TITLE: Is there an integer $y>0$ such that for all $x>y$, at least $4$ of $x+1, x+2, x+3, x+4, x+5$ are divisible by a prime greater than $5$?
QUESTION [6 upvotes]: As I understand it, the answer to this question is yes.
I would appreciate it if anyone could provide a counter example or provide details on how to resolve this question.
My thinking is that this is true for $y > 80$
Here's my thinking:

We can assume that there is at least on integer in the sequence $x+i$ such that $x+i = 2^a3^b5^c$ where $a,b,c \ge 0$ and $a+b+c > 0$ and $1 \le i \le 5$
So, I am trying to prove that there is no solution for $x>24$ where $2^a3^b5^c + d = 2^e3^f5^g$ and $1 \le d \le 5$
Let $f=gcd(2^a3^b5^c,2^e3^f5^g)$ and we need only deal with the following forms:

Case 1:  $2^a \pm 1 = 3^b5^c$ where $a,b,c > 0$

Since $a > 1$, we know that $b$ is even since $(-1)^b(1) \equiv \pmod 4$
Since $a > 2$, we know that $c$ is even since $(1)^{\frac{b}{2}}(-3)^c \equiv 1 \pmod 8$
But then $(3^{\frac{b}{2}}5^{\frac{c}{2}} + 1)(3^{\frac{b}{2}}5^{\frac{c}{2}} - 1)  = 2^a$ which is impossible for $a > 1$
Let's look at $2^a-1 = 3^b5^c$
There is no solution for $a$ is even because $(2^{\frac{a}{2}} - 1)(2^{\frac{a}{2}} + 1)= 3^b5^c$ which is impossible since $b,c > 1$
There is no solution for $a$ is odd because $(-1)^a - 1 \not\equiv 0 \pmod 3$

Case 2:  $3^a \pm 1 = 2^b5^c$ where $a,b,c > 0$

Let's start $3^a+1 = 2^b5^c$
$a$ cannot be odd because only $(-2)^{4x+2} + 1 \equiv 0 \pmod 5$
$a$ cannot be even since $(-1)^a + 1 \not\equiv 0 \pmod {10}$
Let's look at $3^a - 1 = 2^b5^c$
$a$ cannot be even since $(3^{\frac{a}{2}}-1)(3^{\frac{a}{2}}+1) = 2^b5^c$ is impossible for $b > 5,c>0$ or for $b >0,c>1$
$a$ cannot be odd since only $(-2)^(4x) - 1 \equiv 0 \pmod 5$

Case 3:  $5^a \pm 1 = 2^b3^c$ where $a,b,c > 0$

Let's look at $5^a + 1 = 2^b3^c$
We know if there's a solution that $b=1$ since $(1)^a + 1 \equiv 2 \pmod 4$
Since $c>1$, it follows that $a=6k+3$ since $5^a \equiv -1 \pmod 9$
$5^{6k+3}+1\equiv 0 \pmod 7$
But $2^b3^c \not\equiv 0 \pmod 7$
Let's look at $5^a-1 = 2^b3^c$
$a$ cannot be even since $(5^{\frac{a}{2}}-1)(5^{\frac{a}{2}}+1) = 2^b3^c$ is impossible for $b>1,c> 0$
$a$ cannot be odd since $(-1)^a-1 \not\equiv 0 \pmod 6$

Case 4:  $2^a \pm 1 = 3^b$ where $a,b > 0$
Case 5:  $2^a \pm 1 = 5^b$ where $a,b > 0$
Case 6:  $3^a \pm 1 = 5^b$ where $a,b > 0$
Case 4,5, and 6 are impossible for $x > 24$ by the proof behind Catalan's conjecture.

Edit: Eric Towers has pointed out that $\{80,81\}$ do not have a prime greater than $5$.  It seems that my logic in Case 2 did not account for $(3^2 - 1)(3^2 + 1) = 2^4*5$
I've updated my assumption to $y > 80$.  This removes the counter example.  I will spend time reading up on smooth numbers and spending time verifying that $80$ is the correct value for $y$.  If anyone has a precise formulation, that will be greatly appreciated.  :-)

REPLY [3 votes]: If you unravel the definitions, you'll find as Eric Towers said that this is really about the minimum gap between $r$-smooth (aka friable) numbers.  This other recent MSE question cites a survey of Tijdeman (link) containing a relevant excerpt:

...and in 1945 by Skolem to give an algorithm for determining all
  solutions of certain equations of the form $a_1^{m_1} \cdots a_k^{m_k} -  b_1^{n_1} \cdots b_l^{n_l}= c$ in positive integers $m_1,\ldots,m_k,n_1,\ldots,n_l$ where
  $a_1,\ldots,a_k,b_1,\ldots,b_l,c$ are given positive integers. The finiteness
  of the number of solutions had been proved by Pólya in 1918 who
  applied a third type of methods.

One minor quibble is that he says "positive integers" rather than nonnegative, but this could be adjusted by inflating the sets $\{a_1,\ldots,a_k\}$ and $\{b_1,\ldots,b_l\}$ and multiplying $c$ by a constant.
Let me take this opportunity to give a fuller justification of the specific claim in this question, with a few pointers that should help OP narrow down the exact value.  Namely, for any $d \ne 0$ there are only finitely many solutions to $2^{m_1} 3^{m_2} 5^{m_3} - 2^{n_1} 3^{n_2} 5^{n_3} = d$.  Applying this to $d=1,2,3,4$, this means that all sufficiently large $5$-friable numbers must be at least $5$ apart, which I believe directly answers the original question.
The key tool is Thue's theorem which replaces the role of Pell equations in the proof of Størmer's theorem, combined with the "dumb trick" I mentioned in the other thread.  Let $S$ be the finite set $\{2^a 3^b 5^c: 0\le a,b,c \le 2\}$ of $27$ elements.  Then every solution to $2^{m_1} 3^{m_2} 5^{m_3} - 2^{n_1} 3^{n_2} 5^{n_3} = d$ is also a solution to $r x^3 - s y^3 = d$ for some $r,s \in S$ (where $rx^3 = 2^{m_1} 3^{m_2} 5^{m_3}$ and $sy^3 = 2^{n_1} 3^{n_2} 5^{n_3}$).
So it suffices to show that $rx^3 - sy^3 = d$ has only finitely many solutions for a fixed $r,s \in S$ and $d \ne 0$.  By our choice of $S$, if $r \ne s$ then $r/s$ does not have a rational cube root, thus $rx^3-sy^3$ is irreducible (being of degree $3$ with no roots) and Thue's theorem gives us this.  If $r = s$, then this reduces to showing that $x^3 - y^3 = d$ has finitely many solutions, which is clear from the fact that $(x+1)^3 - x^3 \to \infty$ as $x$ increases.
Computationally, Thue's theorem is effective (so is the much easier bound in the previous sentence) and Pari/GP contains a Thue equation solver.  If you have a programming bent (or know someone who does), you could just churn through the $4\cdot 27^2$ (at most) possibilities to find the exact value of $y$ in your question.