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Statement

A positive integer $n$ is $r$-powerful if $p^r\mid n$ for every prime $p\mid n$. Theorem (the manuscript's single theorem, printed as Theorem 1, p. 1). Fix an integer $r\ge6$. Infinitely many tuples $(a_1,\ldots,a_{r-2},N)$ of pairwise distinct positive $r$-powerful integers satisfy both

$$a_1+\cdots+a_{r-2}=N \quad\text{and}\quad \gcd(a_1,\ldots,a_{r-2})=1 .$$

Coprimality is joint: the gcd of all summands is $1$, and the summands need not be pairwise coprime. The formal-conjectures declaration erdos_939.variants.infinite_of_six_le (∀ r ≥ 6, (Erdos939Sums r).Infinite, tagged research solved with proof sorry, catalog main as of commit 6fbb54f2, 2026-09-18) states the same fact for the catalog's Finset formulation with positive summands; the decoded playground theorem infinite_rpowerful_sums states it with an injective summand tuple, and it is kernel-checked as part of the Conjectures.io file.

Proof sketch

Let $J=\{j:1\le j\le r,\ j\ \text{odd}\}$, so $|J|=\lceil r/2\rceil$, and $t=r-2-|J|=\lfloor r/2\rfloor-2$; $r\ge6$ gives $t\ge1$.

Splitting the cubic coefficient. Put $C=2\binom r3=r(r-1)(r-2)/3$ and $v_i=i$ for $1\le i<t$, $v_t=C-t(t-1)/2$. These are distinct and positive with $v_1+\cdots+v_t=C$: distinctness and positivity need $v_t\ge t$, that is $C\ge t(t+1)/2$, and since $t\le r/2$ one has $t(t+1)/2\le r(r+2)/8\le C$ because $8(r-1)(r-2)\ge3(r+2)$ for $r\ge6$.

The identity. Expanding $(X+Y)^r-(X-Y)^r$ cancels the even-$j$ terms and doubles the odd ones, so

$$(X+Y)^r=(X-Y)^r+\sum_{j\in J\setminus\{3\}}2\binom rj X^{r-j}Y^j +\sum_{\ell=1}^{t}v_\ell X^{r-3}Y^3 ,$$

with $1+(|J|-1)+t=r-2$ summands on the right, all positive when $X>Y>0$.

Choice of $X$ and $Y$. Let $P$ be the set of primes dividing some $v_\ell$ or some $2\binom rj$ with $j\in J\setminus\{3\}$, let $B=\prod_{p\in P}p$, choose a prime $q>B$, and set $X=q^r$, $Y=B^r$. Then $X>Y>0$ and $\gcd(X,Y)=1$.

Powerfulness. $(X+Y)^r$ and $(X-Y)^r$ are $r$-th powers of positive integers. Every other summand is $cX^aY^b$ with $b\ge1$ and every prime of $c$ in $P$; such a prime divides $Y^b=B^{rb}$ to exponent at least $r$, and $q$ occurs only through $X^a=q^{ra}$. So every summand and the sum are $r$-powerful.

Joint coprimality. A prime dividing all summands divides $(X-Y)^r$, hence $X-Y$, and divides some $cX^aY^b$, hence $XY$; but $\gcd(X-Y,XY)=1$ because $\gcd(X,Y)=1$.

Infinitude. There are infinitely many primes $q>B$, and distinct $q$ give distinct $X$ and distinct totals $(X+Y)^r$.

Distinctness of the summands (asserted in the theorem but not argued in the manuscript's proof; supplied here). The $q$-adic valuation of $2\binom rj X^{r-j}Y^j$ is $r(r-j)$, distinct for distinct $j$; the $t$ split terms share valuation $r(r-3)$ but have distinct coefficients $v_\ell$; $(X-Y)^r$ has valuation $0$, as does the $j=r$ term $2Y^r$ when $r$ is odd, and $(X-Y)^r=2Y^r$ is impossible since $2$ is not an $r$-th power of a rational. The total exceeds every summand.

The full argument, with every deduction written out and the distinctness step labeled as supplied, is reconstructed on the Theorem 1 reconstruction page of the research folder for Problem 939 (author-recorded; not a review).

Depends on. The binomial theorem and the infinitude of primes only.

Bears on. Problem 939: answers the second question (at most finitely many solutions?) in the negative for every $r\ge6$, and gives instances of the first question for every $r\ge6$, with "coprime" read jointly as above (the summands need not be pairwise coprime); it does not reach $r=4$ or $r=5$.