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Source. Bajpai--Bennett--Chan, accepted author manuscript (June 26, 2023), Section 5.3 and Proposition 5.2, pp. 12--19.

Statement. There are infinitely many four-term arithmetic progressions of positive, pairwise-coprime squarefull numbers. More precisely, infinitely many have, up to reversing the order, the form

$$x^2,\quad y^2,\quad z^2,\quad 73^3w^2$$

with $x,y,z,w$ pairwise coprime.

Dependency. Proposition 5.1.

Proof. Begin with coprime integers $a,b$ of opposite parity and put

$$N_0=a^2-b^2+2ab,\qquad r=a^2+b^2,\qquad s=a^2-b^2-2ab,$$ $$N=N_0^2,\qquad d=4ab(b^2-a^2).$$

The standard parametrization of $N_0^2+s^2=2r^2$ gives

$$N+d=r^2,\qquad N+2d=s^2,$$

while direct expansion gives

$$N+3d=F(a,b) =a^4-8a^3b+2a^2b^2+8ab^3+b^4. \tag{1}$$

It remains to produce infinitely many admissible $a,b$ for which $F(a,b)=73^3c^2$.

First impose $F(a,b)=73z^2$. With $X=a/b$ and $Y=z/b^2$, the quartic is

$$X^4-8X^3+2X^2+8X+1=73Y^2. \tag{2}$$

Here is the full birational calculation used in the source. Set $X=-2+1/u$ and define

$$v=u^2Y,\qquad H=u^2-\frac{64}{73}u+\frac{653}{73^2}.$$

After multiplying (2) by $u^4/73$ and completing the square, one obtains

$$v^2=H^2-\frac{1680}{73^3}u-\frac{37392}{73^4}. \tag{3}$$

Put $T=H-v$ and $S=uT$. Substitution in (3) gives

$$2S^2-\frac{128}{73}ST-\frac{1680}{73^3}S =T^3-\frac{1306}{73^2}T^2+\frac{37392}{73^4}T.$$

The scaling

$$x=2\cdot73^2T,\qquad y=4\cdot73^3S$$

therefore produces

$$E:y^2-128xy-3360y=x^3-2612x^2+149568x. \tag{4}$$

Conversely,

$$T=\frac{x}{2\cdot73^2},\qquad u=\frac{y}{146x},\qquad v=H-T,$$

and then $X=-2+1/u$ and $Y=v/u^2$. Thus (2) and (4) are birational away from the finitely many zeros or poles in these formulas.

The source computes

$$E(\mathbb Q)\cong(\mathbb Z/2\mathbb Z)^2\times\mathbb Z^2$$

and lists four generators: $T_1=(-1176,-73\cdot1008)$ and $T_2=(-300,-73\cdot240)$, both of order $2$, and

$$P_1=(-976,-49344),\qquad P_2=(-408,-30192).$$

Only the infinite order of $P_1$ is needed here, and it has a short independent certificate. The discriminant of (4) is $2^{20}3^2 73^6$. At the good primes $17$ and $23$, direct group-law calculation gives orders $5$ and $8$ for the reductions of $P_1$. If $P_1$ had finite rational order $M$, injectivity of good reduction on prime-to-$p$ torsion would force both $M=5\cdot17^\alpha$ and $M=8\cdot23^\beta$, which is impossible. We use the standard reduction theorem in Silverman, The Arithmetic of Elliptic Curves, 2nd ed., Proposition VII.3.1; an accessible statement appears on slide 64 of Silverman's 2006 Wyoming lectures.

