Powerful integers and consecutive-product tests

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evidence/: Exact integer checks for the retained square quotients, Pell interval and product witnesses; all domains and limitations are stated on the owning research page.

These author-recorded arguments by Plasma AI distinguish finite certificates from unbounded conclusions in several problems about powerful integers and products of consecutive integers. A positive integer is powerful when every prime divisor has exponent at least two; this includes 1. Each argument states the parameter range it reaches. No complete catalog resolution or independent verification is claimed.

Fixed square quotients at length two

For Problem 686, fix a square $N=a^2$, $a\ge2$, and length $k=2$. The equation is $(m+1)(m+2)=a^2(n+1)(n+2)$ with $n\ge0$ and $m\ge n+2$. Put $X=2m+3$, $Y=2n+3$. Then both are odd, $Y\ge3$, $X\ge Y+4$, and $X^2-a^2Y^2=1-a^2$. Since $X<aY$, the positive factors $u=aY-X$, $v=aY+X$ satisfy

Conversely every divisor $1\le u\le\sqrt{a^2-1}$, with $v=(a^2-1)/u$, gives a solution precisely when these formulas give odd integers satisfying the stated inequalities. Recover $n=(Y-3)/2$, $m=(X-3)/2$. This proves an exhaustive finite test for this length, without an arbitrary height cutoff.

The evidence exhausts those seven divisor domains and checks each returned pair in the original equation. Empty rows do not exclude other lengths. A settling finite test for a square still needs a proved bound on all possible lengths, or a uniform construction at some length.

Adjacent and separated blocks

In Problem 389, put $s=n-1$ and $t=s+k$. The adjacent-block quotient is $Q=s!(2t-s)!/(t!)^2$. Legendre's formula gives the exact integrality test

The sums are finite. Direct multiplication checks $(n,k)=(1,1),(2,5),(3,4)$. This formulation tests a given $k$; it does not construct or bound one for arbitrary $n$. The theorem text of Ulas's A note on Erdős–Straus and Erdős–Graham divisibility problems, IJNT 9 (2013), 583–599, doi:10.1142/S1793042112501497, remains uninspected here. Its title alone does not determine whether it covers this universal quantifier.

For Problem 930, separate nonsquare block products need not have a nonsquare product:

The disjoint blocks each have length three. This defeats the inference from two separate one-block exclusions, not the existence of a larger threshold $K(2)$. A successful extension must control each prime's total exponent across the blocks.

A certified Pell interval

For Problem 938, the construction in van Doorn's paper uses $x^2-343y^2=2$ and the powerful progression $(x-2)^2,(x-1)^2,x^2-2=7^3y^2$. Its common difference is $2x-3$. The seed $x=11427$, $y=617$ gives the closed interval $[130530625,130576327]$.

Every powerful integer has a unique representation $a^2b^3$ with $b$ squarefree. Hence the complete enumeration in a positive interval $[L,H]$ has domain

The upper bound on $b$ for the seed is 507. The exact output is:

The four intruders disprove consecutiveness only at this seed. Van Doorn's Theorem 4 supplies infinitely many good indices for each fixed squarefree $b\ge648560$; it does not exchange those quantifiers to avoid every unwanted core at one index. A useful continuation needs simultaneous avoidance and a uniform tail estimate, or an obstruction for every index. No such estimate or index cutoff is supplied by this finite check.

Smooth parts of neighbors

Let $A(m)=2^{v_2(m)}3^{v_3(m)}$, as in Problem 933. At $n=1487503359=311\cdot3^{14}$ one has $v_2(n+1)=15$. The inequalities $n<2^{31}$ and $32768>3\cdot311\cdot31$ prove exactly, without floating-point logarithms,

This is a counterexample to a uniform bound by that constant, not proof of an infinite limsup. On the family $n=2^t$ with odd $t$, lifting the exponent gives $v_3(2^t+1)=1+v_3(t)$ and the ratio is $3^{1+v_3(t)}/(t\log2)\le3/\log2$, with equality for $t=3^j$. For even $t$ the ratio is $1/(t\log2)$. Thus this entire pure-power family is bounded. A variable-cofactor construction must prove unbounded normalized values, beyond finitely many records.

