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Source. Bajpai--Bennett--Chan, accepted author manuscript (June 26, 2023), Theorem 1.2 and Sections 5.1--5.3, pp. 3 and 11--19.

Statement. For each

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

there are infinitely many positive integers $N,d$ for which the $m$-term progression $N,N+d,\ldots,N+(m-1)d$ consists of $k$-full integers with $\gcd(N,d)=1$. In the four-term squarefull case the construction makes the terms pairwise coprime, so this case resolves Problem 937 unconditionally.

Dependency for $(4,2)$. Proposition 5.2.

Proof. For $(m,k)=(3,2)$, the identities

$$X^2+Y^2=2Z^2$$ $$X=a^2-b^2-2ab,\quad Y=a^2-b^2+2ab,\quad Z=a^2+b^2,$$

give a family of solutions whenever $\gcd(a,b)=1$ and $a,b$ have opposite parity. Taking $b=1$ and any sufficiently large even $a$ gives infinitely many positive triples with $X<Y$. They are pairwise coprime: a common odd prime of any two of $X,Y,Z$ would, from the displayed formulas, divide both $a$ and $b$, while $2$ is excluded because all three values are odd. Then

$$N=X^2,\qquad d=Z^2-X^2$$

gives the progression $X^2,Z^2,Y^2$.

For $(m,k)=(3,3)$, start from

$$(X_0,Y_0,Z_0)=(37,17,7),\qquad X_0^3+Y_0^3=2\cdot3^4Z_0^3,$$

and iterate

$$\begin{aligned} X_{i+1}&=X_i(X_i^3+2Y_i^3),\\ Y_{i+1}&=-Y_i(2X_i^3+Y_i^3),\\ Z_{i+1}&=Z_i(X_i-Y_i)(X_i^2+X_iY_i+Y_i^2). \end{aligned} \tag{1}$$

Direct expansion shows that every triple still satisfies $X_i^3+Y_i^3=2\cdot3^4Z_i^3$, while

$$|Z_{i+1}|=|X_i-Y_i|(X_i^2+X_iY_i+Y_i^2)|Z_i|>|Z_i|,$$

so the triples are distinct. Coprimality is preserved: if a prime divided both new $X$ and new $Y$, it cannot divide either old $X$ or old $Y$; it must therefore divide both $X^3+2Y^3$ and $2X^3+Y^3$. Their linear combinations force the prime to be $3$, but $X\equiv-Y\pmod3$ and $3\nmid XY$ make those factors nonzero modulo $3$. The defining equation then also makes $Z$ coprime to $X$ and $Y$.

There are infinitely many all-positive triples in this orbit. Swapping $X,Y$ and changing all three signs preserve the equation, coprimality, and the recurrence orbit up to the same symmetries. Thus the only unresolved sign pattern may be arranged as $X>0$, $Y<0$, $Z>0$, and $X>|Y|$. Put

$$R=\frac{X}{|Y|}>1.$$

If $R>2^{1/3}$, the formulas in (1) make the next $X,Y$ have the same sign. Otherwise $1<R<2^{1/3}$; after one recurrence step and the permitted swap and sign normalization, the new ratio is

$$R'=\frac{2R^3-1}{R(2-R^3)}.$$

It satisfies

$$R'-1 =(R-1)\frac{(R+1)^3}{R(2-R^3)} >8(R-1), \tag{2}$$

because $(R+1)^3>8$ and $R(2-R^3)<1$ on this interval. If the signs never agree, iterating (2) eventually forces $R\geq2^{1/3}$, a contradiction. Thus every starting index is followed by a later same-sign pair. Applying this argument after each such index, while $|Z_i|$ strictly increases, supplies infinitely many positive triples.

For any positive triple, order $X^3,Y^3$ so that the smaller is first. The identity says that $3^4Z^3$ is their average, so these three positive cubefull integers form a progression. They are pairwise coprime: neither $X$ nor $Y$ is divisible by $3$, and any prime common to $Z$ and one of them would divide the other through the defining equation.

Finally, Proposition 5.2 proves the $(4,2)$ case and verifies the stronger pairwise-coprime conclusion.

Method. The two three-term cases come from elementary conic and cubic recurrences. The four-term case is the elliptic-curve construction carrying the substantive content for Problem 937.

Bears on. #937.