<s>[INST] Ordered pairs of positive integers $2^{2x} - y^2 = 60$ How many ordered pairs of positive integers are there that satisfy $2^{2x} - y^2 = 60$? 
This can be rewritten as $(2^x)^2-y^2 = 60$ and then $(2^x+y)(2^x-y) = 60$. Then since $2^x$ is always positive and so $2^x+y$ and $2^x-y$ are both positive and the first is always bigger than the latter. This means I only have to account for the positive factors of $60$, and so I had the following pairs $(60,1), (30,2), (20,3), (15,4), (12,5)$, and I plugged them in and solved the systems. Of the 5 possible cases, only $(30,2)$ worked and yielded $x=4, y=14$. 
Am I right in having done so? They don't have an answer, and when I tried to graph this, it didn't exactly work out well (I didn't get an integer value for $x$). 
 [/INST] Basic things to note about $(2^x + y)(2^x - y) = ab = 60; a = 2^x + y; b=2^x-y)$.
1) $2^x + y > 0$ so $2^x - y > 0$.
2) $a= 2^x + y > 2^x -y=b > 0$ so $a > \sqrt{60} > 7$ and $b < \sqrt{60} < 8$.
3) $a$ and $b$ can't both be odd so they are both even.
So $b = 2,\color{red}4,6$ and $a=30,\color{red}{15},10$
4) $a+b = 2^{x+1} = 32, 16$ so $x=4,3$ and $y = \frac {a-b}2 = 14, 2$.
So the two possible solutions are $(x,y)=(4,14): (2^4+14)(2^4 -14)=30*2=60$ and $(x,y)=(3,2): (2^3+2)(2^3-2)=10*6=60$.
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