Question 1019846
The question (as you asked it) allows for a really cheap solution - just take all integers of the form 1, 11, 101, 1001, 10001, etc. They are palindromes, so the reverse is the same. Also, their squares are palindromes, and this is easy to show in general.


If the pairs must consist of *distinct* positive integers (such as (13,31)), then the following construction also works:


(13,31) = (11+2, 11+20)
(103,301) = (101+2, 101+200)
(1003,3001) = (1001+2, 1001+2000)
etc.


since each of the squares is equal to 10...060...9 and 90...060...1 respectively.