../

---

Source. Infinite $r$-Powerful Sums, the one-page manuscript whose author line reads GPT-5.5 Pro and which Liam Price shared in the erdosproblems.com forum thread for Problem 939 on 24 May 2026: Theorem 1, its proof, and the display numbered (1), all on physical p. 1 (the only page) of the PDF held by its library source card, Price (2026). The held PDF is the card's local typesetting of the downloaded TeX source, as the card records, so the page reference is to that artifact. The card's result page states the theorem with a proof sketch.

Standing. This is an author-recorded reconstruction of the manuscript's proof. It is not an independent review, changes no status of Problem 939, and assigns no tier. The external inputs are the binomial theorem, unique factorization in $\mathbb Z$ (used through $p$-adic valuations), and the infinitude of primes; each is invoked in the elementary form stated where it is used. One step that the theorem asserts and the proof omits, the distinctness of the summands, is supplied below and labeled as a compilation fill.

Definitions

Let $r\ge2$ be an integer. A positive integer $n$ is $r$-powerful if $p^r\mid n$ for every prime $p\mid n$; equivalently, $v_p(n)=0$ or $v_p(n)\ge r$ for every prime $p$, where $v_p$ is the $p$-adic valuation. In particular $1$ is $r$-powerful, and so is $m^r$ for every positive integer $m$: if $p\mid m^r$ then $p\mid m$, so $v_p(m^r)=r\,v_p(m)\ge r$. A finite family of positive integers is jointly coprime if the greatest common divisor of all its members is $1$; the members need not be pairwise coprime.

Statement

Theorem 1. Let $r\ge6$ be an integer. There are infinitely many tuples $(a_1,\ldots,a_{r-2},N)$ of positive integers such that

$$a_1+\cdots+a_{r-2}=N ,$$

the $r-1$ numbers $a_1,\ldots,a_{r-2},N$ are pairwise distinct and each is $r$-powerful, and $\gcd(a_1,\ldots,a_{r-2})=1$.

The theorem is stated for each fixed $r$ separately: "infinitely many" refers to the tuples for that one exponent.

Proof

The number of summands

Let $J=\{j:1\le j\le r,\ j\ \text{odd}\}$. The odd integers in $[1,r]$ number $\lceil r/2\rceil$, so $|J|=\lceil r/2\rceil$, and since $\lceil r/2\rceil+\lfloor r/2\rfloor=r$,

$$t:=r-2-|J|=\lfloor r/2\rfloor-2 .$$

For $r\ge6$ one has $\lfloor r/2\rfloor\ge3$, so $t\ge1$. Also $3\in J$, since $r\ge3$. The identity (1) below has $|J|$ summands from the odd part of a binomial expansion, one of which is split into $t$ pieces, plus one more; $|J|+t=r-2$ is the number of summands the theorem requires.

Splitting the cubic coefficient

Set

$$C:=2\binom r3=\frac{r(r-1)(r-2)}3 ,$$

a positive integer. Define

$$v_i:=i\quad(1\le i<t),\qquad v_t:=C-\frac{t(t-1)}2 .$$

When $t=1$ the first clause is empty and $v_1=C$. Then

$$v_1+\cdots+v_t=\frac{t(t-1)}2+C-\frac{t(t-1)}2=C .$$

The $v_i$ with $i<t$ are the distinct positive integers $1,\ldots,t-1$, so the whole family is distinct and positive as soon as $v_t\ge t$, that is, as soon as

$$C\ge\frac{t(t-1)}2+t=\frac{t(t+1)}2 .$$

Since $t=\lfloor r/2\rfloor-2\le r/2$,

$$\frac{t(t+1)}2\le\frac{(r/2)(r/2+1)}2=\frac{r(r+2)}8 ,$$

and $r(r+2)/8\le r(r-1)(r-2)/3=C$ is equivalent to $3(r+2)\le8(r-1)(r-2)$, which holds for $r\ge6$: at $r=6$ the two sides are $24$ and $160$, and the difference $8(r-1)(r-2)-3(r+2)=8r^2-27r+10$ increases for $r\ge2$. Hence $v_1,\ldots,v_t$ are distinct positive integers with sum $C$.

