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Find all integers $x,y \in \Bbb N$ such that $2^x = 3^y+5$.

I found solutions $(x,y)=(3,1)$ and $(x,y)=(5,3)$ and now I'm trying to prove that there are no solutions for $x  \ge 6, y \ge 4$.

I've found little examples of these kind of exponential diophantine equations and they often consider the order of an element of some kind of cyclic behavior w.r.t to some modulo, but I don't know how to approach this. May I have some hints on what should I do?

Edit: I noted that if I subtract $2$ from both sides the equation is as $2^x-2=3^y+3 \iff 2(2^{x-1}-1)=3(3^{y-1}+1)$ this makes $2^{x-1}-1$ a multiple of $3$ and $3^{y-1}+1$ a multiple of $2$. So we have that $$2^{x-1} \equiv 1 \pmod 3 $$ and that $$3^{y-1} \equiv 1 \pmod 2$$ now the order of $2$ modulo $3$ is $2$ and the order of $3$ modulo $2$ is one $1$ that is $x-1$ is a multiple of $2$ and well $y-1$ is a multiple of $1$. The latter doesn't give anything nice, but if $x-1$ is a multiple of $2$ we have that $$x-1=2k \implies 2(2^{x-1}-1)=2(2^{2k}-1)=2(2^2-1)(2^{2(k-1)} +2^{2(k-2)}+ \dots + 2^2+1)$$ i.e the rhs is a multiple of $3$ also?

Mike
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Johansen
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  • I think you meant $(5,3)$. – lulu Jun 16 '22 at 15:58
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    Indeed, these are the only pairs $(k,k+5)$ less than $10^{18}$where both members are perfect powers. See this. I assume a deeper search than this has been conducted. – lulu Jun 16 '22 at 16:00
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    $(3,,1)$ is disallowed because $1$ isn't prime. – J.G. Jun 16 '22 at 16:18
  • @J.G. Damn I had a typo. It was supposed to be integers instead of primes... – Johansen Jun 16 '22 at 16:21
  • You have too many mistakes. For example, subtracting $2$ from both sides gives $2^x-2=3^y+3$ (the RHS is a sum, not a difference), and of course, no power of $2$ is ever $\equiv 0 \pmod 3$ nor is any power of $3$ ever $\equiv 0 \pmod 2$. – Geoffrey Trang Jun 16 '22 at 16:44
  • Yeah I see the mistake with the sign there. However it doesn't actually affect the result that much since we would then have that $3^{y-1} \equiv -1 \pmod{2}$ which is equivalent to $3^{y-1} \equiv 1 \pmod{2}$. Also the zeros should be ones. I edited. – Johansen Jun 16 '22 at 16:50
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    Does this answer your question? Solve for integer $m,n$: $2^m = 3^n + 5$ - found using an Approach0 search. FYI, there's also Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$ where the OP asked this as part of solving your problem, as said in their comment there. – John Omielan Jun 16 '22 at 20:32
  • @Johansen Welcome to Math SE. FYI, there's also quite a few AoPS threads that deal with your problem, e.g., Cute number theory, Equation Solve and 2^x=3^y+5. – John Omielan Jun 16 '22 at 20:37
  • Hi John! I saw the threads, but none seemed to have a "solid" answer for the problem. Also the linked dupe seems to be pulling modulo $64$ from a magicians hat so not very informative. – Johansen Jun 16 '22 at 20:46
  • @Johansen I wasn't pinged regarding your comment reply since you didn't use @ with my username but, instead, just happened to check back. Regarding the linked dupe's answer, the modulo $64$ comes from the previous line that stated $x \ge 6$, which means since $64 = 2^6$ it can be used as a modulo there to be able to state the line of $3^y \equiv -5 \pmod{64}$. I agree the answer doesn't give every detail, but I hope that with this explanation you can determine the rest. If not, you may wish to comment there to ask the OP (Original Poster) of that answer for a more detailed explanation. – John Omielan Jun 16 '22 at 20:54

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