I apologize if this is a naive question about greatest prime factors (gpf). I was thinking about the sequence of integers where $\mathrm{gpf}(x) \le p$ where $p$ is any prime.


Clearly, as $x$ increases, the distance $d$ between an integer where $\mathrm{gpf}(x) \le p$ and $\mathrm{gpf}(x+d) \le p$ increases at a seemingly every increasing rate.


For all primes $p$, does there exist an integer $C$ where if $x \ge C$, then there is at most one integer in the sequence $x+1, x+2, \dots, x+p$ has $\mathrm{gpf}(x) \le p$


For example, if $p=2$, $C = 2$ since for any $x \ge 2$, either $x$ or $x+1, \mathrm{gpf}(x) > 2$



For any prime $p$, is there $C$ such that if $x\ge C$, then all but one integer among $x+1, x+2, \dots, x+p$ has Greatest Prime Factor $> p$
This is a summary of Gerhard Paseman's comment.


Suppose that $A$ and $B$ are two numbers with $A-B = c$ small such that $A$ and $B$ are only
divisible by prime numbers less than or equal to $p$. Then one can write


$$A = a x^3, \qquad B = b y^3,$$


where all the prime factors of $a$ and $b$ are less than or equal to $p$, and all the exponents of each prime number is at most $2$ (the higher factors get absorbed into the cube). Note that if $p$ is the $n$th prime number, then there are $3^n$ possible values of $a$ and $b$. The answer to your problem will be positive provided that one can show that each of the ($3^{2n}$) equations


$$a x^3 - b y^3 = c$$


Has only finitely many solutions. Yet a theorem of Thue (1908) guarantees that if $f(x,y)$ is irreducible of degree $\ge 3$, then $f(x,y) = c$ has only finitely many integral solutions.


This leaves the reducible cases corresponding to the ratio $[a:b]$ being a perfect cube. Yet these are trivial; after absorbing the coefficients and multiplying through, one is left with an equation


$$X^3 - Y^3 = C$$


to solve in integers $X$ and $Y$. Yet this also has finitely many solutions for a trivial reason - the consecutive cubes become further apart.


Thue's results are actually effective, one can also prove this result by various other means, like Baker's method. 


This problem is actually the baby case of the $S$-unit equation.
In particular, it ultimately boils down to solving the equation


$$A + B = C$$


where $A$, $B$, and $C$ have a fixed set of prime factors dividing $S$ (Take $S$ to be the product of primes less than or equal to the maximum of $p$ and the range corresponding to $c$ in the Thue equations above). In other words, one is looking for a solution in units 
to the equation $A+B=C$ in the ring 
$\mathbf{Z}[1/S]$.
The methods above can be generalized to applied to similar problems where
$\mathbf{Z}[1/S]$ is replaced by the $S$-units in a number field, and one can even increase 
the number of terms (providing that one is careful only considers primitive solutions, ruling out trivialities like $A-A+B-B=0$.)


