Which pairs of positive integers (,) satisfy $^2−2^=153$? My attempt: Rearrange to $x^2=2^n + 153$ and with $2^n\geq 2\ $ it follows $x^2 \geq 155\ $.
The next square number is 169, so $x = 13$ and $n = 4$. A first solution. Since $2^n$ is even and 153 is odd, $x^2$ will be odd. So any candidate solution will have an even distance of $2m$ from a previous solution and the difference between these solutions is $(x+2m)^2 - x^2 = 4mx + 4m^2$. This difference can be expressed as a difference between two powers of 2, $4mx + 4m^2 = 2^p - 2^n$. My idea was to show that this doesn´t work so that the first solution is the only one. 

Suppose $x \geq 0$. Studying the equation modulo $2$ reveals that $x$ must be odd. Studying it modulo $3$ reveals that $n$ must be even $n=2k$. 
$$x^2-2^n=(x-2^k)(x+2^k)=153$$
So $x-2^k$ and $x+2^k$ are two integers that average to $x$ but multiply to $153$. But $153=51(3)=1(153)=17(9)$, so either $x=26$ or $x=77$ or $x=13$. But $x$ is odd, so we must have $x=13$ or $x=77$. But $x=77$ doesn’t work.

Modulo 2: $x^2 \equiv 1$ so $x \equiv 1$.
Modulo $3$: $x^2-(-1)^n \equiv 0$ so $x^2 \equiv (-1)^n$. But since any square is equivalent to $0$ or $1$ modulo $3$. Either $(-1)^n \equiv 0$ (which is obviously impossible) or $(-1)^n \equiv 1$ (which is only possible if $n$ is even).
