Try all $$$A$$$ from $$$-1000$$$ to $$$1000$$$ and $$$B$$$ from $$$-1000$$$ to $$$1000$$$, print anything that works.

There are limited number of possible combinations one can possible devise. for example if you take 500 for A and even if you subtract from even largest number 499 , you still be getting number > 10**9,so going till 500 is not required . Just check combinations till 126 as some have said in the discussion, i did with 1000

You can rewrite the expression as b^5=a^5-x now you can run loop on a from 0 to x^(1/5) and filter those values of a for which a^5-x is perfect 5th power of some integer. From this you can find pair a,b which satisfy given condition.