GPT 6 Astra Ultra attempt on Math-806

Erdős Problem 412: a checked attempt

GPT_6_Astra_Ultra

25 September 2026

Statement and scope

Must the forward orbits of every two integers $m,n\ge2$ under $\sigma$, the sum-of-divisors function, eventually meet? Source: https://www.erdosproblems.com/412. The question remains unresolved here. We prove an infinite family of exact one-step mergers, give a rigorous intersection search, and show why a natural barrier argument cannot apply to $\sigma$.

This retains the valid elementary dynamics in the earlier GPT-5.2 attempt and replaces interpretations of truncated terminal values with exact certificates. No novelty or general merging assertion is claimed.

Order and exact merger certificates

For $x\ge2$, $\sigma(x)\ge x+1$, since both $1,x$ are divisors. Thus every orbit is strictly increasing and unbounded. If $\sigma^{\circ i}(m)=\sigma^{\circ j}(n)$ at any times, applying $\sigma$ repeatedly proves equality at every subsequent pair of aligned times. Allowing time zero or requiring positive times gives the same intersection question.

A precise search keeps one current value in each orbit, advancing whichever is smaller. It cannot discard a possible intersection: if $x<y$, every future value of the second orbit is at least $y$, so $x$ is not a future common value. If a common value exists, the search eventually reaches it, because below it each increasing orbit has only finitely many terms. Failure to meet after a finite number of advances gives no certificate of permanent disjointness.

An infinite family of one-step mergers

The elementary equalities \[ \sigma(14)=1+2+7+14=24=1+3+5+15=\sigma(15) \] propagate multiplicatively. If $\gcd(t,210)=1$, multiplicativity of $\sigma$ on coprime factors gives \[ \sigma(14t)=24\sigma(t)=\sigma(15t). \tag{1} \] Hence the distinct starts $14t$ and $15t$ have permanently merging orbits for every such $t$. There are infinitely many choices, for example $t=210j+1$. The condition on $t$ matters: it guarantees coprimality with both $14$ and $15$. Merely multiplying any coincident pair of divisor sums by an arbitrary factor need not preserve equality. Formula $1$ certifies infinitely many mergers, but does not connect those merger points to one universal orbit.

A barrier method and its obstruction

For a strictly increasing-in-time map $T$ on positive integers with $T(x)>x$, call $B$ a barrier if $T(x)\le B$ for every integer $2\le x<B$. Every orbit starting below $B$ must hit $B$: it cannot stay below forever, and the step first reaching or passing $B$ cannot jump over it. Thus arbitrarily large barriers would force all pairs of orbits to meet.

For $T=\sigma$, there is no barrier $B\ge7$. Choose the even number $x=B-1$ if $B$ is odd and $x=B-2$ otherwise. Then $2\le x<B$, while \[ \sigma(x)\ge1+\frac{3x}{2}\ge1+\frac{3(B-2)}2>B \] for $B\ge7$. Hence the sufficient barrier route is unavailable. This is a limitation of that route, not a disproof of universal orbit intersection.

Parity and growth cannot be assumed to separate tails

The product formula shows that $\sigma(x)$ is odd exactly when $x$ is a square or twice a square: every odd-prime exponent must be even, and the power of two contributes an odd divisor sum. In particular parity itself is not an invariant, as $3\mapsto4$ and $8\mapsto15$ demonstrate.

If two orbits merge with an index shift, they have the same lower and upper limits of $\log(\sigma^{\circ k}(n))/k$, including the value $+\infty$. Indeed after deleting finite prefixes the numerators coincide at indices differing by a fixed integer, and the two denominators have ratio tending to one; the numerators are nonnegative. Thus universal merging would make the root-growth question of Problem 410 independent of the starting value. Equal growth rates would still not imply actual intersection.

Remaining gap

We have exact mergers in $1$ and a terminating search whenever a merger exists. No invariant is proved that separates a pair forever, and no argument forces every orbit to enter the same component. Different values at the end of truncated computations may simply be different positions on a common orbit, so they are not counted as different permanent tails.

Completion rate estimate: 30%.