Write the multiples of $P_1$ as

$$nP_1=\left(\frac{\phi_n}{\psi_n^2}, \frac{\Omega_n}{\psi_n^3}\right).$$

For $x_0=-976$, the source's division polynomials start with

$$\psi_0=0,\quad\psi_1=1,\quad \psi_2=2^5\cdot5\cdot11\cdot13=22880.$$

Using

$$\begin{aligned} b_2&=5936,& b_4&=729216,\\ b_6&=11289600,& b_8&=-116185227264, \end{aligned}$$

their next values are defined by

$$\psi_3 =3x_0^4+b_2x_0^3+3b_4x_0^2+3b_6x_0+b_8 =-861920436224$$

and

$$\begin{aligned} \psi_4=\psi_2(&2x_0^6+b_2x_0^5+5b_4x_0^4+10b_6x_0^3 +10b_8x_0^2\\ &+(b_2b_8-b_4b_6)x_0+b_4b_8-b_6^2)\\ &=-19111064818388639416320. \end{aligned}$$

The addition recurrence is

$$\psi_{r+s}\psi_{r-s} =\psi_{r+1}\psi_{r-1}\psi_s^2 -\psi_{s+1}\psi_{s-1}\psi_r^2. \tag{5}$$

The remaining coordinates satisfy

$$\phi_n=-976\psi_n^2-\psi_{n-1}\psi_{n+1}, \tag{6}$$ $$\Omega_n= \frac{\psi_{2n}+\psi_n^2(128\phi_n+3360\psi_n^2)}{2\psi_n}, \tag{7}$$

or, for $n\geq2$ and without dividing by $\psi_n$,

$$\Omega_n= \frac{\psi_{n+2}\psi_{n-1}^2-\psi_{n-2}\psi_{n+1}^2}{2\psi_2} +\psi_n(64\phi_n+1680\psi_n^2). \tag{8}$$

These are the exact formulas from the paper. General division-polynomial background is in Silverman, The Arithmetic of Elliptic Curves, Exercise III.3.7.

Tracing the birational map back to $a/b$ gives

$$\frac{2b}{a+2b}=\frac{\Omega_n}{73\psi_n\phi_n}, \qquad \frac ab=\frac{146\psi_n\phi_n}{\Omega_n}-2. \tag{9}$$

A rational solution $(X,Y)$ of (2), with $X=a/b$ in lowest terms, does give an integral $z$. Indeed,

$$F(a,b)=73(b^2Y)^2$$

is an integer. If $b^2Y$ had a denominator $h>1$ in lowest terms, then $h^2$ would divide the squarefree integer $73$, which is impossible.

We now impose the additional square factor $73^2$. A finite computation from (5)--(8) modulo $73$ gives the exact minimal periods

$$2628,\qquad1314,\qquad876$$

for $\psi_n,\phi_n,\Omega_n$, respectively. This is certified from a six-value state, not inferred from one shifted block. For $j\geq5$, (5) with second index $2$ gives

$$\psi_j\psi_{j-4} =\psi_2^2\psi_{j-1}\psi_{j-3}-\psi_3\psi_{j-2}^2,$$

and for $j\geq6$, the version with second index $3$ gives

$$\psi_j\psi_{j-6} =\psi_3^2\psi_{j-2}\psi_{j-4} -\psi_4\psi_2\psi_{j-3}^2.$$

At every transition modulo $73$, at least one denominator is nonzero, and the two routes agree when both are available. The six-value state after $2628$ steps equals the initial state. Equations (6) and (8) then certify their periods on every residue. Testing one complete $2628$-integer window gives

$$\psi_n\phi_n\equiv2\Omega_n\pmod{73},\qquad 73\nmid\Omega_n$$

exactly for the $36$ indices $n\equiv39\pmod{73}$. Therefore (9), in lowest terms, has $73\nmid b$ and

$$\frac ab\equiv290\pmod{73^2}. \tag{10}$$

Direct multiplication in $(\mathbb Z/73^2\mathbb Z)[X]$ gives

$$X^4-8X^3+2X^2+8X+1 \equiv (X-290)(X-2738)(X-2896)(X-4742)\pmod{73^2}. \tag{11}$$

