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

Question

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

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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

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

for the Link $(3)$ was posted on the M.S.E.

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Thanks and welcome

Answer 1

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.