# Tag Archives: sum of powers

## Sums of equal powers

Let $k$ be a fixed positive integer. Let

$S_k(n):=1^k+\cdots+n^k$.

Let us consider expressions of the form

$\pm 1^k\pm\cdots\pm n^k=\displaystyle\sum_{j=1}^nj^ke_j=: S(e_1,\dots,e_n)$

where $e_j=\pm 1\,\forall j$.

Note that $S$ takes $2^n$ possible values in $[-S_k(n),S_k(n)]$ which are all congruent modulo $2$. Hence the number of distinct values that $S$ takes is at most

$\displaystyle\left\lceil\frac{2S_k(n)+1}{2}\right\rceil=S_k(n)+1$

Since this is $O(n^{k+1})$, it is eventually dominated by $2^n$. Thus for all sufficiently large $n$, we can find $e_j,e_j'\in\{\pm 1\}$ such that

$\displaystyle\sum_{j=1}^nj^ke_j=\sum_{j=1}^nj^ke_j'\Leftrightarrow\sum_{j=1}^n(e_j-e_j')j^k=0$

with $e_j\neq e_j'$ for some $j$.

Note that $e_j-e_j'\in\{0,\pm 2\}\,\forall j$. Let $A=\{j:e_j-e_j'=2\}$ and $B=\{j:e_j-e_j'=-2\}$. Then $A$ and $B$ are non-empty and disjoint, $A\cup B\subseteq\{1,\dots,n\}$, and

$\displaystyle\sum_{j\in A}j^k=\sum_{j\in B}j^k$.

Thus:

Proposition. Let $k$ be a fixed positive integer. Then one can find distinct positive integers $x_1,\dots, x_r,y_1,\dots,y_s$ such that

$x_1^k+\cdots+x_r^k=y_1^k+\cdots+y_s^k$.

In particular, one can choose $x_i,y_j\le\min\{n\in\mathbb N: 2^n>S_k(n)+1\}\,\forall i,j$.

Example. Let $k=4$. Note that $2^n>S_4(n)+1$ for all $n\ge 20$. So there exist distinct integers $1\le x_i,y_j\le 20$ such that $\sum_ix_i^4=\sum_jy_j^4$. In particular, one has

$1^4+2^4+9^4=3^4+7^4+8^4$

as well as

$7^4+8^4+10^4+13^4=3^4+9^4+14^4$.

Likewise, for fifth powers we can find such $x_i$‘s and $y_j$‘s in $[1,26]$.

This wikipedia article contains many results related to sums of powers.

Filed under Number theory

## A ‘sum of squares’ problem

Recently I came to know from a friend that the only non-trivial value of $n$ satisfying the Diophantine equation

$1^2+2^2+\dots+n^2=m^2$

is $n=24$. And apparently, although the problem looks fairly simple, especially since we have the formula for the left hand side

$\displaystyle 1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}$

the proof is hard. In fact, I was told that all proofs of this result use advanced machinery. Nevertheless, given the simple nature of the problem I was intrigued to give it a try.

So let us assume that

$n(n+1)(2n+1)=6m^2$.

Noting that $2n(n+1)-n(2n+1)=n$, we have that $n(n+1)$ and $2n+1$ are coprime. Indeed, if $g$ is a common factor, then $g$ has to divide both $n$ and $2n+1$, forcing $g=1$. So we deduce that

$\{n(n+1),2n+1\}=\{a^2,6b^2\}$ or $\{2a^2,3b^2\}$

for some integers $a,b$.

Note that $2n+1\not\in \{2a^2,6b^2\}$ so $2n+1\in\{a^2,3b^2\}$. Hence the possibilities are:

(i) $n(n+1)=2a^2$, $2n+1=3b^2$.

(ii) $n(n+1)=6b^2$, $2n+1=a^2$.

First suppose that $n(n+1)=2a^2$. Then $n(n+1)/2=a^2$ is a square triangular number, so

$n=\displaystyle\frac{H_{2k}-1}{2}$

for some $k$, where $P_0,P_1,\dots$ are the Pell numbers. and $H_0,H_1,\dots$ are the half-companion Pell numbers, and they are given by $(1+\sqrt 2)^k=H_k+P_k\sqrt 2$ for each $k$. So we have to find all $k$ such that

$2n+1=\boxed{H_{2k}=3b^2}$.

On the other hand, if $2n+1=a^2$, then $a=2c-1$ for some integer $c$. Then $n=2c(c-1)$. Hence $n(n+1)=6b^2$ implies

$c(c-1)(2c^2-2c+1)=3b^2$.

Now $\gcd(c(c-1),2c^2-2c+1)=1$, so $\{c(c-1),2c^2-2c+1\}=\{u^2,3v^2\}$ for some integers $u,v$. Clearly $c(c-1)\neq u^2$, so

$c(c-1)=3v^2$ and $2c^2-2c+1=u^2$.

So $\{c,c-1,u\}$ is a near isosceles Pythagorean triple, i.e.

$c=\displaystyle\frac{H_{2k-1}+1}{2}$.

Hence

$\boxed{H_{2k-1}^2-1=12v^2\Leftrightarrow P_{2k-1}^2-1=6v^2\Leftrightarrow P_{2k}P_{2k-2}=6v^2}$.

The two boxed cases are what only remain to check, and they show why this actually is a difficult problem! (Refer to the articles about Fibonacci numbers in the ‘Notes and Articles’ tab above to see how little information we have on divisors of sequences of this kind.) Nevertheless, I’ll leave it here and maybe come back to it another time.

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Filed under Number theory