I would like to know if it is possible to find an integer solution to
$2^a + 3^b = 2^{a+b} +1$
with $a,b > 1$
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Any thoughts about the problem? – Hetebrij Oct 03 '15 at 17:17
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a solution would imply a counterexample for the collatz conjecture, maybe. I just have no idea if there is some way to solve this equation. – gw653 Oct 03 '15 at 17:22
1 Answers
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We have $$2^a(2^b - 1) = (3-1)(1 + 3 + 3^2 + \ldots + 3^{b-1}),$$ so $$2^{a-1}(2^b - 1) = (1 + 3 + 3^2 + \ldots + 3^{b-1}).$$
Considering the second equation mod $3$ we thus see that $2^{a-1}$ and $2^b - 1$ aren't zero. As $2^b$ isn't zero mod $3$ either, it must be $2$ mod $3$, i.e. $b$ must be odd.
But this implies that $1 + 3 + 3^2 + \ldots + 3^{b-1}$ is $2$ mod $4$ and hence (from the first equality in this answer) that $a = 2$.
But there are no solutions with $a = 2$, since then we have $4 - 1 + 3^b = 2^{a+b}$ and the two sides are different modulo $3$.
hunter
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the rhs of the second equation is 1 mod 3 and $2^b - 1$ divides the lhs – hunter Oct 03 '15 at 17:51
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the inverseness doesn't matter, it was confusing before, sorry. i cleaned up a bit just now. – hunter Oct 03 '15 at 21:29
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