The medium-modulus gap in divisor pairing

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This author-recorded argument by Plasma AI concerns Problem 975 at the formulation and standing recorded there. It supplies the deduction or obstruction below, without independent acceptance or a change to the problem's standing.

The target is an asymptotic with a fixed positive leading constant, not just order $X\log X$, for an irreducible integer polynomial eventually positive. Erdős's held 1952 introduction and root-count definitions were read; its general result gives the order, whereas the exact asymptotic is recorded for degrees at most two.

Let $d=\deg f$, discard a fixed initial interval so that $f$ is positive and increasing, and let $\rho_f(q)$ count roots modulo $q$. Pairing divisors gives

$$\tau(f(n))=2\sum_{q\le\sqrt{f(n)}}1_{q\mid f(n)} -1_{f(n)\text{ is a square}}.$$

After interchanging the finite sums, the inner count lies on an interval $t(q)\le n\le X$, where $t(q)$ is the first allowed integer with $f(n)\ge q^2$. Counting each root progression by length gives its expected length times $\rho_f(q)/q$ with error $O(\rho_f(q))$. The moduli run to $Y\asymp X^{d/2}$. Even granting the favorable average $\sum_{q\le Y}\rho_f(q)=O(Y)$, summing these individual errors gives only $O(X^{d/2})$. For $d=2$ this is $O(X)$, small against $X\log X$; for $d\ge3$ this estimate is too large. In degree three the range $X<q\ll X^{3/2}$ is already beyond one full period of the counting interval.

This is a limitation of the stated proof method, not a lower bound on the actual error and not a refutation of the requested asymptotic. A finite sample of polynomial values cannot certify the cancellation of these endpoint errors. A discriminating continuation: a theorem controlling their sum in the cubic medium-modulus range, or a divisor decomposition that avoids it. Stop if only individual $O(\rho_f(q))$ errors are available.

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