Letting $k$ be a natural number, can we solve the following $k$-th degree equation ? $$x^k+(x+1)^k+(x+2)^k+\cdots+(x+k-1)^k=(x+k)^k \tag{$\star$}.$$

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 conjecture: 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.

I crossposted to MO.


  • $\begingroup$ You should find an inequality between LHS and RHS. For example: artofproblemsolving.com/Forum/… $\endgroup$ – Tengu Sep 5 '13 at 16:26
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    $\begingroup$ For cubes, the next best thing is that it can be shown there is an infinite number of solutions to $x^3+(x+1)^3+(x+2)^3+\dots+(x+n)^3 = y^3$ where $y \neq x+n+1$ such as $11^3+12^3+13^3+14^3 = 20^3$, etc. $\endgroup$ – Tito Piezas III Sep 9 '13 at 4:01
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    $\begingroup$ I just recalled that for squares, there is also an infinite number for $x^2+(x+1)^2+(x+2)^2+\dots+(x+n)^2 = y^2$ where $y \neq x+n+1$ such as the famous cannonball stacking problem $1^2+2^2+3^2+\dots+24^2 = 70^2$ See Kevin Brown's short article. $\endgroup$ – Tito Piezas III Sep 9 '13 at 5:05
  • $\begingroup$ @TitoPiezasIII: I've known the last relational equation you wrote. Thank you for your information. $\endgroup$ – mathlove Sep 9 '13 at 15:11
  • $\begingroup$ @harrypham: Thank you for your information. It must be a good hint, but I haven't got any non-trivial result. $\endgroup$ – mathlove Sep 12 '13 at 15:27

This is more a comment; it shows another approach but does not (yet?) provide a solution

Using the logic of the Bernoulli-polynomials, which allow a polynomial expression for sums of consecutive like powers, we can generate polynomials depending on x and whose order depends on k for which we can then search for an integer root.

First I rewrite the problem as a function, for which a result with a valid x is a root: $$ f_k(x) = \sum_{j=0}^{k-1} (x+j)^k - (x+k)^k $$ such that for some k in $ f_k(x_k) = 0 $ the $x_k$ indicates the root, and for instance
1. $\qquad f_2(3)=0 \qquad \to x_2=3 $ is the first solution (for $k=2$) and
2. $\qquad f_3(3)=0 \qquad \to x_3=3 $ is the second solution (for $k=3$).

Fiddling a bit with the adaption of the Bernoulli-polynomials we get the family of polynomials $$ \small \begin{array} {r|l} k & f_k(x) \\ \hline \\ 2 & x^2-2 x-3 \\ 3 & 2 x^3-12 x-18 \\ 4 & 3 x^4+8 x^3-12 x^2-112 x-158 \\ 5 & 4 x^5+25 x^4+50 x^3-250 x^2-1355 x-1825 \\ 6 & 5 x^6+54 x^5+285 x^4+180 x^3-4755 x^2-20106 x-26141 \\ 7 & 6 x^7+98 x^6+882 x^5+3430 x^4-4410 *x^3-96726 x^2-353346 x-446782 \\ 8 & 7 x^8+160 x^7+2128 x^6+15232 x^5+40600 x^4-210560 x^3-2165072 x^2-7174784 x-8869820 \\ ... & ... \end{array} $$

The roots for that polynomials are now $$ \small \begin{array} {r|lll} k & real & roots & complex & roots \\ \hline 2 & -1.00000 & 3.00000 & . & . \\ 3 & . & 3.00000 & -1.50000+0.866025 i & ... \\ 4 & -1.96857 & 3.32947 & -2.01378+1.99502 i & ... \\ 5 & . & 3.73445 & -2.46439-0.816002 i & ... \\ 6 & -2.95510 & 4.16379 & -2.96548-1.73889 i & ... \\ 7 & . & 4.60171 & -3.46914-2.80013 i & ... \\ ... & ... \end{array}$$

so we see (in teh second column), that after the two relevant real roots $x_2=3$ and $x_3=3$ all other roots $x_k$ in the second column are non-integer, and thus have a small list of where indeed there is and where is not a solution...

The final answer can now only come from the knowledge of the general form of the polynomials and whether then polynomials of that form can have positive integer roots.
There are some criteria which sometimes allow to determine before actually computed, whether there do exist integer/rational roots for certain polynomials (for instance Eisenstein-criterion), unfortunately I'm not firm with that.

[update] I don't really know the significance of this but after the comment of @minar I evaluated the real roots up to index 128.
If we define the minimum of the $\Gamma(x)$ near $x=1.461$ as $\mu \approx 0.8856...$ then the scaling of the positive roots seem to converge to the integer index plus some fixed fractional $$ \small {\rho_k \cdot 2 / \mu =... 10.3922 , 11.3881 , 12.3865 , 13.3858 ,...\\ ...,73.3707 , 74.3705 , 75.3702 , 76.3700 ,...\\ ...,127.3577, 128.3574, 129.3572, 130.3570, ...\\} $$

[update2] Hmm, after I've tested some odd k it seems that the following is true:

Let $k$ be an odd number (prime or composite) and $w$ its squarefree primefactor-kernel then the sum $$f_k(x) \equiv \sum_{j=1}^{k-1} (x+j) \pmod k$$ is zero only for such $x$ which are multiples of $w$: $ x=t \cdot w \qquad t \in (1,2,3...)$ .

If this is true, then the above estimate of the real root of the sum-quation might come into the play...

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    $\begingroup$ In fact, $f_4$ to $f_{400}$ are even irreducible. $\endgroup$ – minar Sep 14 '13 at 11:29

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