Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$

Question

Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$.

I noticed that $x \equiv 5 \bmod 6$ and thus $2^x \equiv 4 \bmod 7$, but that doesn't seem to help me since $3 \cdot 9^4 +5 \equiv 4 \bmod 7$. Pretty much any other mod I use doesn't seem to work so I think proof by contradiction or something may work better.

For $m \geq 2$, we have $3 \cdot 9^m+5 \equiv 5 \pmod {81}$, so $2^x \equiv 5 \pmod {81}$, meaning $x=23+54z$ for $z \in \Bbb{N}$. Don't know if this really helps, though... — Noble Mushtak, May 12 '16 at 01:07
@alphacapture I was solving the question of finding positive integer solutions to $2^x = 3^y+5$. It is easily checked that $y$ must be odd. Thus, we may transform this equation into precisely the equation in my question, and since the only positive integer solutions to $2^x = 3^y+5$ are $(x,y) = (3,1),(5,3)$, this result must be true. — user19405892, May 12 '16 at 01:25
See https://en.wikipedia.org/wiki/Catalan%27s_conjecture; Given the context in which this problem appeared, I wouldn't be surprised if this turned out to be as difficult — alphacapture, May 12 '16 at 01:26

score 4 · Answer 1 · edited Jun 12 '20 at 10:38

Any solutions must be fairly small - either $m \le 11$ or $x \le 20$. I assume you have checked all those cases and found no solutions but the two you're already aware of; indeed, there aren't any.

Let's prove this - though with a warning: there's lots of computation and black magic ahead.

EDIT: It turns out we can substantially reduce the amount of computation. See edit at the bottom.

Suppose $m \ge 12$, and work modulo $3^{12}$. Since $2$ is a primitive root modulo $3^{12}$, we have a congruence restriction on $x$ modulo $\phi(3^{12}) = 2\cdot3^{11} = 354294.$ In particular, we must have $$x \equiv 199283\pmod{354294}.$$

Now suppose $x \ge 21$, and work modulo $2^{21}$. $3$ is not quite a primitive root modulo $2^{21}$ - it generates an index-$2$ subgroup of $(\mathbb Z/2^{21}\mathbb Z)^{\times}$, but nonetheless we obtain the restriction $$m \equiv 15627\pmod{524288}.$$

Now let's look for small primes divisible by both 354294 and 524288. It turns out that $$p = 1+11\cdot\mathrm{lcm}(354294,524288) = 1021636509697$$ is prime, and $11$ is the smallest number for which this is true.

Now consider the sets $$S_2 = \left\{2^{199283+354294\cdot k}\pmod{p}\right\}, \qquad S_3 = \left\{3^{15627+524288\cdot k}\pmod{p}\right\}.$$

It's clear from definitions that $$|S_2| = \mathrm{ord}_p\left(2^{354294}\right) = 720896, \qquad |S_3| = \mathrm{ord}_p\left(3^{524288}\right) = 649539.$$

So we expect approximately $$\frac{720896\cdot649539}{1021636509697} = 0.458\cdots$$ elements in $S_2 \cap (S_3 + 5)$. If we have no such elements, then there can't be any solutions to $2^x = 3^m + 5$ with $m \ge 12$ and $x \ge 21$.

We get lucky, and there aren't any elements in the intersection! TA-DA - there are no solutions other than the two you know!

In all seriousness, this shouldn't have worked. $S_2$ is one-fourth the size we expect; $S_3$ is one-third the size. And it was still about a fifty-fifty shot if there was anything in their intersection at all. And $p$ was a factor of two smaller than expected. We should have had to find much bigger congruences and use those. But black magic happened, and everything worked out.

One more remark: the reason we chose $p$ divisible by both 354294 and 524288 was that it forced the sizes of $S_2$ and $S_3$ to be fairly small relative to $p$. But their product should have been roughly $11/2$ the size of $p$ in the worst possible case; as I've said, we got lucky.

EDIT: Here's how to make the numbers involved much smaller.

Work modulo $3^5$; then $x \equiv 23\pmod{162}$. Work modulo $2^2$; $m$ must be odd. Then work modulo $163$. The only possible residue of $2^{23+162k}\pmod{163}$ is $139$, and $134$ is not a possible residue for $3\cdot9^m$ modulo $163$. Hence we're done.

Do you think we could get the chance that the two sets will intersect to be as small as we like if we choose the powers large enough? — alphacapture, May 12 '16 at 10:47
Your second solution (after the edit) is neat and elegant. Well done. — Ian Miller, May 17 '16 at 14:40
@IanMiller Given that there are 81 possible residues for $3\cdot 9^m\pmod{163}$, I'm not sure I agree with "elegant" - more like "lucky". Thank you for the compliment, though. — Thurmond, May 18 '16 at 02:04

Reinaldo R. · Answer 2 · 2016-05-12T00:58:24.743

-1

Suppose that $x$ is a solution, then: $ 2^{x} = 3.9^{m} + 5 = 3.9^{m} + 3 + 2$, so: $$2(2^{x-1} - 1) = 3( 9^{m} + 1) \Rightarrow (3.(9^{m} + 1)) | (2^{x-1} - 1)$$ For $m\geq 2$. But $2^{x-1} - 1$ is odd and $3.(9^{m} + 1)$ is pair.

edited May 12 '16 at 00:58

answered May 12 '16 at 00:50

Reinaldo R.

