TITLE: Identities like $4+5+6=7+8$ and $10^2+11^2+12^2=13^2+14^2$ and so on...
QUESTION [14 upvotes]: 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 $3^3+4^3+5^3=6^3$.

REPLY [3 votes]: 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,
$$47^3 + \color{blue}{52}^3 + 57^3 + \color{blue}{62}^3 + 67^3 + \color{blue}{72}^3 + 77^3 + \color{blue}{82}^3 = 87^3 + \color{blue}{92}^3 + 97^3$$