Is this a valid argument to show that for $m > 4$, $2^m - 3^n \ne 7$

Question

Is the follow argument valid? Is there a simpler argument?

Here's the argument:

(1) Assume that there exists for integer $m > 4, n$: $$2^m - 3^n = 7$$

(2) $m$ and $n$ are even since:

$$2^{m} - 3^{n} \equiv 4 - 9 \equiv 7 \pmod {12}$$

(3) So, there exists integers $i,j$ with $m=2i, n=2j$ with:

$$2^m - 3^n = (2^i - 3^j)(2^i + 3^j) = 7$$

(4) But, since $i > 2$, $2^i - 3^j \ne 1$ from the Catalan's Conjecture

(5) Then, both $2^i - 3^j$ and $2^i + 3^j$ are greater than $1$ which is impossible since $7$ is prime. So we reject our assumption in step(1).

It’s much easier to show that $2^i+3^j\neq 7$ than it is to prove Catalan’s conjecture, I think. :) — Thomas Andrews, May 15 '21 at 23:00
@Ripi2 "Catalan's conjecture (or Mihăilescu's theorem) ..." — rtybase, May 15 '21 at 23:01
@Ripi2 Catalan's conjecture is just the name. It was proven by Preda Mihăilescu in 2002. — Larry Freeman, May 15 '21 at 23:02
@rtybase and Larry Freeman: Thanks. A new thing I learned today. — Ripi2, May 15 '21 at 23:03
@LarryFreeman but it is overkill for an elementary question. — Thomas Andrews, May 15 '21 at 23:03

Thomas Andrews · Accepted Answer · 2021-05-15T23:48:16.973

2

Catalan’s conjecture is way overkill.

Since $2^i-3^j<2^i+3^j$ are integer and the second is positive, the only way for their product to be $7$ is if $2^i+3^j=7.$ But $m\geq 6$ means $i\geq 3,$ so $$2^i+3^j>2^3=8>7.$$

However, Catalan might be needed to prove the more general result that there is no prime $p\equiv 7\pmod{12}$ and $m>4,n>0 $ such that:

$$p=2^m-3^n$$

Edit: (From commenter Erick Wong) It turns out, it is pretty easy to show there is no $2^i-3^j=1$ with $i>2$ just by looking modulo $8,$ so you don’t need Catalan for that, either.

edited May 15 '21 at 23:48

answered May 15 '21 at 23:08

Thomas Andrews

177,126

2

Catalan isn't needed for the latter result either. There is an elementary proof that $2^m - 3^n = 1$ has no solutions for $m\ge 3$: just look at the equation mod $8$. Apparently this was already studied in the 14th century :). Reference: https://mathoverflow.net/a/29928 – Erick Wong May 15 '21 at 23:29
Gerson's proof is also given in an answer on this site in 2012. – Bill Dubuque May 16 '21 at 03:14

score 1 · Answer 2 · answered May 15 '21 at 23:30

1

Just a small alternative to Thomas Andrews's answer: If $(2^i-3^j)(2^i+3^j)=7$ then $2^i-3^j=1$ and $2^i+3^j=7$, which implies $2\cdot2^i=1+7=8$, or $i=2$ (corresponding to $m=2i=4$).

answered May 15 '21 at 23:30

Barry Cipra

79,832

Is this a valid argument to show that for $m > 4$, $2^m - 3^n \ne 7$

2 Answers2