Answer
Possible two-digit endings of a square of an integer are,
00,,01,21,41,61,81,04,44,84,24,64,09,69,29,89,49,18,96,76,56,36,25
Work Step by Step
here is the step by step and to the question
Without loss of generality, we can assume that n is non-negative since the square of an integer and square of its negative integer are same.
let +++++ with where l is remainder and an integer between 0 and 9 inclusive. Now calculate the square of all possible cases and observe the pattern of last two digits
(10 k+0)^{2}=100 k^{2}+20 k \times 0+0^{2}
(10 k+1)^{2}=100 k^{2}+20 k \times 1+1^{2}
(10 k+3)^{2}=100 k^{2}+20 k \times 3+3^{2}
(10 k+4)^{2}=100 k^{2}+20 k \times 4+4^{2}
(10 k+5)^{2}=100 k^{2}+20 k \times 5+5^{2}
(10 k+6)^{2}=100 k^{2}+20 k \times 6+6^{2}
(10 k+7)^{2}=100 k^{2}+20 k \times 7+7^{2}
(10 k+8)^{2}=100 k^{2}+20 k \times 8+8^{2}
(10 k+9)^{2}=100 k^{2}+20 k \times 9+9^{2}
for the last two digits, our main concern is only on,
20 k \times 0+0^{2}=00
20 k \times 1+1^{2}=(2 k) 1 bracketed term ( ) is even.
20 k \times 2+2^{2}=(4 k) 4
20 k \times 3+3^{2}=(6 k) 9
20 k \times 1+1^{2}=(8 k+1)6
20 k \times 5 +5^{2}=25
20 k \times 6+6^{2}=(12 k+3)6
20 k \times 7+7^{2}=(14 k+4)9
20 k \times 8+8^{2}=(16 k+6)4
20 k \times 9+9^{2}=(18 k+8)1
hence we ca say possible two digit endings of a square of an integer are,
00,,01,21,18,96,76,56,41,64,09,69,29,89,61,81,04,44,84,24,49,36,25
