Ireland_2017

If at most one term in a sequence $a$ is non-zero, it is immediate that it is $m$-powerful for all $m\in \mathbb{N}$. We will see that there are no other 30-powerful sequences (or indeed, $m$-powerful for any even $m$), but that there are many other sequences that are $m$-powerful for all odd $m$.

We begin our analysis with a simple lemma.

Lemma. For $m\in \mathbb{N}$, let $P_m(x,y)=(x+y{)}^m-x^m-y^m$, $x,y\in \mathbb{R}$. Then $P_m(x,y)=0$ whenever $xy=0$. Moreover, $P_m$ has no other roots if $m$ is even, but $P_m(x,-x)=0$ if $m$ is odd.

Proof. Let $p(t)=(1+t{)}^m-1-t^m$, $t\in \mathbb{R}$. By expansion of $(1+t{)}^m$, we see that $p$ is a polynomial with non-negative coefficients, and so $p(t)>0$ for $t>0$. Suppose additionally that $m$ is even. If $-1\le t<0$, then $(1+t{)}^m-1\le 0$, so $p(t)\le -t^m<0$. If $t<-1$, then $0<(1+t{)}^m<t^m$, so $p(t)<-1<0$. We conclude that $0$ is the only root of $p$ when $m$ is even. If $xy\ne 0$, then $P_m(x,y)=x^{m}p(t)$, where $p$ is as above, and $t=y/x\ne 0$. It follows that $P_m(x,y)\ne 0$ if $m$ is even and $xy\ne 0$:

The statements that $P_m(x,y)=0$ when $xy=0$ (regardless of the parity of $m$) and $P_m(x,-x)=0$ when $m$ is odd, both follow immediately. $\square$

Fix an arbitrary sequence $a=(a_k)$. For $m,n\in \mathbb{N}$, let $$S(m,n)={\left(\sum _{k=1}^n{a}_k\right)}^m-\sum _{k=1}^n{a}_k^m,$$ and let $D(m,n)=S(m,n+1)-S(m,n)$. The condition "a is m-powerful" says that $S(m,n)=0$, $n\in \mathbb{N}$, and so we also have $D(m,n)=0$.

Define $x_n=a_{n+1}$ and $y_n={\sum }_{k=1}^n{a}_k$. The equation $S(m,n)=0$ can be written as $$y_n^m=\sum _{k=1}^n{a}_k^m,$$ so if this equation holds, then the equation $D(m,n)=0$ can be written as $$(x_n+y_n{)}^m-x_n^m-y_n^m=0.$$

Suppose now that $m$ is even and $a$ is $m$-powerful. By the lemma, we conclude that $x_n{y}_n=0$ for all $n\in \mathbb{N}$. We will use these last equations to prove by induction on $i\in \mathbb{N}$, that at most one of the terms $a_1,...,a_i$ is non-zero; we call this property $A_i$.

$A_1$ is trivially true, so suppose that $A_i$ is true for a specific $i=n\in \mathbb{N}$. If $a_1,\dots ,a_n$ are all zero, then $A_{n+1}$ follows immediately, so we may assume that exactly one of these terms is non-zero. But now $y_n\ne 0$, so the equation $x_n{y}_n=0$ implies that $x_n=a_{n+1}=0$, and we again deduce $A_{n+1}$. This finishes the proof that if $m$ is even, then the $m$-powerful sequences are those with at most one non-zero term. Part (a) follows.

The analysis is similar for $m$ odd, but now $P_m$ has other roots in the lemma. By considering these roots, our analysis readily leads us to see that $a=((-1{)}^n{)}_{n=1}^{\infty }$ is $m$-powerful for every $c\in \mathbb{R}$. Taking any non-zero $c$, we get an example with the properties required in (b).