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PaulRS · #5 June 20th, 2008, 09:37 AM

$2^{a + b} \equiv 2^b + 3 \equiv 2^b - 6\left( {\bmod .9} \right)$

Thus: $2^{a + b - 1} \equiv 2^{b - 1} - 3\left( {\bmod .9} \right)$

$2^{a + b - 1} \equiv 2^{b - 1} - 3 \equiv 2^{b - 1} + 6\left( {\bmod .9} \right)$

$2^{a + b - 2} \equiv 2^{b - 2} + 3\left( {\bmod .9} \right)$

In fact (we can also go from bottom to top in the above argument) $2^{a + b} \equiv 2^b + 3\left( {\bmod .9} \right) \Leftrightarrow 2^{a + \left( {b + 2} \right)} \equiv 2^{\left( {b + 2} \right)} + 3\left( {\bmod .9} \right)$ supposing $b\geq{0}$

If $b = 0 \Rightarrow 2^a \equiv 2^0 + 3 \equiv 3\left( {\bmod .9} \right)$ but $2^a \equiv 1;2;4;5;7;8\left( {\bmod .9} \right)$ (one of those options) for all naturals $a$ , thus it is absurd to assume that b is even

If $b = 1 \Rightarrow 2^{a + 1} \equiv 5\left( {\bmod .9} \right)$ (1) which is possible and in fact there are infinitely many $a$ s that satisfy that congruence.

Note that: $2^5 \equiv 5\left( {\bmod .9} \right)$ ,since $2^{\phi \left( 9 \right)}=2^6 \equiv 1\left( {\bmod .9} \right)$ it follows that $a +1= 5 + 6 \cdot k$ satisfies (1) for all natural numbers $k$ (in fact it must be of that form)

In conclusion $b$ must be odd and $a = 4 + 6 \cdot k$ for some natural number $k$

The following users thank PaulRS for this useful post:
sleepingcat
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