Find all positive integers n such that the decimal representation of n^2 consists of odd digits only

if n is even it is not possible

so n is odd

now (10k+m)^2 = 100k^2 + 20kb + m^2

m = 1 => 100k^2 + 20k + 1: k > 1 means tens digit even so k = 1 => n = 1
m= 3 => 100k^2 + 60k + 9 : k > 1 means tens digit even so k = 0 => n = 3
m = 5 => 100k^2 + 100k + 25 tens digit even
m = 7 => 100k^2 + 140k + 49 => tens digit even
m = 9 => 100k ^2 + 180k + 81 => tens digit even

so only choices n =1 and 3 whose square 1 digit numbers 1 and 9

Note: I could have used (10k+ 5), (10k+/-1) , (10k+/- 3) and could have got result with less computation