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 $2b^2 = 3^b-1$.
 A: Subtract $b^3$ from both sides.
$2b^2 + 1 -b^3 = 3^b-b^3$
$\Rightarrow b^2(2-b) = 3^b - (b^3+1)$
Now think about the expression 
$3^b - (b^3+1)$. 
Both $3^b$ and $b^3+1$ are monotonically increasing. Being an exponential function, $3^b$ grows faster than $b^3+1$ for positive b. These two facts ensure that if at any non-negative point $3^b$ dominates $b^3+1$, it will continue to dominate $b^3+1$ at subsequent points. And as a matter of fact, $3^b=b^3+1$ for $b=0$ 
Thus, for $b\geqslant0$ we have $3^b-(b^3+1)\geqslant0$
$\Rightarrow b^2(2-b)\geqslant0$
$\Rightarrow 2 \geqslant b$
A: 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.
A: You've got the right idea about getting an inequality. Since exponential functions grow much faster than polynomials, we should expect $3^b$ to dominate $2b^2+1$ at some point. We can show that it happens at $b=2$ using induction.
Induction hypothesis: Suppose that $2b^2+1<3^b$ for some particular value $b>2$.
Induction step: We want to prove $2(b+1)^2<3^{b+1}$ holds as well.
Let's work on the left side. We have
$$ \begin{array}{ll} 2(b+1)^2+1 & = 2b^2+4b+3 \\ & = [2b^2+1]+[2(2b)+1]+1 \\ & < [2b^2+1]+[2b^2+1]+1 \\ & <3^b+3^b+3^b \\ & =3^{b+1}.  \end{array} $$
That is, we've proved $2(b+1)^2<3^{b+1}$, and so the claim follows.
If you're not very familiar with induction, what we've just shown is that if the inequality holds for a particular value of $b$, then it holds for the next value of $b$ as well. This leads to a chain
$$ \begin{array}{lc} & 2(3^2)+1<3^3 \\ \implies & 2(4^2)+1<3^4 \\ \implies & 2(5^2)+1<3^5 \\ \implies & 2(6^2)+1<3^6 \\ \implies & \vdots \end{array}$$
Thus, $2b^2+1<3^b$ holds for all $b>2$.
