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I am attempting to generalize an answer given by Will Jagy here.

I may have made a mistake since the conclusion is stronger than I typically have seen and my argument is pretty much the same as Will's. If that is the case, then I would appreciate if someone could call out my mistake.

Let:

  • $m>2, n>1, x>0, y>0, c>0$ be integers with:

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

Does it now follow that:

$$2^{m+x} - 3^{n+y} \ne c$$

Here's my thinking:

(1) Assume that $2^{m+x} - 3^{n+y} = 2^{m} - 3^n = c$

(2) $2^m\left(2^x - 1\right) = 3^n\left(3^y - 1\right)$

(3) $3^y - 1 \equiv 0 \pmod {2^m}$

(4) From Euler's Theorem, $2^{m-1} | y$ since from Euler's Totient Function $\varphi(2^m) = 2^{m}\left(1 - \dfrac{1}{2}\right) = 2^{m-1}$

Note: I believe that Carmichael's Theorem can be used to establish that $\varphi(2^m) | y$ since $3$ is an odd prime.

(5) So there exists an integer $a > 0$ with:

$$2^m\left(2^x - 1\right) = 3^n\left(3^{2^{m-1}a} - 1\right) = 3^n(3^{2^0a} - 1)(3^{2^0a} + 1)(3^{2^1a}+1)(3^{2^2a}+1)\dots(3^{2^{m-2}a}+1)$$

(6) For each $0 \le i \le m-2$, $2 | (3^{2^ia} \pm 1)$ so $2^m | (3^y - 1)$

(7) Regardless of whether $a$ is even or odd, $8 | (3^a - 1)(3^a + 1)$

(8) But then $2^{m+1} | (3^y - 1)$ which is impossible since $2^{m+1} \nmid 2^m(2^{x} - 1)$

Edit: Added note about Carmichael's Theorem based on comment.

Edit 2: I applied Carmichael's Theorem incorrectly. For $2^{m}$, Carmichael's Theorem says $\dfrac{1}{2}\varphi(2^n)$

For example:

$$8(2^2 - 1) = 3(3^2 -1 )$$

In this case, $\varphi(8)=4 \nmid 2$ but $\dfrac{1}{2}\varphi(8) = 2 | 2$

This mistake invalidates the argument.

Larry Freeman
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  • I think your proof is mostly correct, it is just that I cannot be sure as to if y must be a whole number multiple of $\phi(2^{m})$. Also I think it is $0 \leq i \leq m-2$ for line 6 – person May 21 '21 at 17:57
  • @person Thanks for the correction on line 6. Now, fixed. Your point about $y$ being a multiple is a great point. As I understand it, Langrange's Theorem says either $\phi(2^m) | y$ or $y | \phi(2^m)$. So, maybe that is the gap. – Larry Freeman May 21 '21 at 18:07
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    @person If I understand right, Carmichael's Theorem confirms that $y|\varphi(2^m)$. This is true for odd primes. I'll add an update to my question. – Larry Freeman May 21 '21 at 18:19
  • I am not very familiar with this theorem, but I think this can be manipulated to help. Although I have not worked out how to. – person May 22 '21 at 01:10
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    In (5): $2^m | (3^{2^{m-2}a}-1)$ : the power of $2$ in the exponent of the $3^{...}-1$-expression is $m-2$, not $m-1$ for $m \ge 3$ . (However, I didn't check the consequences of this yet) – Gottfried Helms May 29 '21 at 01:55

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