<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] 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 $b>2$.
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