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I have this question: can we deduce directly using the Catalan conjecture that the equation $$5\times 2^{x-4}-3^y=-1$$ has or no solutions, or I must look for a method to solve it. Thank you.

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    $\begingroup$ Catalan does not apply because $5\cdot 2^{x-4}$ is not a perfect power. $\endgroup$
    – Peter
    Jul 21, 2018 at 18:30
  • $\begingroup$ $(8/4)$ seems to be the only solution, but I do not know a proof yet. No further solution for $x\le 10^5$ $\endgroup$
    – Peter
    Jul 21, 2018 at 18:33
  • $\begingroup$ maybe we have to solve an equation of form $3^y\equiv 1( \mod 2^{x-4} )$?? $\endgroup$ Jul 21, 2018 at 18:38
  • $\begingroup$ Usually, such equations are difficult to solve completely, but it is well possible that someone on this site finds a complete answer. $\endgroup$
    – Peter
    Jul 21, 2018 at 18:40
  • $\begingroup$ compare various (elementary) answers at math.stackexchange.com/questions/1941354/… They take a few steps but are just the same observations repeated.. Also, this method is not much changed by the extra factor of $5$ $\endgroup$
    – Will Jagy
    Jul 21, 2018 at 19:25

5 Answers 5

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Elementary proof. I learned the method at Exponential Diophantine equation $7^y + 2 = 3^x$

We think that the largest answer is $5 \cdot 16 = 81 - 1. $ Write this as $5 \cdot 16 \cdot 2^x = 81 \cdot 3^y - 1.$ Subtract $80$ from both sides, $ 80 \cdot 2^x - 80 = 81 \cdot 3^y - 81.$ We reach $$ 80 (2^x - 1) = 81 (3^y - 1). $$ This is convenient; we will show that both $x,y$ must be zero. That is, ASSUME both $x,y \geq 1.$ From $2^x \equiv 1 \pmod {81}$ we get $$ x \equiv 0 \pmod {54}. $$ It follows that $2^x - 1$ is divisible by $2^{54} - 1.$ $$ 2^{54 } - 1 = 3^4 \cdot 7 \cdot 19 \cdot 73 \cdot 87211 \cdot 262657 $$ Next, $3^y - 1$ is divisible by the large prime $262657$ From $3^y \equiv 1 \pmod {262657}$ we find $$ y \equiv 0 \pmod {14592} $$ and especially $$ y \equiv 0 \pmod {2^8}. $$ We do not need as much as $2^8 = 256,$ we really just need the corollary $$ y \equiv 0 \pmod 8 $$ Next $3^y - 1$ is divisible by $3^8 - 1 = 32 \cdot 5 \cdot 41.$ This is the big finish, $3^y - 1$ is divisible by $32.$ Therefore $80 (2^x-1)$ is divisible by $32,$ so that $2^x - 1$ is even. This is impossible if $x \geq 1,$ and is the contradiction needed to say that, in $$ 80 (2^x - 1) = 81 (3^y - 1) \; , $$ actually $x,y$ are both zero.

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  • $\begingroup$ Thank you very much. frankly, I just started this method like you. $\endgroup$ Jul 22, 2018 at 0:07
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With the two substitutions: $$x-4=w$$ $$y=2z$$ We rearrange to:

$$5(2^w)=(3^z-1)(3^z+1)$$ So we want $z$ such that $$\bigg[3^z-1=5(2^a)\bigg] \text{ and } \bigg[3^z+1=2^b\bigg]$$ or $$\bigg[3^z+1=5(2^a)\bigg] \text{ and } \bigg[3^z-1=2^b\bigg]$$

We can see $z=2$ (and thus $y=4$) satisfies the second line here. (which leads to $x=8$ as others conclude).

See if you can take it from here.

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  • $\begingroup$ Can we deduce that is a unique solution? $\endgroup$ Jul 21, 2018 at 19:02
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From $3^y\equiv1$ mod $5$, we see that $4\mid y$, so writing $y=4z$, we have

$$5\cdot2^{x-4}=3^{4z}-1=(3^{2z}+1)(3^{2z}-1)$$

Since $3^{2z}+1\equiv2$ mod $8$, we can only have $3^{2z}+1=2$ or $10$. We can dismiss the first possibility (since it implies $z=0$, which gives $3^{2z}-1=0$). Thus $3^{2z}+1=10$, so $z=1$ and thus $5\cdot2^{x-4}=10\cdot8$ implies $x-4=4$. We find $(x,y)=(8,4)$ as the only solution.

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Try $x=8$ and $y=4$. Not sure if there are any other integer solutions.

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  • $\begingroup$ Yes $(x,y)=(8,4)$ is a solution. How did you get the solution and is it unique? $\endgroup$ Jul 21, 2018 at 18:32
  • $\begingroup$ Guess and check. Sorry =) $\endgroup$
    – user443369
    Jul 21, 2018 at 18:34
  • $\begingroup$ This is more a comment, it's good you found a solution but now you need to solve the question in full generality. $\endgroup$ Jul 21, 2018 at 20:33
  • $\begingroup$ Can't comment since my rep is below 50 but you're right $\endgroup$
    – user443369
    Jul 21, 2018 at 21:05
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Solving for $y$ yields no solutions for $-3000\le x \le 3000$ but solving for $x$ does. WolframAlpha has a solution of $x$ in terms of $y$ here and plugging this into a spreadsheet does yield one solution in this range. $$5\times 2^{x-4}-3^y+1=0\implies x = \frac{\log(\frac{16(3^y - 1)}{5})}{\log(2)}\implies x=8\land y=4$$

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