consecutive identities. Are there any consecutive identities except $3^3+4^3+5^3=6^3$ for exponents greater than $2?$

The identities should be built upon additions only.

For consecutive identities $m^k+\cdots+(m+n-1)^k=(m+n)^k+\cdots+(m+2n-2)^k$ there is a gap $\Delta^k$ between last number in one identity sequence to the first number in the next possible consecutive identity for k=1 and for k=2: $\Delta^1=1$ and $\Delta^2=2n+3,\;n=1,2,\dots$ :

$1+2=3$
$4+5+6=7+8$
$9+10+11+12=13+14+15$
$16+17+18+19+20=21+22+23+24$

$1=4-3=9-8=16-15 \dots$

$3^2+4^2=5^2$
$10^2+11^2+12^2=13^2+14^2$
$21^2+22^2+23^2+24^2=25^2+26^2+27^2$
$36^2+37^2+38^2+39^2+40^2=41^2+42^2+43^2+44^2$

$10-5=2\cdot 1+3$
$21-14=2\cdot 2+3$
$36-27=2\cdot 3+3$

I don't think there is a straight continuation for $k>2$, but there might be consecutive identities of a slightly different kind, e.g:

$(1)\quad m^k+\cdots+(m+n-1)^k=(m+n)^k+\cdots+(m+2n-i)^k$, where $i>2?$

Examination of $(1)$ for $k=3$ and $1\leq m,n,i\leq 5000$ only yield the solution $3^3+4^3+5^3=6^3$.

Lehs
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  • The counter-examples to Euler's conjecture are nice ones using fourth powers and powers with even larger exponents. – Peter Aug 28 '17 at 20:57
  • @Peter: which is that? – Lehs Aug 28 '17 at 20:59
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    Look here for some amazing equations : https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture – Peter Aug 28 '17 at 21:01
  • @Lehs ; may be this is interesting for you: https://math.stackexchange.com/questions/2408850/how-to-prove-that-12-3-456-78-ad-infinitum – Davood Aug 28 '17 at 21:17
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    Suppose that $S_k(n) = \sum_{i=1}^n i^k$ is the sum of all the $k$th powers up to $n$. (Note that there's an explicit polynomial expression for $S_k$ as a $(k+1)$-th degree polynomial in $n$). Then any sum of consecutive powers is of the form $S_k(n)-S_k(m)$; you're essentially asking about solutions of $S_k(n)-S_k(m) = S_k(p)-S_k(q)$ (or, equivalently, $S_k(n)+S_k(q) = S_k(m)+S_k(p)$. This should at least make searching (e.g. in the $k=3$ case) much easier; checking all cases with $m,n,p,q\lt 10^6$, say, should be comfortably within a day of CPU time. – Steven Stadnicki Aug 28 '17 at 21:30
  • @StevenStadnicki Also I'd add Lehs that $S_k(n)$ can be express explicitly in terms of Bernoulli polynomials, which there are a lot of algorithms to calculate or if you're using a CAS likely a built in function for. – Ethan Splaver Sep 22 '17 at 23:02

3 Answers3

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MORE EDIT. For cubes, it reduces to the problem of finding three triangular numbers whose squares are in arithmetic progression. Recalling that $$1^3+2^3+\cdots+r^3=\left({r(r+1)\over2}\right)^2=T_r^2$$ an equation of the form $$(m+1)^3+(m+2)^3+\cdots+(m+n)^3=(m+n+1)^3+(m+n+2)^3+\cdots+(m+n+k)^3$$ becomes $$T_{m+n}^2-T_m^2=T_{m+n+k}^2-T_{m+n}^2$$ which says the squares of $T_m$, $T_{m+n}$, and $T_{m+n+k}$ are in arithmetic progression.

E.g., we have $T_2=3$, $T_5=15$, $T_6=21$, and the squares $9,225,441$ are in arithmetic progression, and this corresponds to $3^3+4^3+5^3=6^3$. Whether there are any more examples, I do not know.

Everything below relates to a misunderstanding of the question, and can safely be ignored.

There are (infinitely) many examples of consecutive cubes summing to a cube. About 20 are listed here, starting with
$$11^3+12^3+13^3+14^3=20^3$$ $$3^3+\cdots22^3=40^3$$ $$15^3+\cdots+34^3=70^3$$ An answer here links to Dave Rusin's "known math" pages, but Rusin's site seems to have vanished from the web. Much of what Rusin had for the problem of sum of consecutive cubes a cube is reproduced here.

EDIT: Discussion of related questions at MathOverflow.

Gerry Myerson
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The formula for the sum of $n$ cubes of numbers in arithmetic progression $d$ is given by,

$$\begin{align}F(d,a,n) &=a^3+(a+d)^3+(a+2d)^3+\cdots+(a+dn-d)^3\\ &= (n/4)(2a-d+dn)(2a^2-2ad+2adn-d^2n+d^2n^2)\end{align}$$

Consecutive numbers is just the special case $d=1$ and we have the well-known,

$$3^3+4^3+5^3 = 6^3$$

A computer search within a reasonable radius doesn't yield other solutions, even if the RHS is a sum of cubes. But if we allow more general $d$, then for arithmetic progression $d=5$ we also have the nice,

$$47^3 + \color{blue}{52}^3 + 57^3 + \color{blue}{62}^3 + 67^3 + \color{blue}{72}^3 + 77^3 + \color{blue}{82}^3 = 87^3 + \color{blue}{92}^3 + 97^3$$

Tito Piezas III
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$$n^2+(n+1)^2+...+(n+k)^2=(n+k+1)^2+...+(n+2k)^2$$

Solutions are easy. $n=(2k+1)k$

individ
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