The choice of X and Y

Let $P$ be the set of primes dividing at least one of $v_1,\ldots,v_t$ or at least one of the integers $2\binom rj$ with $j\in J\setminus\{3\}$, and let

$$B:=\prod_{p\in P}p .$$

Since $1\in J\setminus\{3\}$ and $2\binom r1=2r$, the prime $2$ lies in $P$, so $B\ge2$; the manuscript's convention that an empty product is $1$ is never needed for $r\ge6$. Choose a prime $q>B$; one exists because there are infinitely many primes. Set

$$X:=q^r ,\qquad Y:=B^r .$$

Then $X>Y>0$ because $q>B\ge1$. Every $p\in P$ divides $B$, so $p\le B<q$ and $p\ne q$; hence $q\notin P$, $q\nmid B$, and $\gcd(X,Y)=\gcd(q^r,B^r)=1$.

The identity

The binomial theorem gives

$$(X+Y)^r=\sum_{j=0}^r\binom rjX^{r-j}Y^j ,\qquad (X-Y)^r=\sum_{j=0}^r(-1)^j\binom rjX^{r-j}Y^j ,$$

and subtracting cancels the even-$j$ terms and doubles the odd ones:

$$(X+Y)^r=(X-Y)^r+\sum_{j\in J}2\binom rjX^{r-j}Y^j .$$

Replacing the $j=3$ term by $t$ terms, with its coefficient split as $2\binom r3=C=v_1+\cdots+v_t$, gives the manuscript's display (1):

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

The right side has $1+(|J|-1)+t=|J|+t=r-2$ summands. Every summand is a positive integer: $X-Y\ge1$, so $(X-Y)^r\ge1$, and each remaining summand is a product of positive integers. Name the summands $a_1,\ldots,a_{r-2}$ in the order displayed, with $a_1=(X-Y)^r$, and put $N:=(X+Y)^r$; then $a_1+\cdots+a_{r-2}=N$.

Every summand and the sum are r-powerful

$N=(X+Y)^r$ and $a_1=(X-Y)^r$ are $r$-th powers of positive integers, hence $r$-powerful (Definitions). Every other summand has the form $cX^aY^b$ with $a\ge0$, $b\ge1$, and $c$ a positive integer all of whose prime divisors lie in $P$: the binomial summand with index $j$ has $c=2\binom rj$, $a=r-j$ and $b=j\ge1$, and each split summand has $c=v_\ell$, $a=r-3$ and $b=3$. Let $p$ be a prime dividing $cX^aY^b$. Then $p\mid c$, $p\mid X$ or $p\mid Y$.

• If $p\mid c$ or $p\mid Y$, then $p\in P$: every prime of $c$ is in $P$, and the primes of $Y=B^r$ are those of $B$, which are exactly $P$. Then $p\mid B$, so $v_p(Y^b)=rb\,v_p(B)\ge rb\ge r$, and $v_p(cX^aY^b)\ge r$.
• If $p\mid X=q^r$, then $p=q$. Since $q\notin P$, $q\nmid c$ and $q\nmid B$; so $q\mid cX^aY^b$ forces $a\ge1$, and $v_q(cX^aY^b)=v_q(X^a)=ra\ge r$.

So every prime divisor of $cX^aY^b$ occurs to exponent at least $r$: the summand is $r$-powerful.

The summands are jointly coprime

Suppose a prime $p$ divides every one of $a_1,\ldots,a_{r-2}$. From $p\mid a_1=(X-Y)^r$ it follows that $p\mid X-Y$. Since $r-2\ge4$ there is a second summand, of the form $cX^aY^b$ above; $p$ divides it, so $p\mid c$, $p\mid X$ or $p\mid Y$, and in the first case $p\in P$, so $p\mid B\mid Y$. Thus in every case $p\mid XY$, so $p\mid X$ or $p\mid Y$. If $p\mid X$ then $p\mid X-(X-Y)=Y$; if $p\mid Y$ then $p\mid(X-Y)+Y=X$. Either way $p\mid\gcd(X,Y)=1$, a contradiction. This is the manuscript's step "$\gcd(X-Y,XY)=1$ since $\gcd(X,Y)=1$". Hence $\gcd(a_1,\ldots,a_{r-2})=1$.

