Need some help regarding the equation $$2^a-3^b=(2^c-1)\cdot d >0$$ where $a,b,c,d$ are integers; $a,b$ are fixed; and $c>2$.
Can we show that $c,d$ exist? Thank you!
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So, what you're asking is whether every integer of the form $2^a-3^b$ has a divisor of the form $2^c-1$? – Bart Michels May 02 '15 at 15:01
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No. A counterexample is $2^6-3^3=37$, which is prime and not of the form $2^c-1$.
wythagoras
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Thank you wythagoras. What can we say rearranging the equation with 2^c+1, with sufficiently large a,b? – Enric May 02 '15 at 15:27
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I think for neither 2^c-1 nor 2^c+1 it is going to hold for all sufficiently large a,b, but that is pure conjecture. – wythagoras May 02 '15 at 15:32
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There exist sufficiently large a,b such that it holds. For example 8-3=5 is a divisior of 8^n-3^n=2^3n-3^n for all n. – wythagoras May 02 '15 at 15:36
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How we should proceed in order to see if, with sufficiently large and fixed a,b, there always exist c? Any idea...? – Enric May 02 '15 at 15:56
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I have no idea to be honest, that is exactly why I conjecture that is not true. – wythagoras May 02 '15 at 15:58