What the Pell construction proves about a powerful part

Let $Q_2(m)$ retain every prime power in $m$ whose exponent is at least two. The existing Pell solution of the middle part of Problem 935 has an explicit parameter choice. Put $\alpha=3+\sqrt8=(1+\sqrt2)^2$ and $\alpha^k=x_k+y_k\sqrt8$. Then $n_k=8y_k^2$ and $n_k+1=x_k^2$ are powerful.

For a prime $p\equiv5\pmod8$, Frobenius in $\mathbb Z[\sqrt2]/(p)$ sends $\sqrt2$ to $-\sqrt2$, giving $\alpha^{(p+1)/2}\equiv-1\pmod p$. Write $m=(p+1)/2$, which is odd. Raising $\alpha^m=-1+p\delta$ to the power $p$ gives $\alpha^{mp}\equiv-1\pmod{p^2}$. At the positive integer

one has $\alpha^{2k}\equiv-\alpha$ and $\alpha^{-2k}\equiv-\alpha^{-1}\pmod{p^2}$. Since $4x_k^2=\alpha^{2k}+\alpha^{-2k}+2$ and $\alpha+\alpha^{-1}=6$, it follows that $p^2\mid n_k+2$. Prime valuations give $Q_2(AB)\ge Q_2(A)Q_2(B)$, so

Dirichlet's theorem supplies infinitely many such primes. Both $k(p)$ and the positive Pell sequence increase, so the inputs are distinct and tend to infinity. Adding factors through $n_k+\ell$, $\ell\ge2$, preserves the lower bound. This reconstructs the known middle conclusion; it supplies no upper estimate and changes no source claim's standing.

For each fixed $\ell\ge2$, the first proposed upper estimate would imply the third conclusion: choose $\epsilon=(\ell-1)/2$ to bound the ratio to $n^{\ell+1}$ by $n^{-(\ell-1)/2}$. For $\ell=1$ the first estimate is already immediate from $Q_2(n(n+1))\le n(n+1)$. The remaining upper problem begins at $\ell=2$.

Conditional finiteness keeps its premise

Assume the abc conjecture in the form $c\le C_\epsilon\operatorname{rad}(abc)^{1+\epsilon}$ for coprime positive integers $a+b=c$. If $B=2^n\pm1$ is powerful, then $\operatorname{rad}(B)\le\sqrt B$ and, with $\epsilon=1/2$,

Thus both signs occur powerfully only finitely often. For powerful $B=n!\pm1$, the elementary primorial bound $\prod_{p\le n}p\le4^n$ similarly gives $B^{1/4}\le C_{1/2}8^n$. Since $B\ge n!/2$ for $n\ge3$ and $n!\ge\lfloor n/2\rfloor^{\lfloor n/2\rfloor}$, this too is possible only finitely often. All four cases of Problem 936 follow under abc. No effective value of its constant or unconditional bound is supplied.

The orientation in Problem 366 is also essential: it asks for powerful $n$ followed by 3-full $n+1$. Then $\operatorname{rad}(n(n+1))\le(n+1)^{5/6}$, and abc with $0<\epsilon<1/5$ implies finiteness. Reversing the two types changes the question; known reverse-oriented pairs are not witnesses here.

The loss when cube factors vary

L24 gives $R(N)\le N^{2/5+o(1)}$ for the ordered sum count in Problem 943. For each fixed pair of squarefree cube factors its imaginary-quadratic norm argument is uniformly subpower. Summing those factors is the remaining loss. When all four parameters $a,b,c,d$ in $a^2b^3+c^2d^3=N$ are of order $N^{1/5}$, each of the three dyadic bounds used in the proof is of order $N^{2/5}$ up to subpower factors. Repeating the fixed-core count cannot by itself remove this loss. A sharper estimate in that balanced range, or subpower control of the core pairs occurring for each $N$, would change the research decision. L24 does not prove the desired $N^{o(1)}$ bound.

Evidence and standing

The finite input is finite_cases.json. The checker uses exact integers, exhaustive divisors and the full powerful-number domain in the displayed closed interval. From the repository root, run

uv run --no-sync python wiki/research/powerful_numbers_and_products/evidence/main.py

Runtime is a few seconds. The checker covers the seven square roots, one Pell interval, the smooth-part counterexample, three divisibility witnesses and the separated-block witness. It certifies no recurrence tail or unbounded catalog statement. The mathematical reductions and the finite replay are author work; no verification tier is asserted by this note.