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

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  • $\begingroup$ The counter-examples to Euler's conjecture are nice ones using fourth powers and powers with even larger exponents. $\endgroup$
    – Peter
    Aug 28, 2017 at 20:57
  • $\begingroup$ @Peter: which is that? $\endgroup$
    – Lehs
    Aug 28, 2017 at 20:59
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    $\begingroup$ Look here for some amazing equations : en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture $\endgroup$
    – Peter
    Aug 28, 2017 at 21:01
  • $\begingroup$ @Lehs ; may be this is interesting for you: math.stackexchange.com/questions/2408850/… $\endgroup$
    – Davood
    Aug 28, 2017 at 21:17
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    $\begingroup$ 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. $\endgroup$ Aug 28, 2017 at 21:30

3 Answers 3

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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.

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    $\begingroup$ However those are not consequtive identities in the sense put forth in the question. $\endgroup$
    – skyking
    Aug 29, 2017 at 7:59
  • $\begingroup$ @sky, yes, I missed that the right side was meant to start where the left side ended. $\endgroup$ Aug 29, 2017 at 9:17
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    $\begingroup$ Here's a close one: $$4^3+5^3+\dots+28^3 = 30^3+31^3+\dots+34^3$$ $\endgroup$ Sep 24, 2017 at 16:22
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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,

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

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$$n^2+(n+1)^2+...+(n+k)^2=(n+k+1)^2+...+(n+2k)^2$$

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

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