The numbers $24^2 = 576$ and $56^2 = 3136$ are examples of perfect squares that have a units digits of $6.$
If the units digit of a perfect square is $5,$ then what are the possible values of the tens digit?
If 5 is the units digit of a perfect square.....the original squared integer must end in 5
Let the other part of the original integer be some non-negative multiple of 10...call this part "a"
Then
(a + 5) (a + 5) =
a^2 + 10a + 25
a^2 will either = 0 or it will end in at least two 0's
10a will = 0 or it will end in at least two 0's
Then the tens digit of the perfect square wil always = 2