<s>[INST] The diophantine equation $5\times 2^{x-4}=3^y-1$ 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.
 [/INST] 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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