2. Find all pairs of positive integers $n, m$ that satisfy the equation $3 \cdot 2^{m}+1=n^{2}$.
2. We have $n^{2} \equiv 1(\bmod 3)$. Thus $n=3 k+1$ or $3 k+2$ for some nonnegative integer $k$.
(i) $n=3 k+1$. After simplifying, we have $2^{m}=3 k^{2}+2 k=k(3 k+2)$. Thus $k$ and $3 k+2$ are both powers of 2 . It is clear that $k=2$ is a solution and $k=1$ is not. If $k=2^{p}$, where $p \geq 2$, then $3 k+2=2\left(3 \cdot 2^{p-1}+1\right)$ is not a power of 2 as $3 \cdot 2^{p-1}+1$ is odd. We have one solution: $n=7, m=4$.
(ii) $n=3 k+2$ : Again we have $2^{m}=3 k^{2}+4 k+1=(3 k+1)(k+1)$ and both $k+1$ and $3 k+1$ must be powers of 2 . Both $k=0,1$ are solutions. When $k=0, m=0$, which is not admissible. For $k>1$, we have $3 k+1=2 k+(k+1)>2 k+2$ and therefore $4(k+1)>3 k+1>2(k+1)$. Hence if $k+1=2^{p}$ for some positive integer $p$, then $2^{p+2}>3 k+1>2^{p+1}$ and we conclude that $3 k+1$ cannot be a power of 2 . Thus there is one solution in this case: $(n, m)=(5,3)$.
et $A$ be an $n$-element subset of $\{1,2, \ldots, 2009\}$ with the property that the difference between any two numbers in $A$ is not a prime number. Find the largest possible value of $n$. Find a set with this number of elements. (Note: 1 is not a prime number.)