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. 
 A: $n$ must be even. 
$$ 153 = 3^2 \cdot 17 $$
If $$ x^2 - 2 y^2  $$
is divisible by $3,$ then both $x,y$ are divisible by $3.$  Since this $y$ would be a power of $2$ this is impossible. 
So $n$ is even, $n=2k,$ and we actually have $$ x^2 - (2^k)^2 = 153 \; , $$
$$ (x+ 2^k) (x-2^k) = 153   $$
Umm. $$ (x+ 2^k) - (x-2^k) = 2^{k+1} $$
This leads to a finite set of possible $x,$ we can factor $153$ as (ordered pairs)
$$ 153 \cdot 1  $$
$$ 51 \cdot 3 $$
$$   17 \cdot 9  $$
$$ 153 - 1 = 152 = 8 \cdot 19 $$
$$ 51 - 3 = 48 = 16 \cdot 3 $$
$$ 17 - 9 = 8  $$
A: 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).
