3 It really is correct! (now) I had x and y switched. I've double checked!

It turns out that there are consecutive integers $7^3x^2$ 7^3y^2$and$2^3y^2$2^3x^2$ where

x=1280501243627863240958544690009528057630666112209166210134539296491713575228511562822601

y=195559058652792360721040748497175750756141280467009959592151199119238991562483618043657

are each a product of 9 primes according to http://www.alpertron.com.ar/ECM.HTM , I could check the purported factors in MAPLE but haven't)

The method in brief: Find a prime p with $p^3-1$ or $p^3+1$ of the form $qz^2$ (with q a prime) then the Pell equation $p^3u^2-qv^2=\pm 1$ has solutions and the values of $v \mod q$ are periodic. If $v$ is ever a multiple of $q$ then there is a sequence of solutions to $p^3x^2-q^3y^2=\pm1$ If $x,y$ ever have the same signature, there's the example (as long as y is prime to q and x to p) .

For the example above: There are solutions to $7b^2+1=8a^2$ the first two are $a_0=1,b_0=1$ and $a_1=31,b_1=29$ with,as easy to find,

$b_n=16a_{n-1}+15b_{n-1}$ and $a_n=14a_{n-1}+15b_{n-1}$.

When $n \equiv 3 \mod 7$ then $b_n \equiv 0 \mod 7$

After misses at n=3,7,10,17,31,38,45,52 (and ignoring $n=24 \mod 49$ where $b_n$ divides by 49) success at n=59.

$27a^2-2b^2=1$ is never $27x^2-8y^2=1$ since b is odd but $7\cdot a^2-3^3 \cdot b^2=1$ is $7^3x^2-3^3y^2=1$ every 7th time.

2 previous y had a factor of 7 and 7*(oldy)^2+1=8x^2 vs 7^3*(thisy)^2+1=8*x^2

It turns out that there are consecutive integers $7^3x^2$ and $2^3y^2$ where $$x=1280501243627863240958544690009528057630666112209166210134539296491713575228511562822601$$ $$y=1368913410569546525047285239480230255292988963269069717145058393834672940937385326305599$$

x=1280501243627863240958544690009528057630666112209166210134539296491713575228511562822601

y=195559058652792360721040748497175750756141280467009959592151199119238991562483618043657

are each a product of 9 primes according to http://www.alpertron.com.ar/ECM.HTM , I could check the purported factors in MAPLE but haven't)

The method in brief: Find a prime p with $p^3-1$ or $p^3+1$ of the form $qz^2$ (with q a prime) then the Pell equation $p^3u^2-qv^2=\pm 1$ has solutions and the values of $v \mod q$ are periodic. If $v$ is ever a multiple of $q$ then there is a sequence of solutions to $p^3x^2-q^3y^2=\pm1$ If $x,y$ ever have the same signature, there's the example (as long as y is prime to q and x to p) .

For the example above: There are solutions to $7b^2+1=8a^2$ the first two are $a_0=1,b_0=1$ and $a_1=31,b_1=29$ with,as easy to find,

$b_n=16a_{n-1}+15b_{n-1}$ and $a_n=14a_{n-1}+15b_{n-1}$.

When $n \equiv 3 \mod 7$ then $b_n \equiv 0 \mod 7$

After misses at n=3,7,10,17,31,38,45,52 (and ignoring $n=24 \mod 49$ where $b_n$ divides by 49) success at n=59.

$27a^2-2b^2=1$ is never $27x^2-8y^2=1$ since b is odd but $7\cdot a^2-3^3 \cdot b^2=1$ is $7^3x^2-3^3y^2=1$ every 7th time.

1

It turns out that there are consecutive integers $7^3x^2$ and $2^3y^2$ where $$x=1280501243627863240958544690009528057630666112209166210134539296491713575228511562822601$$ $$y=1368913410569546525047285239480230255292988963269069717145058393834672940937385326305599$$ are each a product of 9 primes according to http://www.alpertron.com.ar/ECM.HTM , I could check the purported factors in MAPLE but haven't)

The method in brief: Find a prime p with $p^3-1$ or $p^3+1$ of the form $qz^2$ (with q a prime) then the Pell equation $p^3u^2-qv^2=\pm 1$ has solutions and the values of $v \mod q$ are periodic. If $v$ is ever a multiple of $q$ then there is a sequence of solutions to $p^3x^2-q^3y^2=\pm1$ If $x,y$ ever have the same signature, there's the example (as long as y is prime to q and x to p) .

For the example above: There are solutions to $7b^2+1=8a^2$ the first two are $a_0=1,b_0=1$ and $a_1=31,b_1=29$ with,as easy to find,

$b_n=16a_{n-1}+15b_{n-1}$ and $a_n=14a_{n-1}+15b_{n-1}$.

When $n \equiv 3 \mod 7$ then $b_n \equiv 0 \mod 7$

After misses at n=3,7,10,17,31,38,45,52 (and ignoring $n=24 \mod 49$ where $b_n$ divides by 49) success at n=59.

$27a^2-2b^2=1$ is never $27x^2-8y^2=1$ since b is odd but $7\cdot a^2-3^3 \cdot b^2=1$ is $7^3x^2-3^3y^2=1$ every 7th time.