338

"is pair" = "is even"? – Semiclassical May 12 '16 at 00:54
2

Perhaps there is an obvious reason, but why can't the following happen? $$\begin{cases}9^m+1=\frac{2(2^{x-1}-1)}{3}\ 3\mid 2^{x-1}-1\end{cases}$$ – May 12 '16 at 00:54
2

Now, you have your division statement backwards. $a \mid b \iff ya=b$, so you should have $(2^{x-1}-1) \mid (3 \cdot (9^m+1))$, except that's not a contradiction since an odd number can certainly divide an even one. – Noble Mushtak May 12 '16 at 00:59
I've edited the answer, this solve your problems? – Reinaldo R. May 12 '16 at 00:59
It is possible for an odd number to divide an even number, so I don't see a contradiction. – user19405892 May 12 '16 at 01:00
A pair number cannot divide an odd number. – Reinaldo R. May 12 '16 at 01:01
@ReinaldoR. What do you mean by a "pair number"? – user19405892 May 12 '16 at 01:03
Sorry, pair numbers are even numbers. – Reinaldo R. May 12 '16 at 01:04
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@user19405892 This answer is still incorrect. They wrote the division statement backwards. $a \mid b \iff ya=b$, not the other way around. – Noble Mushtak May 12 '16 at 01:08
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@ReinaldoR. No, you did not solve the problems of your answer. Let's say I claim that $9^m+1$, being an even number is $9^m+1=2k$ and that it holds (as it must) $$\begin{cases}9^m+1=2k\ 3k=2^{x-1}-1\end{cases}$$ where does – May 12 '16 at 01:11
Ok, but $a = (9^{m} + 1)$ and $b = 2^{x} - 2$, but, $a$ cannot divide $2$, so, following the results. – Reinaldo R. May 12 '16 at 01:13
I don't see what is wrong with this proof. It is true that if $2(2^{x-1}-1) = 3(9^m+1)$, then $9^m+1$ must divide $2^{x-1}-1$. – user19405892 May 12 '16 at 01:14
@G.Sassatelli, all odd numbers are $3k$ ? I think isn't true... – Reinaldo R. May 12 '16 at 01:15
@ReinaldoR. Not all odd number are in the form $2^{x-1}-1$ and not all odd numbers are in the form $\frac{9^m+1}{2}$. – May 12 '16 at 01:16
3

@user19405892 I can see that if $2(2^{x-1}-1)=3(9^m+1)$ then $9^m+1$ must divide $2(2^{x-1}-1)$, but I do not see why it must divide $2^{x-1}-1$. – alphacapture May 12 '16 at 01:17
@alphacapture Because any number $\dfrac{x}{y}$, which is an integer must have $y|x$. – user19405892 May 12 '16 at 01:19
@user19405892 what are your $x$ and $y$ in this case? – alphacapture May 12 '16 at 01:19
@alphacapture $x$ and $y$ are integers. – user19405892 May 12 '16 at 01:20
@user19405892 I mean what are they in this particular case? – alphacapture May 12 '16 at 01:20
@user19405892 If a number $n$ is not prime it may not be true that $n\mid ab\implies n\mid a\vee n\mid b$. In fact, for $m=1$ and $x=5$ it's precisely what happens to $n=9^1+1$, $a=2$, $b=2^{5-1}-1$. – May 12 '16 at 01:28
@abiessu There is a contradiction, though, as Reinaldo R. points out. – user19405892 May 12 '16 at 01:29
@G.Sassatelli The point is that the fraction is an integer. – user19405892 May 12 '16 at 01:31
@user19405892 Can you write out the exact fraction that you are referring to? – alphacapture May 12 '16 at 01:31
You can decompose $2.(2^{x-1} -1)$ in prime number, in $2^{x-1} -1$ decomposition has just odd number, then you can see the result, the answer hasn't nothing wrong. – Reinaldo R. May 12 '16 at 01:34
1

@ReinaldoR. How does $2^{x-1}-1$ being odd imply that $9^m-1$ cannot divide $2(2^{x-1}-1)$? – alphacapture May 12 '16 at 01:36
$9^{m} + 1$ is even numer, and $m\geq 2$ implies $9^{m} + 1 > 2$, then cannot divide 2 and cannot divide any prime (odd number) of the decomposition of $2^{x-1} -1$. – Reinaldo R. May 12 '16 at 01:39
1

@ReinaldoR. Note that when $m=1$, $9^m+1=10$ is still greater than 2. How come your argument does not work in that case? – alphacapture May 12 '16 at 01:40
@alphacapture see my last comment. – Reinaldo R. May 12 '16 at 01:41
Let us continue this discussion in chat. – alphacapture May 12 '16 at 01:41
Yes, this answer is incorrect. The expression $9^m+1$ is even so it will have a factor of $2$ canceled. – user19405892 May 12 '16 at 01:42

Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$

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