Ordered positive integers

$2^x+2^{6-x}=20$ $2^{2x}+2^6=20\times 2^x$ $(2^x)^2-20\times 2^x + 64 =0$ $2^x = \frac{20 \pm \sqrt{20^2-4\times 64}}{2}$ $2^x = 4 \mbox{ or } 16$ $x = 2 \mbox{ or } 4$ $(x,y)=(2,4) \mbox{ or } (4,2)$

$2^x + 2^y = 20 \ x,y \in \mathbb {N}$

Due to symmetry, we can assume, that x ≤ y (and if we have a solution (a,b) , then (b,a) will be another solution.)

$\text {Then, since } 2^x ≤ 2^y \text { and } 2^x > 0 , 2^y > 0 :$

$\frac {20}{2} ≤ 2^y ≤ 20$

This gives us the only integer solution in this case (16 is the only power of 2 between 10 and 20):

y = 4 ,

and after substituting this into the original equation and solving it:

$2^x + 2^4 = 20$

$2^x = 4$

x = 2 ,

which means that we have 2 (x , y) solutions altogether: (4 , 2) and (2 , 4).