I don't know how far Larry went in pursuing this problem, but this touches on a topic I've spent some time on, ie. Lehmer's method.

Let $S_j$ be the maximum $S$ for which the pair {$S, S+j$} is p-smooth, and let $S_m$ be the maximum of $\{S_1, S_2 \ldots S_p\}$. Also let k = $\pi(p)$, ie. the number of primes $\leq p$.

It follows then that the minimal $C$ for which the desired property holds is $C = S_m$.

Determining each $S_j$ is not so straight-forward, apart from the cases $j=1, 2$, which are a direct application of Lehmer's method, which provides for the enumeration of all smooth pairs of the form $\{S, S+1\}$, $\{S, S+2\}$, by solving roughly $2^k$ standard Pell equations, ie. $x^2 - Dy^2 = 1$, for D ranging over all combinations of the k primes $\leq p$. Both sets of pairs can be obtained with a single pass.

For $3 \leq j \leq p$, however, things are not so simple. Lehmer did not address these cases, and perhaps we can understand why. We can in fact extend Lehmer's method to identify smooth pairs $\{S, S+j\}$, but this requires solving $x^2 - Dy^2 = j^2$, again for all $2^k$ values of D.

The good news is that these equations can be solved from the $x^2 - Dy^2 = 1$ solutions, so that the number of continued fractions we have to compute is still the same. See John Robertson's article on the LMM method (Lagrange-Matthews-Mollin) at JPR_Pell.

Note that there can be multiple solution classes for any j.

The bad news is that Lehmer's main achievement, by which he is able to reduce the number of Pell equations from $3^k$ to $2^k$, is not applicable for j \geq 3. For $j = 1, 2$ he showed that any smooth pair that does not turn up as a fundamental solution $(x_1, y_1)$ will be found at some $(x_m, y_m)$ with $m \leq (p+1)/2$. This is because the $y_n$ values form a Lucas sequence, and so $y_1$ divides all $y_n$. Thus, if $y_1$ isn't smooth, neither will be any other $y_n$. And if $y_1$ is smooth, we only need check a limited number of $y_n$.

Sadly, the multiple solutions in any class of solutions to $x^2 - Dy^2 = N$, ($N=j^2$), do not have these Lucasian properties. So we don't know how many $(x_n, y_n)$ to look at, and we can't assume that $y_1$ not being smooth means that $y_2$ isn't either.

We could of course revert to the original Störmer method, where we solve for D being all possible combinations of the $k$ primes to the power $\{0, 1, 2\\\$$, thus requiring roughly $3^k$ equations to be solved. That's very slow, but guarantees that smooth pairs occur only as fundamental solutions.

Alternately, it might well be that $S_1 > S_j$ always, in which case we would avoid all of these complications, solving only the standard equations $x^2 - Dy^2 = 1$. I have not yet done any investigation of this question, but I remember that generally $S_2 < S_1$, so this property can't be ruled out.

Finally, I would like to know if Larry looked into the method described above involving $X^3 - Y^3 = C$, and if so, how it works.

I don't know how far Larry went in pursuing this problem, but this touches on a topic I've spent some time on, ie. Lehmer's method.

Let $S_j$ be the maximum $S$ for which the pair {$S, S+j$} is $p$-smooth, and let $S_m$ be the maximum of $\{S_1, S_2 \ldots S_p\}$. Also let k = $\pi(p)$, ie. the number of primes $\leq p$.

It follows then that the minimal $C$ for which the desired property holds is $C = S_m$.

Determining each $S_j$ is not so straight-forward, apart from the cases $j=1, 2$, which are a direct application of Lehmer's method, which provides for the enumeration of all smooth pairs of the form $\{S, S+1\}$, $\{S, S+2\}$, by solving roughly $2^k$ standard Pell equations, ie. $x^2 - Dy^2 = 1$, for $D$ ranging over all combinations of the $k$ primes $\leq p$. Both sets of pairs can be obtained with a single pass.

