17
$\begingroup$

Question : Is the following conjecture true?

Conjecture : Let $a,b(\ge 2),c,n(\ge 2)$ be natural numbers. If $$\left(\sum_{k=1}^nk^a\right)^b=\sum_{k=1}^nk^c\ \ \ \ \ \cdots(\star)$$ for some $n$, then $(a,b,c)=(1,2,3).$

Remark : This question has been asked previously on math.SE without receiving any answers.

Motivation : This question comes from
$$\left(\sum_{k=1}^nk\right)^2=\sum_{k=1}^nk^3.$$

This got me interested in $(\star)$. I've got the followings :

1. If $(\star)$ for any $n\in\mathbb N$, then $(a,b,c)=(1,2,3).$

We can easily prove this by considering the limitation $n\to\infty$ of the both sides of $$\frac{n^{(a+1)b}}{n^{c+1}}\left\{\sum_{k=1}^n\frac 1n\left(\frac kn\right)^a\right\}^b=\sum_{k=1}^n\frac 1n \left(\frac kn\right)^c.$$

2. If $(\star)$ for $n=2$, then $(a,b,c)=(1,2,3).$

3. If $(\star)$ for $n=3$, then $(a,b,c)=(1,2,3).$

Since both 1 and 2 are easy to prove, I'm going to prove 3.

Proof : Supposing $c\le ab$, since $b\ge 2$, we get $$(1+2^a+3^a)^b=1+2^{ab}+3^{ab}+\cdots\gt 1+2^c+3^c.$$ This is a contradiction. Hence, $c\gt ab$. Supposing $b\ge 3$, we get $c\gt ab\ge 3$.

Here, since $3^c+1\equiv 4,2$ (mod $8$) for any $c\in\mathbb N$, $3^c+1$ is not a multiple of $8$. By the way, since $1+2^a+3^a$ is even, $(1+2^a+3^a)^b$ is a multiple of $8$. Since $2^c$ is a multiple of $8$, this leads that $3^c+1$ is a multiple of $8$, which is a contradiction. Hence, $b=2, c\gt 2a$.

If $a\ge 3$, since $$\left(\frac 23\right)^a+\left(\frac 13\right)^a\le\left(\frac 23\right)^3+\left(\frac 13\right)^3=\frac13,$$ $2^a+1\le \frac{3^a}{3}.$ Hence, $$3^c\lt 1+2^c+3^c=(1+2^a+3^a)^2\le \left(\frac{3^a}{3}+3^a\right)^2=3^{2a}\left(\frac 43\right)^2=3^{2a}\cdot\frac {16}{9}\lt 3^{2a+1}.$$ $3^c\lt 3^{2a+1}$ leads $0\lt c-2a\lt 1$, which means that $c-2a$ is not an integer. This is a contradiction. Hence, we know $a=1$ or $a=2$.

The $(a,b)=(1,2)$ case leads $c=3$.

The $(a,b)=(2,2)$ case leads $c\ge5\Rightarrow 1+2^c+3^c\gt 196$, which is a contradiction. Now the proof is completed.

After getting these results, I reached the above conjecture. Can anyone help?

$\endgroup$
  • $\begingroup$ Perhaps you can try bounding either side of your first equation from below via the Arithmetic Mean-Geometric Mean Inequality, and from above via the Power Mean-Arithmetic Mean Inequality? (See this PDF document by Kiran Kedlaya for more details regarding these inequalities - starting on Chapter 3, page 18 in particular.) The bounds you will obtain may help you in reducing the search space / number of cases you need to consider. $\endgroup$ – Jose Arnaldo Bebita-Dris Oct 11 '13 at 19:26
2
$\begingroup$

This is a partial solution. I've just been able to get the following theorem:

Theorem : If $(\star)$ for some $n=8k-5,8k-4\ (k\in\mathbb N)$, then $(a,b,c)=(1,2,3)$."

I wrote the proof for this theorem on MSE.

PS: This idea (using mod $8$) does not seem to work for the other $n$. Another idea would be needed.

$\endgroup$
  • $\begingroup$ I think 'you' have to explain why you gave a downvote. $\endgroup$ – mathlove Oct 12 '13 at 14:27
  • $\begingroup$ Unfortunately, explanations are a polite convention, and not a guarantee. Some voters are not polite, it seems. On another note, consider the idea that the right hand side is a power only for small c. There may be some work of Erdos in that direction, but I know of no reference. Gerhard "Other Ideas Provided Upon Request" Paseman, 2013.10.12 $\endgroup$ – Gerhard Paseman Oct 12 '13 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.