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Source. Bajpai--Bennett--Chan, accepted author manuscript (June 26, 2023), Theorem 1.1 and its proof, pp. 3 and 9--10.

Statement. Assume the $abc$ conjecture. Let $m\geq3$ and $k\geq2$ be integers, and let $N,d$ be positive integers such that

$$N,N+d,\ldots,N+(m-1)d$$

are $k$-full. For every $\varepsilon>0$,

$$\gcd(N,d)\gg_{\varepsilon,k,m} \max\{N,d\}^{ \frac{m(1-1/k)-2}{m(1-1/k^2)-2}-\varepsilon}, \tag{1}$$ $$d\gg_{\varepsilon,k,m} N^{\frac{m(1-1/k)-1}{m(1-1/k^2)-1}-\varepsilon}, \tag{2}$$

and

$$N\gg_{\varepsilon,k,m} d^{\frac{m(1-1/k)+1/k-2} {m(1-1/k^2)+1/k-2}-\varepsilon}. \tag{3}$$

If $m\geq2k-1$, replacing $1/k^2$ by $1/(2k-1)$ gives the stronger versions of (2) and (3), and also the stronger version of (1) unless $(m,k)=(3,2)$. At that one endpoint, the denominator printed in the paper's strengthened gcd formula (1.5) is zero, so that displayed expression is undefined; the baseline bound (1) remains valid.

The gcd exponent in the baseline bound is positive except for

$$(m,k)\in\{(3,2),(3,3),(4,2)\}.$$

These are conditional restrictions. The construction resolving Problem 937 is the unconditional exceptional case $(m,k)=(4,2)$ in Theorem 1.2.

Dependencies. Lemma 2.1, Lemma 2.2, and Lemma 3.1.

The $abc$ input. For every $\eta>0$, the conjecture supplies a constant $\kappa(\eta)$ such that positive coprime integers $a,b,c$ with $a+b=c$ satisfy

$$c<\kappa(\eta)\operatorname{Rad}(abc)^{1+\eta}.$$

Proof. Put $\ell=m-1$. Lemma 3.1 gives

$$\prod_{\substack{1\leq j\leq m-1\\j\ {\rm odd}}} (N+jd)^{\binom{m-1}{j}} = \prod_{\substack{0\leq j\leq m-1\\j\ {\rm even}}} (N+jd)^{\binom{m-1}{j}} +d^{m-1}G_d(N). \tag{4}$$

Let $t=\gcd(N,d)$. It is $k$-full because $t=\gcd(N,N+d)$ and the gcd of two $k$-full numbers is $k$-full. Write $N_0=N/t$ and $d_0=d/t$. Dividing (4) by $t^{2^{m-2}}$ gives the same identity in $N_0,d_0$. Call its odd- and even-index products $O$ and $E$.

The third term is nonzero and positive. Indeed, the finite-difference integral for the logarithm of their ratio is

$$\log\frac OE =(m-2)!\int_{[0,d_0]^{m-1}} \frac{dt_1\cdots dt_{m-1}} {(N_0+t_1+\cdots+t_{m-1})^{m-1}}>0. \tag{5}$$

Thus $O>E$ and $d_0^{m-1}G_{d_0}(N_0)=O-E>0$. This verifies the positivity hypothesis needed for the $abc$ equation, rather than assuming that $G$ has a fixed sign.

Let $D=\gcd(O,E)$. For each prime $p$, at most one normalized term $N_0+jd_0$ can have $p$-adic valuation greater than $\lfloor\log_p(m-1)\rfloor$: two such terms would make $p^{\lfloor\log_p(m-1)\rfloor+1}$ divide their nonzero index difference. The exceptional term occurs in only one of $O,E$, while each product has total binomial weight $S=2^{m-2}$. Consequently

$$D\mid\operatorname{lcm}(1,\ldots,m-1)^S \leq(m-1)^{(m-1)S}. \tag{6}$$

In particular, $D\ll_m1$. Since $D$ also divides $O-E$, the positive integers

$$a=E/D,\qquad b=d_0^{m-1}G_{d_0}(N_0)/D,\qquad c=O/D$$

are pairwise coprime and satisfy $a+b=c$.

