Finding integer solutions to an equation

Question

The question is what are the integer solutions for $x$ and $y$ which satisfy this equation: $$4^x = y^2+15$$ What I did was sketch the graph of $y = 4^x$ and the graph of $x = y^2 + 15$ and since these graphs don't intersect, there are no solutions to the equation. My understanding is obviously wrong because at least 1 solution is $x = 2, y = 1$, but can someone help me understand why I am wrong.

Answer 1

One approach is noticing that $4^x = \left(2^x \right)^2$ to rewrite the equation as $$ \left(2^x +y\right)\left(2^x -y\right) = 15 $$ If both $x$ and $y$ are integers, then so are $2^x \pm y$, so they must be divisors of $15$. Furthermore, since $(2^x+y) +(2^x -y) = 2^{x+1} >0$ then both pairs of divisors must be positive to add up to a positive number (since the alternative is both being negative because their product is postive). The full list of positive divisors of $15$ is $\left\{1, 3, 5, 15 \right\}$, so checking each of the two pairs that multiply to $15$ and equating one of each pair to one of $2^x \pm y$, gives the complete list of solutions $$ (x,y) \in \{(3,7), (2,1), (2,-1), (3,-7)\} $$

Answer 2

Graphing functions tells us when two same variables of functions are equal. In order to find x and y which satisfies the given the equation, we must know for what value this equation has x and y as integers. $4^x$ is exponential function of x and it has values always greater than zero for $x\in$[-R,R]. So for $4^x$ less than 15, we have y as a complex number. So, 4^x must be greater than 15 for real y. For solution to be integer, 4^x has to be a integer. It can be seen explicity that if $x=n$ or $x=n+\frac{1}{2}$ and $n\geq2$ where $n\in{N}$, then y is integer which satify the equation. For x and y to be integers, ($x\in{N}$ and $x\geq2$).