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Math-756-A3 · revision r1 · on Math-756

gpt pro 5.2 attempt on Math-756

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #313
  2. 1) RESTATE
  3. 2) KNOWN FACTS
  4. 3) PROOF STRATEGY
  5. 4) ATTEMPTED PROOF
  6. Step 1: Force $m=p_1\cdots p_k$.
  7. Step 2: Reformulate as the primary pseudoperfect equation.
  8. Step 3: Prove the chain lemma $2e$.
  9. 5) OBSTACLES
  10. 6) FINAL
  11. 7) WRITE-UP (clean, complete)

Erdős Problem #313

1) RESTATE

Determine whether there are infinitely many tuples of distinct primes $p_1<\cdots<p_k$ and an integer $m\ge 2$ such that \[ \frac{1}{p_1}+\cdots+\frac{1}{p_k}=1-\frac{1}{m}. \] Equivalently, are there infinitely many integers $m$ for which \[ \frac{1}{m}+\sum_{p\mid m}\frac{1}{p}=1, \] with the sum over prime divisors $p$ of $m$?

2) KNOWN FACTS

  1. (Denominator rigidity) If $p_1,\dots,p_k$ are distinct primes then the rational number $\sum_{i=1}^k \frac{1}{p_i}$ is in lowest terms with denominator $p_1\cdots p_k$.

  2. (For a solution, $m$ is forced) In any solution of the original equation one must have $m=p_1\cdots p_k$. Consequently there is at most one solution for each $m$.

  3. (Name of the $m$'s) The integers $m$ satisfying $\frac{1}{m}+\sum_{p\mid m}\frac{1}{p}=1$ are called primary pseudoperfect numbers.

  4. (Computational status) As of January 2026, $8$ primary pseudoperfect numbers are known: \[ 2,\ 6,\ 42,\ 1806,\ 47058,\ 2214502422,\ 52495396602,\ 8490421583559688410706771261086. \] It is unknown whether there are infinitely many, and it is unknown whether any are odd.

  5. (A simple "chain" lemma) If $m$ is primary pseudoperfect and $m+1$ is prime, then $m(m+1)$ is also primary pseudoperfect.

3) PROOF STRATEGY

A natural strategy to prove infinitude is to find a mechanism that generates new solutions from old ones. One such mechanism is the chain lemma $2e$: if $m$ is a solution and $m+1$ is prime, then $m(m+1)$ is a new solution. Thus, an infinite chain of solutions would follow from proving that the "Euclid-like" numbers $m+1$ are prime infinitely often along this recursion. No method is known.

4) ATTEMPTED PROOF

Step 1: Force $m=p_1\cdots p_k$.
Let $D:=p_1\cdots p_k$. Write \[ \sum_{i=1}^k\frac{1}{p_i}=\frac{A}{D},\qquad A:=\sum_{i=1}^k \frac{D}{p_i}. \] Fix $j\in\{1,\dots,k\}$. Modulo $p_j$ we have \[ A \equiv \frac{D}{p_j}\pmod{p_j}, \] because every term $D/p_i$ with $i\neq j$ is divisible by $p_j$. Since $D/p_j=\prod_{i\neq j}p_i$ is not divisible by $p_j$, we get $A\not\equiv 0\pmod{p_j}$. Therefore $\gcd(A,D)=1$ and $\sum_{i=1}^k\frac{1}{p_i}$ is in lowest terms with denominator exactly $D$.

On the other hand, \[ 1-\frac{1}{m}=\frac{m-1}{m} \] is in lowest terms because $\gcd(m-1,m)=1$. Hence its reduced denominator is $m$. Equality of the two fractions forces $m=D=p_1\cdots p_k$.

Step 2: Reformulate as the primary pseudoperfect equation.
Substituting $m=p_1\cdots p_k$ gives \[ \sum_{p\mid m}\frac{1}{p}=1-\frac{1}{m} \qquad\Longleftrightarrow\qquad \frac{1}{m}+\sum_{p\mid m}\frac{1}{p}=1. \]
Step 3: Prove the chain lemma $2e$.
Assume $m$ is primary pseudoperfect, so $\sum_{p\mid m} \frac{1}{p}=1-\frac{1}{m}$. Let $q:=m+1$ be prime. Then the prime divisors of $mq$ are those of $m$ together with $q$. Compute \begin{align*} \frac{1}{mq}+\sum_{p\mid mq}\frac{1}{p} &=\frac{1}{mq}+\left(\sum_{p\mid m}\frac{1}{p}\right)+\frac{1}{q} =\frac{1}{mq}+\left(1-\frac{1}{m}\right)+\frac{1}{q}\\ &=1-\frac{1}{m}+\frac{1}{q}\left(1+\frac{1}{m}\right) =1-\frac{1}{m}+\frac{m+1}{m(m+1)} =1. \end{align*} So $mq$ is primary pseudoperfect.

5) OBSTACLES

  1. The chain lemma reduces infinitude to a primality question for a very sparse sequence of "Euclid-like" integers; no known sieve/analytic method applies.
  2. Known solutions are extremely sparse and grow rapidly; currently only $8$ are known (as of Jan 2026).
  3. Even basic structural questions are open (e.g. existence of any odd primary pseudoperfect number).

6) FINAL

UNRESOLVED.
  1. What I tried: Proved the structural reduction $m=p_1\cdots p_k$ and recast the problem as infinitude of primary pseudoperfect numbers; recorded a standard "chain" construction $m\mapsto m(m+1)$ when $m+1$ is prime.
  2. Where it fails / stuck: Infinitude would follow from producing infinitely many primes of the form $m+1$ along such constructions, or from an entirely different infinite-family construction. No known argument provides this.
  3. What might work next: (a) Search for additional recursive constructions beyond the $m+1$ prime lemma; (b) study congruence restrictions/modular constraints from the equation to narrow candidates; (c) use graph-theoretic/structural formulations from the literature on pseudoperfect numbers.
  4. Confidence: High confidence in the reductions and lemmas proved here; low confidence about any claim toward infinitude (open problem).

7) WRITE-UP (clean, complete)

The equation $\sum_{i=1}^k \frac{1}{p_i}=1-\frac{1}{m}$ forces $m=p_1\cdots p_k$ because the left-hand side is a reduced fraction with denominator $p_1\cdots p_k$ while the right-hand side is reduced with denominator $m$. Thus solutions correspond exactly to primary pseudoperfect numbers $m$ satisfying $\frac{1}{m}+\sum_{p\mid m} \frac{1}{p}=1$. Only finitely many such $m$ are currently known (eight as of Jan 2026), and it is unknown whether there are infinitely many. A simple construction is: if $m$ is primary pseudoperfect and $m+1$ is prime then $m(m+1)$ is primary pseudoperfect. This yields some "chains" of solutions, but proving infinitely many steps in any chain would require infinitely many primes of the form $m+1$ along that chain, which is out of reach.