Is there a positive integer $n$ such that the prime divisors of $n^3 - 1$ are $2$, $3$ and $7$?

Question

For a positive integer $k$ write $\pi(k)$ for the set of all prime divisors of $k$. For example, $\pi(24) = \{2,3\}$ and $\pi(1) = \emptyset$.

Question. Is there a positive integer $n$ for which $\pi(n^3 - 1) = \{2,3,7\}$?

An alternative point of view asks whether there are positive integers $a$, $b$ and $c$ such that $1 + 2^a3^b7^c$ is a perfect cube (of a positive integer).

To be clear, $n^3 - 1$ must be divisible by all three primes $2$, $3$ and $7$, and not divisible by any other prime.

I think the answer is "no", but have so far been unable to prove it. I actually only need to find an answer for prime $n$, but it appears it may be true without assuming that $n$ is prime. Using Maple, I have checked this for all $n \leq 10^9$, and for the first $106000000$-ish primes.

This is actually just one case ($r = 3$) of the problem I am working on which is to find all primes $p$ and integers $r>1$ for which $\pi(p^r - 1) = \{ 2, 3, 7 \}$. I think I have managed to show that $r = 2$ and $p \in \{ 13, 97, 127 \}$, unless $r = 3$. (And, as you might guess, this whole thing is just one very special case of a more general problem I am looking at.)

Answer 1

Given $n\in\mathbf N$ s.t. $n^3 - 1$ is only divisible by primes in the set $S = \{2,3,7\}$ we can write $$n^3 - 1 = 2^{v_2}3^{v_3}7^{v_7}$$ if any of the $v_p \ge 2$ we can factor out a square to write $$n^3 - 1 = \underbrace{2^{v'_2}3^{v'_3}7^{v'_7}}_{=:d} m ^2$$ where each $v_p'$ must be either $0$ or $1$. So any solution of the original problem gives us an integral point on the curve $$d m^2 = n^3 - 1$$ such curves (with one variable squared and the other cubed) are known as elliptic curves, normally the equation looks like $$y^2 = x^3 + ax + b$$ but adding the term $d$ in front of the squared variable is an operation known as quadratic twisting. To "remove" this $d$ and get to a standard form we will rewrite our equation:

If $m,n,d$ are integers satisfying $dm^2= n^3- 1$ we have $d^3 \cdot d m^2 = d^3 n^3 - d^3$ so $$(d^2 m)^2 = (d n)^3 - d^3$$ and letting $y = d^2 m$ and $x = dn$ we have an integer solution of $$y^2 = x^3 - d^3$$ (where in particular $y$ is only divisible by $2,3,7$)

The upshot of this is that we have a finite list of possible $d$s (as they are only divisible by $2,3,7$ and squarefree, $d$ must be one of $1, 2, 3, 7, 6, 14, 21, 42$). Moreover given any elliptic curve, the set of integral points on that curve is finite (this is a theorem due to Siegel), and this set can be computed via Sage or other software. So we have a finite list of curves, and only finitely many points to check for each curve to see if they gave us a solution to the original problem (we have to check that the $(x,y)$ we found has $d|x$ and $d^2|y$ and all that the primes required divide $d m^2$).

By checking all these, we find there are no solutions, here is some Sage code to check these things (the last couple of conditions we can check by hand as there are only 4 candidates where $y$ is not divisible by other primes than those in $S$):

sage: S = [2,3,7]

sage: list(Combinations(S).map(prod))

[1, 2, 3, 7, 6, 14, 21, 42]

sage: for d in Combinations(S).map(prod):
....:    for (x,y,_) in EllipticCurve([0,-d^3]).integral_points():
....:        if y.is_S_unit(S):
....:            print("d:",d," x:",x," y:",y)
....:            print("n:",x/d, " m:",y/(d^2))
....:            print("n^3 - 1 =",(x/d)^3 - 1, "factors as", ((x/d)^3 - 1).factor())
....:            print("--")

d: 7  x: 14  y: 49
n: 2  m: 1
n^3 - 1 = 7 factors as 7
--
d: 7  x: 28  y: 147
n: 4  m: 3
n^3 - 1 = 63 factors as 3^2 * 7
--
d: 6  x: 10  y: 28
n: 5/3  m: 7/9
n^3 - 1 = 98/27 factors as 2 * 3^-3 * 7^2
--
d: 6  x: 33  y: 189
n: 11/2  m: 21/4
n^3 - 1 = 1323/8 factors as 2^-3 * 3^3 * 7^2
--

Note that similar code if you include $-1$ in the set $S$ this does find Barry Cipra's example, so this should lend some confidence that it is doing the right thing. Even if we vary the set of primes here there doesn't seem to be solutions for many sets of 3 or more primes, so perhaps there is a deeper reason too, or maybe its just unlikely!

Answer 2

Partial? answer Way too long for a comment.

Since $2,3$ must divide $n^3-1$ we have $n$ odd and $3|n-1$. Therefore $$ n=6k+1 $$ for some positive integer.

Then $$ n^3-1=(6k+1)^3-1=18k(12k^2+6k+1) $$

Since $12k^2+6k+1$ is relatively prime to $2,3$ it follows that we must have $$ 12k^2+6k+1=7^l $$ for some positive integer $ l \geq 1$.

This immediately implies that $7 \nmid k$ and hence there exists non-negative integers $a,b$ such that $$ k=2^a3^b $$

This leads to the diophantine equation $$ 2^{2a+2}3^{2b+1}+2^{a+1}3^{b+1}+1=7^l \\ a,b,l \in \mathbb Z \\ a,b \geq 0 \\ l\geq 1 $$

Here I got stucked.

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Note that there are good chances $$ 12k^2+6k+1=7^l $$ already does not have positive solutions.

Note that if this quadratic in $k$ has a solution $$ \Delta=36-48(1-7^l)=48 \cdot 7^l-12 $$must be a perfect square. This implies that there exists a positive integer $c$ such that $$ 12 \cdot 7^l-3=c^2 $$

Since $3|c$ there exists some integer $d$ such that $c=3d$ and hence $$ 4 \cdot 7^l=3d^2+1 $$ ??