# Can a sum of consecutive $n$th powers ever equal a power of two?

I have already post this claim in Mathematics Stack Exchange (25sep 19) but not get complete solution Link

Edit(6 jun 20): change notation $$n_{u,m}$$ to $$S_n^m(u)$$

Let $$n,u,m\in \mathbb{N}$$

$$S_n^m(u)$$ is a number defined as

$$S_n^m(u)= n^m+(n+1)^m+(n+2)^m+...+(n+u)^m= \sum_{i=0}^{u}(n+i)^m$$

example: $$S_3^4(2)=3^4+(3+1)^4+(3+2)^4=962$$

Question: Is the following claim true?

Can it be shown that $$2^t$$ cannot be written in $$S_n^m(u)$$

$$S_n^m(u) \ne 2^t \ \ \ \ \ \forall n,u,m,t\in \mathbb{N}$$

Check solution by Andrea Marina, Good progress over above claim link

Proved for $$S_n^1(u)$$ and $$S_n^2(u)$$ never equals a power of two.

Proof for $$S_n^1(u)\ne 2^t$$

Suppose $$S_n^1(u) =\frac{(u+1)(2n+u)}{2}= 2^t$$ So $$(u+1)(2n+u)= 2^{t+1}$$

Case$$1$$: $$u$$ is $$odd$$, then $$u+1= even$$ and $$2n+u = odd$$ it implies $$even×odd \ne 2^{t+1}$$ because $$2^{t+1}$$ content only $$even$$ multiples except $$1$$ and $$2n+u>1$$.

Case$$2$$: $$u$$ is $$even$$, then $$u+1= odd$$ and $$2n+u = even$$ it implies $$odd×even \ne 2^{t+1}$$ similarly as case1

Both cases shows complete proof for $$S_n^1(u) \ne 2^t$$

Note

By using Newton's interpolation method, we can calculate formula for $$n_{u,m}$$. I write the general formula at bottom of the post.

$$S_n^2(u)=n^2(u+1)+(2n+1)\frac{(u+1)u}{2} +\frac{(u+1)u(u-1)}{3} \ \ \ \ \ \ ...eq(1)$$

Proof for $$S_n^2(u)\ne 2^t$$

Suppose $$S_n^2(u) = 2^t$$ We can write $$eq(1)$$ as

$$(u+1)(6n^2+3(2n+1)u+2u(u-1))= 3×2^{t+1}$$

Case$$1$$: $$u =even$$ implies $$u+1=3$$ and $$3n^2+3(2n+1)+2=2^{t}=even$$. But whenever $$n$$ is $$even$$ either $$odd$$ gives $$3n^2+3(2n+1)+2\ne even$$

Case$$2$$: $$u =odd, u+1=even=2^x$$ for some $$x$$. Then $$6n^2+3(2n+1)u+2u(u-1)= even=3×2^y$$ for some $$y$$. Where $$2^x2^y=2^{t+1}$$ implies $$2n+1= even$$, which is not true.

Both cases shows complete proof for $$S_n^2(u)\ne 2^t$$

General formula for $$S_n^m(u)$$

$$S_n^m(u)=\sum_{i=0}^{m} \binom{u+1}{i+1} \sum_{j=i}^{m}\binom{m}{j}n^{m-j}\sum_{k=0}^{i}(i-k)^j(-1)^k\binom{i}k$$

Where $$n\in \mathbb{R}$$ and $$u,m\in \mathbb{Z^*}$$ and $$0^0=1$$

Moreover if we put $$n=0$$ then

$$S_0^m(u)=\sum_{l=0}^{u}l^{m}=\sum_{i=0}^{m}\binom{u+1}{i+1}\sum_{k=0}^{i}(i-k)^i(-1)^k\binom{i}k$$

Related questions

also have claimed

(1) Let $$d$$ be any odd positive integer then can it be shown that (Extending of above claim)

$$\sum_{q=0}^{u}(n+qd)^{m}\ne 2^t \ \ \ \ \forall n,u,m,t\in\mathbb{N}$$

(2) Can it be shown that

$$\sum_{q=0}^{u}(n+qd)^{2m}\ne 3^t \ \ \ \ \forall n,d,u,m,t\in\mathbb{N}$$

Solution for $$m=1$$

(3) let $$p$$ be the odd prime then

$$\sum_{q=0}^{u}(n+qd)^{(p-1)m}\ne p^t\ \ \ \ \forall n,u,m,d,t\in\mathbb{N}$$

Thanks and welcome

• maybe, bernoulli numbers would help here – vidyarthi Dec 12 '19 at 8:55
• @vidyarthi probably referss to the fact that the sum of the first $k$ integer $m$-th powers always has the form $q_{m+1}(k)$ where $q_{m+1}(x)$ is a rational polynomial of degree $m+1$ in $x$. – Geoff Robinson Dec 12 '19 at 10:57
• If m is odd and u is even, the sum of consecutive powers is a multiple of u+1, so for positive even u and odd m the answer is no. It may be possible to adjust this for m even. For u odd, there may be some tight necessary conditions on u to have the sum a power of 2. Gerhard "Using Symmetry Can Help Here" Paseman, 2019.12.12. – Gerhard Paseman Dec 12 '19 at 20:15

For m odd, one can get a quick proof for many u, specifically for u+1 not a power of two.

If u is even, the sum modulo (u+1) is the sum of mth powers, one from each residue class of 1+u. As m is odd, we match up negative with positive residue classes to get the sum is 0 modulo (u+1), so the sum is a multiple of an odd number. If the sum is also positive, it can't be a power of two.

If u is odd, and u+1 is not a power of two, then u+1 has an odd factor k. Split the sum into consecutive sums each with k terms, and apply the previous paragraph to get the final sum a multiple of k.

If one can get a similar result for even u and even m, one can apply a similar reduction to apply to most odd u.

When u+1 is a power of two, some other idea is needed to analyze this case.

Gerhard "Prefers Applying Rather Simple Arguments" Paseman, 2019.12.12.

• Indeed, -1,0,1,2^(2k+1) contains several examples when u+1 is a small power of two. When m is even, 1,0,1 sums to 2. Gerhard "Probably Not Many More Examples" Paseman, 2019.12.12. – Gerhard Paseman Dec 13 '19 at 6:01