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.<br><br>
Let $S_j$ be the maximum $S$ for which the pair {$S, S+j$} is <i>p</i>-smooth, and let $S_m$ be the maximum of $\{S_1, S_2 \ldots S_p\}$. Also let <i>k</i> = $\pi(p)$, ie. the number of primes $\leq p$.<br><br> 
It follows then that the minimal $C$ for which the desired property holds is $C = S_m$. <br><br>
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 <i>D</i> ranging over all combinations of the <i>k</i> primes $\leq p$. Both sets of pairs can be obtained with a single pass.<br><br>
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 <i>D</i>.<br><br>
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][1].<br><br>
Note that there can be multiple solution classes for any <i>j</i>.<br><br>
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 <i>j \geq 3</i>. 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 $y_1$ <i>is</i> smooth, we only need check a limited number of $y_n$.<br><br>
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.<br><br>
We could of course revert to the original Störmer method, where we solve for <i>D</i> 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.<br><br>
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.<br><br>
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.

  [1]: http://www.jpr2718.org/pell.pdf