How many two digits positive integers N have the property that sum N and the natural obtained reversing the order of the digits of N is perfect square ?

							
 
We have to find how many 2 digit numbers which on adding with their mirror images (obtained by reversing the order of the digits) give perfect squares, are there. Positive integers are just whole numbers greater than zero. So we are hunting for the number (quantity) of the numbers, between 9 and 100, which on adding with their mirror images give perfect squares.
Right? Let me continue (please inform me if I'm wrong).
We can express any double digit number by 10x+y where x and y are 1, 2,….,8, 9, 0 (x can't be 0 as the number would be a one digit number). Clearly the mirror image would be 10y+x. And the sum would be,
(10x+y) + (10y+x) = 10(x+y)+(x+y) = 11(x+y).
In order to 11(x+y) be a perfect square (x+y) need to be 11, right? (Other than 11, (x+y) can be 44, 99, 176,… to make 11(x+y) a perfect square but then the final number would be 484, 1089, 1936,… where the sum of any two double digit numbers can't be more than 198(99+99=198), so only 11!)
Now, how many single digit pairs are there to give 11 as answer on adding?
4 (9+2, 8+3, 7+4, 6+5).
So, there must be 4 double digit numbers fulfilling our criteria. Actually not. They're 8 such numbers as such a number and it's mirror image both meet the criteria (For clearer picture: 29 and 92 both are such numbers)
So, there are 8 two digit positive integers which on adding with their mirror images (obtained by reversing the order of the digits) give perfect squares.