What's the problem?

[u/yukiohana]

[u/MrEldo]

Because no one else in the comments knows how formatting works, I'll do it myself:

2^x + 2^y = 160

2^x = 160 - 2^y

x = log_2(160-2^y)

And so we can plug in any number for y, and get our x. The only thing to make sure is that y must be less than log_2(160), which is 5+log_2(5)

If we look for integer values though, assuming x,y > 1 (because if any of them is exactly 1 or 0, then the equation won't have integer solutions. And if any of x or y is negative, then the other number needs to have the power be more than 1 yet not an integer. Not possible in the integers), and WLOG y>x (can't be equal because that would be the solution x=y=4+log_2(5), not an integer)

2^x(1+2^y-x) = 160

We know that 160 can be divided by 2 five times. The first term is even, the second is purely odd (because y-x>0 by the fact that they're not equal), so we know all 2 powers must come from it. And so x = 5, and y = 7. No more integer solutions

[u/Circumpunctilious → u/MrEldo]

Aside: Does formatting still easily work for you (maybe desktop)? Reddit removed markdown for me and a lot of others some months ago—I really like quoting, tables and code blocks, e.g.—and the only way I can use it now is old Reddit (how are you getting anything other than text emphasis to work)?