a) Solve for $(x, n)$ natural integers the equation

$$
2^{n}=x^{2}+1
$$

b) Solve for $(x, n)$ natural integers

$$
2^{n}+1=x^{2}
$$
a) ( $\mathrm{x}=0, \mathrm{n}=0$ ) and ( $\mathrm{x}=1, \mathrm{n}=1$ ) are the only solutions for $n \leq 1$. For $n \geq 2$, the equation does not admit any solution modulo 4. Therefore, the solutions are the pairs $(0,0)$ and $(1,1)$

b) The equation can be rewritten as

$$
2^{n}=(x-1)(x+1)
$$

$(x-1)$ and $(x+1)$ are two powers of 2 with a difference of 2. Therefore, $x-1=2$ and $x+1=4$. Thus, the only solution to this problem is $(x=3, n=3)$