First suppose $d\leq N$. Every normalized term is at most $mN/t$, and Lemma 2.1 gives

$$\operatorname{Rad}(N_0+jd_0)\ll_m \frac{N^{1/k}}{t^{1/k^2}}.$$

The radical of $abc$ is bounded by the product of the radicals of the $m$ normalized progression terms, one factor $d/t$, and the absolute value of $G_{d_0}(N_0)$. Since $G$ has degree $S-m+1$, $|G_{d_0}(N_0)|\ll_m(N/t)^{S-m+1}$. Also $O\geq(N/t)^S$. Applying $abc$ with exponent $1+\eta$ gives

$$\frac{N^S}{t^S D} \ll_{\eta,m} \left\{ \left(\frac{N^{1/k}}{t^{1/k^2}}\right)^m \frac dt \frac{N^{S-m+1}}{t^{S-m+1}} \right\}^{1+\eta}. \tag{7}$$

Write the expression in braces as $B$. For fixed $m,k$, every one of its factors is bounded by a fixed power of $X=\max\{N,d\}$, uniformly in $t$, so $B\ll_{m,k}X^C$ for some $C=C(m,k)$. Hence $B^{1+\eta}\ll B X^{C\eta}$. Choose $\eta$ sufficiently small in terms of the desired $\varepsilon,m,k$. Rearranging (7), with the resulting power $X^{C\eta}$ absorbed into $N^\varepsilon$ because $X=N$ in this case, gives

$$N^{m(1-1/k)-1-\varepsilon} \ll d\,t^{m(1-1/k^2)-2}. \tag{8}$$

Using $d\leq N$ in (8) gives (1) with $\max\{N,d\}=N$; using $t\leq d$ gives (2), after reducing $\eta$ once more to account for the fixed positive denominator in the final exponent. The exponent in (3) lies in $[0,1]$, so (3) is automatic in this case from $d\leq N$.

Now suppose $d>N$. In the odd product every index is positive, so $O\geq(d/t)^S$. Among the normalized terms, the $j=0$ term contributes the sharper radical $N^{1/k}/t^{1/k^2}$ and the other $m-1$ terms contribute $d^{1/k}/t^{1/k^2}$. Homogeneity gives $|G_{d_0}(N_0)|\ll_m(d/t)^{S-m+1}$. Thus

$$\frac{d^S}{t^S D} \ll_{\eta,m} \left\{ \left(\frac{d^{1/k}}{t^{1/k^2}}\right)^{m-1} \frac{N^{1/k}}{t^{1/k^2}} \frac dt \frac{d^{S-m+1}}{t^{S-m+1}} \right\}^{1+\eta}. \tag{9}$$

The same uniform-power argument, now with $X=d$, lets us choose $\eta$ so that rearrangement yields

$$d^{m(1-1/k)+1/k-2-\varepsilon} \ll N^{1/k}t^{m(1-1/k^2)-2}. \tag{10}$$

Using $N<d$ in (10) gives (1), now with maximum $d$, and using $t\leq N$ gives (3). The exponent in (2) lies in $(0,1)$, so (2) is automatic from $d>N$.

Suppose finally that $m\geq2k-1$. In Lemma 2.2's notation,

$$\prod_{i=k}^{2k-1}a_{i,j} \leq (N+jd)^{1/k}, \tag{11}$$

because every $a_{i,j}\geq1$ and every exponent $i\geq k$. Thus that lemma replaces the product contribution $t^{-m/k^2}$ in (7) and (9) by $t^{-m/(2k-1)}$, with the same numerator bounds. The two rearrangements give the strengthened formulas, subject to the zero-denominator qualification in the statement.

Source qualifications. At $(m,k)=(3,2)$ the denominator in the accepted manuscript's displayed formula (1.5) vanishes; the proof and statement above retain (1) and make no claim for that undefined strengthening. Also, the manuscript prints $D\leq(m-1)^{(m-1)^2}$, which does not track the binomial weights in (4). The valuation argument leading to (6) supplies the needed $m$-dependent bound. These are explicit compilation clarifications; no author-issued correction is asserted.

Method. A binomial product identity turns all $m$ terms into one $abc$ equation. The powerfulness hypothesis makes its radical small, while normalizing by $t=\gcd(N,d)$ tracks exactly how a primitive progression can escape the resulting bound.

Bears on. #937.