For $3 \leq j \leq p$, however, things are not so simple. Lehmer did not address these cases, and perhaps we can understand why. We can in fact extend Lehmer's method to identify smooth pairs $\{S, S+j\}$, but this requires solving $x^2 - Dy^2 = j^2$, again for all $2^k$ values of $D$.

The good news is that these equations can be solved from the $x^2 - Dy^2 = 1$ solutions, so that the number of continued fractions we have to compute is still the same. See John Robertson's article on the LMM method (Lagrange-Matthews-Mollin) at JPR_Pell.

Note that there can be multiple solution classes for any $j$.

The bad news is that Lehmer's main achievement, by which he is able to reduce the number of Pell equations from $3^k$ to $2^k$, is not applicable for $j \geq 3$. For $j = 1, 2$ he showed that any smooth pair that does not turn up as a fundamental solution $(x_1, y_1)$ will be found at some $(x_m, y_m)$ with $m \leq (p+1)/2$. This is because the $y_n$ values form a Lucas sequence, and so $y_1$ divides all $y_n$. Thus, if $y_1$ isn't smooth, neither will be any other $y_n$. And if $y_1$ is smooth, we only need check a limited number of $y_n$.

Sadly, the multiple solutions in any class of solutions to $x^2 - Dy^2 = N$, $(N=j^2)$, do not have these Lucasian properties. So we don't know how many $(x_n, y_n)$ to look at, and we can't assume that $y_1$ not being smooth means that $y_2$ isn't either.

We could of course revert to the original Störmer method, where we solve for $D$ being all possible combinations of the $k$ primes to the power $\{ 0, 1, 2 \}$, thus requiring roughly $3^k$ equations to be solved. That's very slow, but guarantees that smooth pairs occur only as fundamental solutions.

Alternately, it might well be that $S_1 > S_j$ always, in which case we would avoid all of these complications, solving only the standard equations $x^2 - Dy^2 = 1$. I have not yet done any investigation of this question, but I remember that generally $S_2 < S_1$, so this property can't be ruled out.

Finally, I would like to know if Larry looked into the method described above involving $X^3 - Y^3 = C$, and if so, how it works.

I don't know how far Larry went in pursuing this problem, but this touches on a topic I've spent some time on, ie. Lehmer's method.

Let $S_j$ be the maximum $S$ for which the pair {$S, S+j$} is p-smooth, and let $S_m$ be the maximum of ${S_1, S_2 \ldots S_p}$}. Also let k = $\pi(p)$, ie. the number of primes $\leq p$.

It follows then that the minimal $C$ for which the desired property holds is $C = S_m$.

Determining each $S_j$ is not so straight-forward, apart from the cases $j=1, 2$, which are a direct applications of Lehmer's methods, which provide for the enumeration of all smooth pairs of the form {$S, S+1$}, {$S, S+2$}, by solving roughly $2^k$ standard Pell equations, ie. $x^2 - Dy^2 = 1$, for D ranging over all combinations of the k primes $\leq p$. Both sets of pairs can be obtained with a single pass.

For $3 \leq j \leq p$, however, things are not so simple. Lehmer did not address these cases, and perhaps we can understand why. We can in fact extend Lehmer's method to identify smooth pairs {$S, S+j$}, but this requires solving $x^2 - Dy^2 = j^2$, again for all $2^k$ values of D.

The good news is that these equations can be solved from the $x^2 - Dy^2 = 1$ solutions, so that the number of continued fractions we have to compute is still the same. See John Robertson's article on the LMM method (Lagrange-Matthews-Mollin) at link text.

Note that there can be multiple solution classes for any j.

The bad news is that Lehmer's main achievement, by which he is able to reduce the number of Pell equations from $3^k$ to $2^k$, is not applicable for j \geq 3. For $j = 1, 2$ he showed that any smooth pair that does not turn up as a fundamental solution $(x_1, y_1)$ will be found at some $(x_m, y_m)$ with $m \leq (p+1)/2$. This is because the $y_n$ values form a Lucas sequence, and so $y_1$ divides all $y_n$. Thus, if $y_1$ isn't smooth, neither will be any other $y_n$. And if $y_1$ is smooth, we only need check a limited number of $y_n$.

