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

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:

1. 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 $1 \le i \le 5$

2. 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 $1 \le d \le 5$

3. 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. :-)

Answer 1

tl;dr: If $x+1 \geq 321$, then the five consecutive integers starting at $x+1$ have at most one 5-smooth member.

It is sufficient to ask how often 5-smooth numbers are nearby. If there are a pair of 5-smooth numbers with difference < 5, then no consecutive 5 integers containing the pair can satisfy the proposed divisibility pattern. A 5-smooth number is a number with no prime factor greater than 5.

Størmer has shown that there are only a finite number of consecutive 5-smooth numbers. More on these kinds of questions here. Pillai conjectured that for fixed $A$, $B$, and $C$, $A x^n - B y^m = C$ has only a finite number of solutions $(x,y,m,n)$. Erick Wong, in his answer to this problem, has demonstrated a reduction of the posted question to this form, such reduction being a (large-ish) list of such polynomials with exponent greater than three. In fact, these are irreducible binary forms of degree three, so Thue's theorem applies and each has at most a finite number of solutions. (Strictly, not all the forms on the raw list are irreducible, but this is no obstruction.) Thue has shown that each equation has at most a finite number of solutions. Wong's reduction thereby shows that there are a finite number of pairs of 5-smooth numbers that differ by less than 5.

Update: 20140410 Added extreme summary at the top. The following paragraphs and code with results of PARI/GP solution of many Thue equations, replaces the subsequent paragraphs.

Rather than work through a reduction of the number of equations, we use very fast code in PARI-GP (2.7.0) to solve the problem quickly. The code

{ for (a2 = 0, 2,     
    for (a3 = 0, 2, 
      for (a5 = 0, 2, 
        for (b2 = 0, 2, 
          for (b3 = 0, 2, 
            for (b5 = 0, 2, 
              ti = thueinit(2^a2*3^a3*5^a5*x^3 - 2^b2*3^b3*5^b5, 1);
              for (d = 1, 4, 
                str = Strprintf("[% d, % d, % d] :  % s ", 
                  2^a2*3^a3*5^a5, 2^b2*3^b3*5^b5, d, thue (ti, d));
                print (str);
                write ("some-output-file.txt", str);
 ) )))))) }

solves the Thue equations we are interested in in a few seconds. (Note that by default, the thueinit() function sets up internal state to search for solutions assuming the generalized RH. We have given the option to not make this optimistic assumption. Consequently, the set of solutions found is unconditionally exhaustive.) Of the 2916 equations, only 304 have solutions and the union of those sets contains 334 pairs (not all distinct). Examples (ordered pairs are $(x,y)$): $$ 1 x^3 - 1 y^3 = 1 : \{(0,-1),(1,0)\} $$ $$ 1 x^3 - 3 y^3 = 3 : \{(0,-1),(3,2)\} $$ The pairs of nearby 5-smooth integers produced by these solutions are frequently repeated. Collecting pairs, sorting, and ignoring negative integers, the many equations give the following set of candidates (some of which are spurious):

$\{ (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (4,8), (5,6), (5,8), (5,9), (6,8), (6,9), (6,10), (8,9), (8,10), (8,12), (9,10), (9,12), (10,12), (12,15), (12,16), (15,16), (15,18), (16,18), (16,20), (18,20), (20,24), (24,25), (24,27), (25,27), (27,30), (30,32), (32,36), (36,40), (45,48), (48,50), (50,54), (60,64), (72,75), (80,81), (96,100), (125,128), (160,162), (240,243), (320,324), (6859,6860), (13718,13720), (20577,20580), (27436,27440) \}$

Except for the last four pairs, this agrees with the previous list (see below) which was generated by brute force calculation of 5-smooth numbers with exponents less than 200. The last four pairs are spurious solutions because $6859=19^3$, $6860=2^2 \cdot 5 \cdot 7^3$ and the subsequent three pairs are this pair multiplied by $2$, $3$, and $4$. (Interestingly, only 7 and 19 appear as spurious factors in any of the solutions generated by the mass of polynomials.)

The PARI-GP code's output was parsed by the Mathematica 9.0 code

Module[{ inFile,   lines, nonempty, instances  },
  inFile = StringSplit[ Import["some-output-file.txt"], "\n" ];
  lines = ToExpression[ StringReplace[
     "{" <> # <> "}", {"[" -> "{", "]" -> "}", ":" -> ","}]] & /@ inFile;
  Print["Total equations: ", Length[lines]];

  nonempty = Select[lines, #[[2]] =!= {} &];
  Print["Satisfiable equations: ", Length[nonempty]];

  (* Switch from equation indexed to per-solution lists. *)
  Print["Total solutions: ", Length[Flatten[nonempty[[All, 2]], 1]]];

  instances[{eqn_, pairs_}] := 
     {eqn[[1]]*#[[1]]^3, eqn[[2]]*#[[2]]^3} & /@ pairs;

  Select[Split[Sort[
     Sort /@ Flatten[instances /@ nonempty, 1]
     ]][[All, 1]],
     #[[1]] > 0 &
  ]
]

generating only mildly syntactically different output from that listed above.

We therefore ...

• ... know there are only a finite number of solutions to the exponential Diophantine equations which are satisfied by counterexamples to the 5-consecutive 5-smooth numbers problem.
• ... have generated the complete set of solutions of polynomial relaxations to the exponential equations. This introduced only a handful of spurious solutions.
• ... have eliminated the spurious solutions.
• ... showed the complete solution set is that found by previous direct (but not provably complete) generation of small 5-smooth integers.

Discarded 20140410 starts here: Unfortunately, although we can show that each item in the list of equations has at most a finite number of solutions, part of the general solution method boils down to laboriously checking which of some set of potential solutions are actual solutions. The bounds on the sizes of these sets are rather large, so this becomes impractical even for modest sizes of the coefficients. Just $45x^n - y^n = 1$ is not easily disposed (in Mathematica 9.0). The equations with coefficients $>100$, e.g., $900x^n - y^n = 4$, are likely infeasible at present.

A few small 5-smooth number pairs that differ by less than 5, skipping the near continuum up to 27: {27,30}, {30,32}, {32,36}, {36,40}, {45,48}, {48,50}, {50,54}, {60,64}, {72,75}, {80, 81}, {96,100}, {125, 128}, {160, 162}, {240, 243}, {320, 324}. There may be more pairs. There might not. (There is no other with powers of 2, 3, and 5 less than 200.)

Edit 20140326: Removed the first, wrong, sentence, "The question of even whether $2^m - 3^n = 5$ only finitely many times, a specialization of the generalized Tijdeman's conjecture, is still open.", and the rest of the first paragraph, largely replacing it with the second paragraph, above. Tijdeman has shown that the quoted sentence is false and Erick Wong was right to insist that this is the case.

Answer 2

...no such $y$ exist. Suppose otherwise, and let $p>5$ be such a prime, then $p|x+i$ and $p|x+(i+1)$ for some i, $1<=i<5$, hence $p|(x+i+1)-(x+i)$ or $p|1$, a contradiction...

Answer 3

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.