If $(x-8)\cdot (x-10) = 2^y$, where $x,y\in \mathbb{Z}$. then $(x,y)$ is (1) If $(x-8)\cdot (x-10) = 2^y$, where $x,y\in \mathbb{Z}$. Then the no. of ordered pairs of $(x,y)$
(2) If $x^4-6x^2+1 = 7\cdot 2^y$,where $x,y\in \mathbb{Z}$. Then the no. of ordered pairs of $(x,y)$
$\underline{\bf{My\; Try}}::$ for (1) one , Given $(x-8)\cdot (x-10) = 2^y$. and $x,y\in\mathbb{Z}$
So $(x-8)=2^m\Rightarrow x= 8+2^m$ and $(x-10) = 2^n\Rightarrow x=10+2^n$and $m+n=y$
Now $10+2^n-8-2^m = x-x = 0\Rightarrow 2^m-2^n = 2$
Now I did not understand How can i solve after that
Help me
Thanks

for $2$, you have $x^4-6x^2+(1-7.2^y)=0 \Rightarrow x^2= \frac{6\pm\sqrt{32-28.2^y}}{2}$
do you now see what should be $y$???
