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.

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