Sadly, the multiple solutions in any class of solutions to $x^2 - Dy^2 = N$, ($N=j^2$), do not have these Lucasian properties. So we don't know how many $(x_n, y_n)$ to look at, and we can't assume that $y_1$ not being smooth means that $y_2$ isn't either.

We could of course revert to the original Störmer method, where we solve for D being all possible combinations of primes to the power {$0, 1, 2$}, thus requiring roughly $3^k$ equations to be solved. That's very slow, but guarantees that smooth pairs occur only as fundamental solutions.

Alternately, it might well be that $S_1 > S_j$ always, in which case we would avoid all of these complications, solving only the standard equations $x^2 - Dy^2 = 1$. I have not yet done any investigation of this question, but I remember that generally $S_2 < S_1$, so this property can't be ruled out.

Finally, I would like to know if Larry looked into the method described above involving $X^3 - Y^3 = C$, and if so, how it works.

I don't know how far Larry went in pursuing this problem, but this touches on a topic I've spent some time on, ie. Lehmer's method.

Let $S_j$ be the maximum $S$ for which the pair {$S, S+j$} is p-smooth, and let $S_m$ be the maximum of $\{S_1, S_2 \ldots S_p\}$. Also let k = $\pi(p)$, ie. the number of primes $\leq p$.

It follows then that the minimal $C$ for which the desired property holds is $C = S_m$.

Determining each $S_j$ is not so straight-forward, apart from the cases $j=1, 2$, which are a direct application of Lehmer's method, which provides for the enumeration of all smooth pairs of the form $\{S, S+1\}$, $\{S, S+2\}$, by solving roughly $2^k$ standard Pell equations, ie. $x^2 - Dy^2 = 1$, for D ranging over all combinations of the k primes $\leq p$. Both sets of pairs can be obtained with a single pass.

For $3 \leq j \leq p$, however, things are not so simple. Lehmer did not address these cases, and perhaps we can understand why. We can in fact extend Lehmer's method to identify smooth pairs $\{S, S+j\}$, but this requires solving $x^2 - Dy^2 = j^2$, again for all $2^k$ values of D.

The good news is that these equations can be solved from the $x^2 - Dy^2 = 1$ solutions, so that the number of continued fractions we have to compute is still the same. See John Robertson's article on the LMM method (Lagrange-Matthews-Mollin) at JPR_Pell.

Note that there can be multiple solution classes for any j.

The bad news is that Lehmer's main achievement, by which he is able to reduce the number of Pell equations from $3^k$ to $2^k$, is not applicable for j \geq 3. For $j = 1, 2$ he showed that any smooth pair that does not turn up as a fundamental solution $(x_1, y_1)$ will be found at some $(x_m, y_m)$ with $m \leq (p+1)/2$. This is because the $y_n$ values form a Lucas sequence, and so $y_1$ divides all $y_n$. Thus, if $y_1$ isn't smooth, neither will be any other $y_n$. And if $y_1$ is smooth, we only need check a limited number of $y_n$.

Sadly, the multiple solutions in any class of solutions to $x^2 - Dy^2 = N$, ($N=j^2$), do not have these Lucasian properties. So we don't know how many $(x_n, y_n)$ to look at, and we can't assume that $y_1$ not being smooth means that $y_2$ isn't either.

We could of course revert to the original Störmer method, where we solve for D being all possible combinations of the $k$ primes to the power $\{0, 1, 2\\\$$, thus requiring roughly $3^k$ equations to be solved. That's very slow, but guarantees that smooth pairs occur only as fundamental solutions.

Alternately, it might well be that $S_1 > S_j$ always, in which case we would avoid all of these complications, solving only the standard equations $x^2 - Dy^2 = 1$. I have not yet done any investigation of this question, but I remember that generally $S_2 < S_1$, so this property can't be ruled out.

Finally, I would like to know if Larry looked into the method described above involving $X^3 - Y^3 = C$, and if so, how it works.