Thus $F(a,b)\equiv0\pmod{73^2}$. Since the quartic point already gives $F(a,b)=73z^2$, it follows that $73\mid z$ and

$$F(a,b)=73^3c^2. \tag{12}$$

It remains to force $a,b$ to have opposite parity. Take $n=16q+4$ and write $R=4q+1$. Proposition 5.1 yields

$$\nu_2(\psi_{n-1}\psi_{n+1})=52R^2,$$ $$\nu_2(\psi_n)=26R^2+4,\qquad \nu_2(\psi_{2n})=104R^2+5.$$

Since $\nu_2(976\psi_n^2)>52R^2$, equation (6) gives $\nu_2(\phi_n)=52R^2$. On expanding (7), the three summands are

$$\frac{\psi_{2n}}{2\psi_n},\qquad 64\psi_n\phi_n,\qquad 1680\psi_n^3,$$

with respective valuations

$$78R^2,\qquad78R^2+10,\qquad78R^2+16.$$

The first is uniquely smallest, so

$$\nu_2(\Omega_n)=78R^2,\qquad \nu_2\!\left(\frac{\Omega_n}{\psi_n\phi_n}\right)=-4. \tag{13}$$

Consequently $146\psi_n\phi_n/\Omega_n$, the fraction used in the second formula in (9), has $2$-adic valuation $5$, while $2$ has valuation $1$. Thus the reduced ratio $a/b$ has valuation $1$: its numerator is even and its denominator is odd.

The simultaneous congruences

$$n\equiv39\pmod{73},\qquad n\equiv4\pmod{16}$$

are exactly $n\equiv404\pmod{16\cdot73}$. Since $P_1$ has infinite order, these indices give infinitely many distinct points. Removing the finitely many exceptional points of the birational maps leaves infinitely many. Moreover, the map to $X=a/b$ has fibers of size at most two, because (2) determines at most two values of $Y$ for each $X$. Thus infinitely many reduced coprime opposite-parity pairs $a,b$ satisfy (10)--(13). Zeros of $N_0,r,s,c$, or $d$ exclude only finitely many ratios, so all four terms may be taken nonzero and the common difference nonzero. A fixed pair $(N,d)$ comes from only finitely many integer pairs $(a,b)$, since

$$a^2+b^2=\sqrt{N+d}.$$

Hence the construction gives infinitely many progressions. Their four terms are squares or $73^3$ times a square and are positive; if the formula gives $d<0$, reverse their order.

Finally, verify the claimed pairwise coprimality rather than only $\gcd(N,d)=1$. For coprime opposite-parity $a,b$, reduction modulo every prime dividing $a$, $b$, or $b^2-a^2$ shows

$$\gcd(N_0,4ab(b^2-a^2))=1.$$

Also $3\nmid N_0$: this is immediate if $3$ divides $a$ or $b$, and if neither does, substitute $b\equiv\pm a\pmod3$. Thus $\gcd(N_0,6d)=1$. Every term in the progression is odd and coprime to $d$. A prime common to terms with indices $i<j$ must therefore divide $j-i\in\{1,2,3\}$. The prime $2$ is excluded by oddness, and the only pair at distance $3$ includes $N=N_0^2$, which is not divisible by $3$. Thus all six pairwise gcds are $1$. This proves Problem 937 unconditionally.

Verification. The verification script checks the two reduction orders proving that $P_1$ has infinite order, the complete six-state period certificate modulo $73$, the coefficient identity in (11), and the accepted manuscript's explicit 190-digit example. From the repository root, uv run --no-sync python library/diophantine_problems/bajpai_2024_arithmetic_progressions_squarefull_numbers/evidence/verify_937_bajpai_examples.py runs every named obligation in well under one second and exits nonzero on any failed check, including under python -O.

Source qualification. This proof uses the repaired last induction branch recorded on Proposition 5.1's page. The repair applies the paper's own recurrence (5) at different indices and is stated explicitly there; no author-issued correction is asserted.

Bears on. #937.