Finding the number of integer solutions, why is this wrong? The question is to find the number of solutions such that $(x, y)$ are integers: $(x-8)(x-10)=2^y$. Here's what I did: $u(u-2)=2^y$. From the quadratic formula, $u=1+\sqrt{1+2^y}$. This is where I think this question was meant to be solved by modular arithmetic here, but I don't know it so please try not to use it in your answers. I now constructed two right angle triangles. Triangle 1 has sides $1$, $2^{y/2}$,a hypotenuse of $\sqrt{1+2^y}$, and an angle $t$ such that $\sin(t)=\dfrac {1}{\sqrt{1+2^y}}$. Triangle 2 has sides of  $1$, $w$, a hypotenuse of $u$, and an angle $t$ such that $\cos t=\dfrac{1}{u}$. Now we want $\cot(t)+\dfrac1u=1$, or $\cot(t)+\cos(t)=1$. The problem now just reduces to finding the number solutions to that equation. However, this gives me an infinite number of solutions. Why is this wrong? Thanks!
 A: Using your idea of substitution but without applying quadratic equations stuff:
$$u,y\in\Bbb Z\;,\;\;u(u-2)=2^y\iff u, u-2\;\;\text{are powers of}\;\;2, $$
and the only possibility I see is
$$u=4\,,\,u-2=2\ldots\ldots$$
A: Let us continue your calculation at the point you reached $u=1+\sqrt{1+2^y}$. The calculation after that was not a good idea: introducing transcendental functions is unlikely to help in a problem of elementary number theory. 
The solutions are $u=1\pm\sqrt{1+2^y}$. We want $\sqrt{1+2^y}$ to be an integer. So we want $1+2^y$ to be a perfect square, say $t^2$. We have arrived at the equation $1+2^y=t^2$ or equivalently
$$2^y=t^2-1=(t-1)(t+1).$$
Now we can argue precisely as in the other solutions: $t-1$ and $t+1$ are each a power of $2$. That forces $t=\pm 3$. Since we are using $1\pm\sqrt{1+2^y}$, the choice $t=3$ is enough. 
To finish, we have $u=1\pm 3$. So $u=-2$ or $u=4$. That gives $x=u+8=6$ or $x=12$. Each has $y=3$. 
Remark: The OP mentions modular arithmetic. That cannot quite work. The reason is that the equation does have a solution. A pure congruential argument would show there are no solutions at all. We can use congruences to rule out certain types of solutions, but certainly not all. 
A: Hint:
$(x-8)(x-10)=2^y \implies x-8=2^m$ and $x-10=2^k$. Only possible way such that both are powers of two and differ by $2$?
$2^m-2^k=2$
$2(2^{m-1}-2^{k-1})=2 \implies 2^{m-1}-1=1 \implies 2^{m-1}=2$, $m=2$ and $k=1$
