1. Every perfect square ends with 0, 1, 4, 5, 6 and 9.
2. No perfect square ends with 2, 3, 7 and 8 or odd number of zeroes
3. No perfect square can be of the form of 
   \(4k + 2, 4k + 3,3k + 2, 5k + 2\) and  \(5k + 3\).
4. Last two digits of \({n^2},{\rm{ }}{(50\; - \;n)^2},{\rm{ }}{\left( {50\; + \;n} \right)^2},{(100 - n)^2},{\rm{ }}{\left( {100{\rm{ }} + n} \right)^2}\) etc are same. For example last two digits of \(12^2\), \(38^2\) , \(62^2\), \(88^2\) (and so on) are same i.e. 44.
5. Last two digits of any square are last two digits of one of the squares from 1 to 24
6. Square of a perfect square ends only with 1, 5, 6 or \(4n\) zeroes.

Example 5: A four digit perfect square is of the form aabb, where a and b are the single digits. How many such numbers are possible?

Solution: Since last two digits are same, hence \(b\) can be 0 or 4.  Now \(b\) can not be  0 , because in that first two digits are not same. So \(b = 4\) or number is ending with 44 which means number is of the form of \({(50 \pm 12)^2}\) or \({\left( {100\;-\;12} \right)^2}\). Since aabb is also multiple of 11. so the number can be only \({\left( {100\;-\;12} \right)^2} = {\rm{ }}{88^2} = 7744\). There is only one such number.

Example 6: There is a five digit perfect square in which last two digits are same and non zero, also the number formed by the first three digits is a perfect cube. How many such numbers exist?

Solution: Since last two digits are same, hence last digit can be only 4. So the number can be a square of 112, 138, 162, 188, 212, 238, 262, 288, 312. Out of these number first three digits of only \({112^2}\)\(\) is a perfect  cube. \({112^2} = {\rm{ }}12544\).\(\)