Infinitely many tuples

For the fixed $r$, the data $J$, $t$, $v_1,\ldots,v_t$, $P$ and $B$ are fixed, and the construction depends only on the choice of the prime $q>B$. There are infinitely many primes, hence infinitely many primes $q>B$. Distinct primes $q<q'$ give $X=q^r<q'^r=X'$ and, with the same $Y$, $N=(X+Y)^r<(X'+Y)^r=N'$. So the values $N$ are pairwise distinct, and therefore so are the tuples. This gives infinitely many tuples with the required sum, positivity, powerfulness and joint coprimality.

Distinctness of the summands (compilation fill)

The theorem asserts that $a_1,\ldots,a_{r-2},N$ are pairwise distinct; the manuscript's proof does not argue this. The following argument, which the library's result page also records, supplies it. Compare $q$-adic valuations. For $j\in J\setminus\{3\}$,

$$v_q\Bigl(2\binom rjX^{r-j}Y^j\Bigr)=r(r-j) ,$$

because $q\nmid2\binom rj$ (its primes lie in $P$) and $q\nmid Y$; likewise $v_q(v_\ell X^{r-3}Y^3)=r(r-3)$ for each $\ell$; and $v_q((X-Y)^r)=0$, because $q\mid X$ and $q\nmid Y$ give $q\nmid X-Y$. Hence:

• two binomial summands with different indices $j$ have different valuations;
• a binomial summand (index $j\ne3$) and a split summand have different valuations, $r(r-j)\ne r(r-3)$;
• two split summands $v_\ell X^{r-3}Y^3$ and $v_mX^{r-3}Y^3$ are equal only if $v_\ell=v_m$, that is, only if $\ell=m$;
• $(X-Y)^r$ differs from every summand of positive valuation. The only other summand of valuation $0$ is the binomial summand with $j=r$, present exactly when $r$ is odd, which equals $2Y^r$. If $(X-Y)^r=2Y^r$ then $((X-Y)/Y)^r=2$ with $(X-Y)/Y$ rational, which is impossible: writing $(u/w)^r=2$ with coprime positive integers $u,w$ gives $u^r=2w^r$, so $2\mid u$, so $2^r\mid2w^r$, so $2\mid w$ because $r\ge2$, contradicting coprimality.

Finally $N$ is a sum of $r-2\ge4$ positive integers and so exceeds each of them. Thus all $r-1$ numbers are pairwise distinct, which completes the proof of Theorem 1. $\square$

Illustration (not in the source)

At $r=6$: $J=\{1,3,5\}$, $t=1$, $C=40$, $v_1=40$; the coefficients $2\binom61=2\binom65=12$ and $v_1=40$ have prime set $P=\{2,3,5\}$, so $B=30$; with $q=31$, $X=31^6$ and $Y=30^6$, the identity reads

$$(X+Y)^6=(X-Y)^6+12X^5Y+40X^3Y^3+12XY^5 ,$$

four summands for $r-2=4$. This instance, and the analogous ones for $7\le r\le12$ with the least prime $q>B$, were checked by exact integer arithmetic while writing this page (the sum, the $r$-powerfulness of every term, the joint gcd, and the distinctness); the check is a sanity check and is not retained as evidence.

Boundary

What the proof uses. The binomial theorem, the elementary valuation facts stated in Definitions, and the infinitude of primes. No analytic input and no other result of the manuscript enter.

What the theorem does not give. Nothing at $r=4$ or $r=5$. Before any splitting, the identity has $|J|+1=\lceil r/2\rceil+1$ summands, which is $3>2=r-2$ at $r=4$ and $4>3=r-2$ at $r=5$; splitting a coefficient only adds summands, and merging two monomials would destroy the monomial shape that the powerfulness step relies on. The manuscript claims nothing at $r\le5$, and this page adds nothing there: the finiteness question of Problem 939 stays open at $r=4$ and $r=5$, and the existence question at $r=4$.

Formal counterpart. The tuple form of this statement, with positive, IsPowerful, injective summands and joint coprimality as "no prime divides every summand", is the theorem infinite_rpowerful_sum_tuples of the Lean file that the Conjectures.io card records; the theorem infinite_rpowerful_sums of the same file states the infinitude for the set of sums $N$, which the proof above also gives. Neither states that $N$ differs from the summands, which positivity supplies. Both are kernel-checked by that site and not built in this repository; neither is a native L-claim here, and this reconstruction was not compared with the Lean text line by line.