This question has been asked previously on math.SE without receiving any answers.
https://math.stackexchange.com/questions/479740/solving-xkx1kx2k-cdotsxk-1k-xkk-for-k-in-mathbb-n
Letting $k$ be a natural number, can we solve the following $k$-th degree equation ?
$$x^k+(x+1)^k+(x+2)^k+\cdots+(x+k-1)^k=(x+k)^k\ \ \ \ \cdots(\star).$$
The following two are famous:
$$3^2+4^2=5^2, 3^3+4^3+5^3=6^3.$$
I've tried to find the other integers which satisfy $(\star)$, but I can't find any nontrivial solution. Then, I suspect that the following might be proven true.
My expectation: There is no integer which satisfies $(\star)$ except $(k,x)=(2,-1),(2,3),(3,3)$.
The followings are what I found:
1. In $k=4$ case, there is no integer which satisfies $(\star)$.
Proof: Suppose that there exists an integer $x$ which satisfies $(\star)$
Then, considering in mod $4$, we reach a contradiction in each remainder, so the proof is completed.
2. Supposing that the one's digit of $k$ is $1$, then there is no integer which satisfies $(\star)$.
Proof: In $k=1$ case, it's obvious. Letting $k=10n+1$ ($n$ is a natural number), let's consider in mod $5$. Let $a_l=l^k$ (mod $5$) ($l$ is an integer).
(i) The $n=1,3,5,\cdots$ cases :
$$a_{5m}\equiv 0, a_{5m+1}\equiv 1, a_{5m+2}\equiv 3, a_{5m+3}\equiv 2, a_{5m+4}\equiv 4.$$
(ii) The $n=2,4,6,\cdots$ cases :
$$a_{5m}\equiv 0, a_{5m+1}\equiv 1, a_{5m+2}\equiv 2, a_{5m+3}\equiv 3, a_{5m+4}\equiv 4.$$
Suppose that there exists an integer $x$ which satisfies $(\star)$. Letting $l=x+k-1$, we get
$$(0+1+2+3+4)\times 2n+a_l \equiv a_{l+1}\ \Rightarrow\ a_l\equiv a_{l+1}.$$
in both (i) and (ii). However, this doesn't happen because of the above. Hence, the proof is completed.
3. Suppose that $k+1$ is a prime number. If there exists an integer $x$ which satisfies $(\star)$, then $k=2$.
Proof: Let's consider when $k=p-1$ ($p$ is a prime number which is more than or equal to $5$). We are going to consider in mod $p$. By Fermat's little theorem, note that
$$a^{p-1}\equiv 0 (a\equiv0), 1(a\not\equiv 0).$$
Suppose that there exists an integer $x$ which satisfies $(\star)$.
(i) The $x\not\equiv1$ case (the multiples of $p$ exists in the integers from $x$ to $x+p-2$): we get
$$0+1\times (p-2)\equiv 1.$$
However, this doesn't happen because $p\ge5$.
(ii) The $x\equiv1$ case (there is no multiples of $p$ in the integers from $x$ to $x+p-2$): we get
$$1\times (p-1)\equiv 0.$$
However, this doesn't happen. Hence, the proof is completed.
Best Answer
Here are a few partial results (the most interesting is probably (3)) towards this problem, but I don't have any hope that these will lead to a general solution. Your problem is a little similar to the famous Erdos-Moser question (which still remains unsolved): does there exist a solution to $1^n+2^n+\ldots+k^n= (k+1)^n$ apart from $1+2=3$?
$1$. If $k$ is even and there is a solution to the equation, then $k \equiv 2\pmod{16}$.
Proof: We may suppose that $k\ge 4$, and let $2^a$ exactly divide $k$. Then $x^k+ \ldots+(x+k-1)^k \equiv k/2 \pmod{2^{a+2}}$, while $(x+k)^k$ is either $0$ or $1 \pmod{2^{a+2}}$. It follows that $a=1$ and that $k/2\equiv 1\pmod 8$, or $k\equiv 2\pmod {16}$ as claimed.
$2$. If $k\equiv 1\pmod 4$ then the equation has no solutions. For $(x+1)^k+\ldots+(x+k-1)^k$ must be even, whereas $(x+k)^k-x^k$ is odd.
$3$. If $k\ge 5$ is odd and squarefree then the equation has no solutions.
Proof: Let $p$ be a prime dividing $k$. Note that $(p-1)\nmid k$. We make use of the fact that if $p$ is odd and $(p-1)\nmid k$ then $1^k+2^k+\ldots+(p-1)^k +p^k\equiv 0\pmod {p}$; this follows easily by picking a primitive root $\pmod p$ and summing the resulting geometric series. Therefore $x^k+\ldots+(x+k-1)^k \equiv 0\pmod p$, and so if this equals $(x+k)^k$ then $p$ must divide $x$. Since $k$ was assumed to be square-free, we conclude that $k|x$. But now note that $0^k+1^1+\ldots+(k-1)^k < k^k$ (for $k\ge 5$), and that $(\ell k)^k+\ldots + (\ell k +k-1)^k > ((\ell+1)k)^k$ for all $\ell \ge 1$ and $k\ge 5$.
$4$. If $k$ is odd, and $p$ is an odd prime dividing $k-1$ such that $(k,p-1)=1$ then the equation has no solutions. In particular if $k \equiv 1\pmod{6}$, or $1\pmod {10}$ or $1\pmod{p(p-1)}$ for any odd prime $p$, there are no solutions.
Proof: If $p|(k-1)$ and $(p-1)\nmid k$ then we find from the observation used in (3) that $(x+1)^k+\ldots+(x+k-1)^k \equiv 0 \pmod {p}$. Thus $(x+k)^k \equiv x^k \pmod p$. Since $(k,p-1)=1$ by assumption the map $x\to x^k \pmod {p}$ permutes all the residue classes $\pmod p$. Thus it follows that $(x+k)\equiv x \pmod p$, but this is impossible since $x+k \equiv x+1 \pmod p$.