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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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  • $\begingroup$ Note that for a fixed $k$ there are not too many possibilities for $x$, since if $(x,k)$ satisfy the equation, then $k x^k < (x+k)^k$. Solving the inequality gives $x \leq \frac{k}{k^{1/k}-1} \approx \frac{k^2}{\ln(k)}$. $\endgroup$ – Igor Shinkar Sep 12 '13 at 15:59
  • $\begingroup$ Actually, a similar trick can show that $x = O(k)$. $\endgroup$ – Igor Shinkar Sep 12 '13 at 16:32
  • $\begingroup$ no solutions are known for $k \geq 9.$ There may be something for $k=8,$ don't recall for sure. In general, just an open problem. The part about consecutive values is what kills it off for sure. A little progress is available if you do not insist on consecutive values, some due to our Noam Elkies. $\endgroup$ – Will Jagy Sep 12 '13 at 16:38
  • $\begingroup$ @IgorShinkar: Thanks. We may be able to find every solution for a fixed $k$ by using computer. $\endgroup$ – mathlove Sep 12 '13 at 16:49
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    $\begingroup$ A heuristics suggests, that the problem can be stated modulo $k$ by the simple expression $$ k \cdot B_k \equiv x^k \pmod k$$, where $B_k$ are the Bernoulli-numbers. Because there are a couple of properties of the Bernoulli-numbers are known, some classes of the parameter $k$ can be excluded for a solution, because no selection of $x$ can make the lhs and rhs equivalent. This can also be extended to the consideration modulo $k^2$ - however, I've not yet put my observations into a formula. $\endgroup$ – Gottfried Helms Nov 4 '13 at 11:37
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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 $x+k \equiv x+1 \pmod p$.

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I have 2 comments:

1) Empirically I have a guess for the positive real roots: Let a scaling-factor $c=\frac1{\ln 2}-1 \approx 0.442$ then $$ \rho_k \approx (k+3) \cdot c $$ or $$ k \lt {\rho_k \over c }-3 \lt k+1 \tag {for $k \ge$ 5}$$ I checked this up to $k=5000$ so far using the bernoulli-polynomials and internal float precision of 400 digits in Pari/GP (see table below). The idea behind this is to show that either the roots are irrational or that the approximation of the real root to an integer is too bad, which is supported by the small magnitude of the $\epsilon$ compared with the systematic approximation of $k \cdot c +1.5 $ with increasing $k$ to an integer.

2) looking at the equation modulo $k$ it seems, the lhs of the equation is always equivalent to zero, so solutions can only exist, if also the rhs is zero modulo $k$ - here I assume only odd $k$ so far.
Let $w$ be the squarefree kernel of primefactors (all are odd) of $k$. Then the rhs is equivalent zero only if $x$ is a multiple of $w$ and thus only such $x$ can solve the equality. So if we have some $k=p_1^{b_1} p_2^{b_3} p_3^{b_3} $ then with the squarefree kernel $w$ we can write $$k= (p_1 p_2 p_3) \cdot \left(p_1^{b_1-1}p_2^{b_2-1}p_3^{b_3-1}\right) = w \cdot m$$ and then the root $x$ must have the form $$ x=w \cdot v \qquad \qquad \text{ where also } v \lt m$$ (by the last observation, a nontrivial $k$ cannot be squarefree because $m$ must be greater than $1$: $1 \le v \lt m$)

3) I hoped to combine the two empirical results somehow, such that $v \approx m/c $but don't see any further useful property except the likely irrationality of the root - so also I do not know, whether it is at all worth the effort to actually prove observation 1).


Here is a list of the roots, scaled by the scaling factor and the integer value $3$ and the fractional value separated: $$\small \begin{array} {r|l} k & {\rho_k \over c} & {\rho_k \over c}=k+{1.5 \over c}+ \epsilon \\ \hline 4 & 4 + 4.16684831867 & 4 + 1.5/c + 0.344643086811 \\ 5 & 5 + 3.65764105650 & 5 + 1.5/c + 0.119219557064 \\ 6 & 6 + 3.48606997990 & 6 + 1.5/c + 0.0432658922944 \\ 7 & 7 + 3.4241172248 & 7 + 1.5/c + 0.0158397148376 \\ 8 & 8 + 3.4015117133 & 8 + 1.5/c + 0.00583236702541 \\ 9 & 9 + 3.3933253612 & 9 + 1.5/c + 0.00220830950877 \\ 10 & 10+3.3903889002 & 10 + 1.5/c + 0.000908352824271 \\ 11 & 11+3.3893375033 & 11 + 1.5/c + 0.000442904593529 \\ 12 & 12+3.3889540764 & 12 + 1.5/c + 0.000273163432055 \\ 13 & 13+3.3888054311 & 13 + 1.5/c + 0.000207358901590 \\ 14 & 14+3.3887395158 & 14 + 1.5/c + 0.000178178523631 \\ 15 & 15+3.3887034724 & 15 + 1.5/c + 0.000162222283615 \\ \ldots & \ldots \\ 497 & 497+3.388349245 & 497 + 1.5/c + 0.00000540736229828 \\ 498 & 498+3.388349220 & 498 + 1.5/c + 0.00000539654486254\\ 499 & 499+3.388349196 & 499 + 1.5/c + 0.00000538577062079\\ 500 & 500+3.388349172 & 500 + 1.5/c + 0.00000537503931484\\ \end{array} $$
[update]
Here is another list, where the exponent $k$ is in exponential progression and the remaining ${\epsilon \over c}$ (from the above table) is displayed by its base-$2$-logarithm: $$\small{ \begin{array} {rr|l} j & k=2^j & \log_2 (\epsilon /c) \\ \hline\\ 2 & 4 & -0.361210135710 \\ 3 & 8 & -6.24608789212 \\ 4 & 16 & -11.5139506158 \\ 5 & 32 & -12.4418890639 \\ 6 & 64 & -13.4004277751 \\ 7 & 128 & -14.3795231280 \\ 8 & 256 & -15.3690269163 \\ 9 & 512 & -16.3637677898 \\ 10 & 1024 & -17.3611354651 \\ 11 & 2048 & -18.3598186117 \\ 12 & 4096 & -19.3591600122 \\ 13 & 8192 & -20.3588306692 \\ 14 & 16384 & -21.3586659868 \\ 15& 32768 & -22.3585836430\\ \end{array} }$$ (The range of $k$ was not continuously computed, only at the explicite values)

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