<s>[INST] Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$ 
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$.

Is there some way we can transform the equation in order to get the inequality? We have $2b^2 = 3^b-1$.
 [/INST] Given:  $ \qquad \displaystyle  2b^2+1 = 3^b \tag 1$             
We see immediately $b=1$ and $b=2$ give equalities.
What if $b$ grows over $2$, so $b=2+c$? Let's first denote $ß=\lg3 $ and we know, $ß >1$ . With this we make the ansatz:
$$     2(2+c)^2+1 = 3^{2+c}  \tag 2$$
and develop by expanding and rearranging
$$ \begin{array}{rll} 
     2(2+c)^2+1 &= 9\cdot 3^c  \\
  9+8c+2c^2 &= 9\cdot (1+ßc + ß^2c^2/2! &+ ß^3c^3/3! + ... ) \\
  1+8/9c+2/9c^2 &= 1+ßc + ß^2c^2/2 &+ ß^3c^3/3! + ...\\
 \text{ getting }&\\
  (8/9-ß)c+1/18(4-9ß^2)c^2 &=   &\phantom+ ß^3c^3/3! + ... & \qquad (3)\\
\end{array}$$         


*

*The rhs is positive with $c>0$.         

*Because $ß>1$ the parentheses on the lhs are negative. So for all $c>0$ the lhs is negative.        


So for all $c>0$  in (3) (and thus $b>2$ in (1)) the lhs is smaller than the rhs and the equality (1) does never hold.
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