It's not hard to prove that $$(1+2+3+\ldots+n)^2=1^3+2^3+\ldots+n^3$$ ( for example using induction )

A generalization of this is also known : $$(\sum_{d \mid n} \tau(d))^2=\sum_{d \mid n} \tau(d)^3$$ where $\tau(k)$ is the number of divisors of the number $k$ .

My question is :

If $a$ and $b$ are positive integers , with $a<b$ then are there an infinite number of sets of positive integers $A$ such that : $$(\sum_{x \in A} x)^a=\sum_{x \in A}x^b$$

What I found :

  • Using Holder's inequality (let $A$ with $n$ elements)

$$\sum_{x \in A} x^b \geq (\sum_{x \in A} x)^b \cdot \frac{1}{n^{b-1}}$$

Let $$\sum_{x \in A} x=S$$ Then :

$$n^{b-1} \geq S^{b-a}$$

But because there are distinct elements in $A$ it follows that :$$S \geq 1+2+\ldots+n=\frac{n(n+1)}{2}>\frac{n^2}{2}$$

Rearranging gives :

$$2^{b-a} >n^{b+1-2a}$$

$$b-a > \log_2 n(b+1-2a)$$ $$b(\log_2 n -1)<a(2 \log_2 n-1)+2 \log_2 n $$ which suggests that $b$ can't be that big in comparison with $a$ :

$$b< a (2+\frac{1}{\log_2 n -1})+2+\frac{2}{\log_2 n-1}$$ (which is around $2a$)

This means that things like $b>3a$ would surely have no solutions .

I'm sure I'm not the first to consider this generalization so I think there must be something known about this problem .

Thanks for everyone who can help me .

  • $\begingroup$ While you´re asking a different question, this seems related. $\endgroup$ – leo Dec 23 '15 at 16:51

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