# numbers' pattern

It is known that

$$\begin{array}{ccc}1+2&=&3 \\ 4+5+6 &=& 7+8 \\ 9+10+11+12 &=& 13+14+15 \\\ 16+17+18+19+20 &=& 21+22+23+24 \\\ 25+26+27+28+29+30 &=& 31+32+33+34+35 \\\ldots&=&\ldots \end{array}$$

There is something similar for square numbers:

$$\begin{array}{ccc}3^2+4^2&=&5^2 \\ 10^2+11^2+12^2 &=& 13^2+14^2 \\ 21^2+22^2+23^2+24^2 &=& 25^2+26^2+27^2 \\ \ldots&=&\ldots \end{array}$$

As such, I wonder if there are similar 'consecutive numbers' for cubic or higher powers. Of course, we know that there is impossible for the following holds (by Fermat's last theorem): $$k^3+(k+1)^3=(k+2)^3$$

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Fantastic proof by coffeemath that there is not a single cubic sequence of this form math.stackexchange.com/a/239430/44154 –  sperners lemma Nov 17 '12 at 19:58

We do have \eqalign{6^3 &= 3^3 + 4^3 + 5^3\cr 20^3 &= 11^3 + 12^3 + 13^3 + 14^3\cr 40^3 &= 3^3 + \ldots + 22^3\cr 70^3 &= 15^3 + \ldots + 34^3\cr 37^3 + 38^3 &= 5^3 + \ldots + 25^3\cr 30^3 + 31^3 + 32^3 &= 7^3 + \ldots + 24^3\cr 101^3 + 102^3 + 103^3 &= 61^3 + \ldots + 71^3\cr 15^3 + \ldots + 20^3 &= 11^3 + \ldots + 19^3\cr 681^3 + \ldots + 687^3 &= 566^3 + \ldots + 577^3\cr \cr} and many others.

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It seems that the question is specifically about $$k^n+(k+1)^n+\cdots +(k+r)^n=(k+r+1)^n+\cdots + (k+2r)^n,$$ i.e. $r+1$ consecutive powers equal the next $r$ consecutive powers. –  Hagen von Eitzen Nov 16 '12 at 20:40
OK, but it's still interesting to see what more general equations are possible. –  Robert Israel Nov 16 '12 at 21:58

There is the famous $3^3+4^3+5^3=6^3$. But I don't know if there are others. At least not where the whole sequence is consecutive... One gets the problem of three triangular numbers T1 T2 T3 for which one wants to solve $T1^2+T2^2=2 \cdot T3^2$.

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Here's one that's nearly consecutive: $4^3 + \ldots + 28^3 = 30^3 + \ldots + 34^3$. –  Robert Israel Nov 18 '12 at 7:53

Let's start by proving the basic sequences and then where and why trying to step it up to cubes fails. I don't prove anything just reduce the problem to a two variable quartic Diophantine equation.

Lemma $1 + 2 + 3 + 4 + \ldots + n = T_1(n) = \frac{n(n+1)}{2}$.

Corollary $(k+1) + (k+2) + \ldots + (k+n) = -T_1(k) + T_1(k+n)$

The first sequence of identities is $$-T_1(s(n)) + T_1(s(n)+n+1) = -T_1(s(n)+n+1) + T_1(s(n)+2n+1)$$ so computing

? f(x) = (x*(x+1))/2
? (-f(s)+f(s+n+1))-(-f(s+n+1)+f(s+2*n+1))
% = -n^2 + (s + 1)


we find $s(n) = n^2-1$ and prove it.

Lemma $1^2 + 2^2 + 3^2 + 4^2 + \ldots + n^2 = T_2(n) = \frac{n(n+1)(2n+1)}{6}$.

The second sequence of identities is $$-T_2(s(n)) + T_2(s(n)+n+1) = -T_2(s(n)+n+1) + T_2(s(n)+2n+1)$$ so computing

? f(x) = (x*(x+1)*(2*x+1))/6
? (-f(s-1)+f(s-1+n+1))-(-f(s-1+n+1)+f(s-1+2*n+1))
% = -2*n^3 + (-2*s - 1)*n^2 + s^2


this is a weird quadratic equation in two integers with some solutions (n,s) = (1,3), (2,10), (3,21), (4,36), (5,55), (6,76), ...

the discriminant of the polynomial (as a polynomial in $s$) is $2^2 n^2 (n+1)^2$ so actually we can solve it and that explains where there's one solution for each $n$.

Now lets try for cubes.. but at this point we know it's not going to work

? f(x) = ((x^2+x)/2)^2
? (-f(s-1)+f(s-1+n+1))-(-f(s-1+n+1)+f(s-1+2*n+1))
% = -7/2*n^4 + (-6*s - 3)*n^3 + (-3*s^2 - 3*s - 1/2)*n^2 + s^3


so this is too complicated to actually solve but if anyone proves this doesn't have solutions for positive $n$ that will show there are no such cubic sequences.

For reference $$7n^4 + (12s + 6)n^3 + (6s^2 + 6s + 1)n^2 - 2s^3 = 0$$ is the Diophantine equation that obstructs a cubic sequence from existing.

Maybe you could conclude by the Mordell Conjecture that there's no infinite family of sequences of identities for cubic and higher power sums, if you can show these polynomials are always irreducible.

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Your equation for the squares has the solutions $s=-n$ and $s=2n^2+n$. $s=-n$ corresponds to the identity $(-n)^2 + \ldots + (-1)^2 + 0^2 = 1^2 + \ldots + n^2$. The fact that the "trivial" factor $s+n$ is a monic polynomial of degree $1$ in $s$ implies that the other factor is also going to be a monic polynomial of degree $1$ in $s$, and thus we get a nontrivial polynomial solution. –  Robert Israel Nov 16 '12 at 21:34
@RobertIsrael, thank you very much I would like to add that to my post. –  sperners lemma Nov 16 '12 at 21:35
For every even power you'll also get a trivial solution, but this doesn't reduce the degree enough... –  Robert Israel Nov 16 '12 at 21:36