Prove that a perfect square must end in one of the following

Prove that a perfect square must end in one of the following pairs of digits: 00, 01, 04, 09,16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96. [Because x^2 = (50 + x)^2 (mod 100) and.x^2 = (50 - x)^2 (mod 100), it suffices to examine the final digits of x^2 for the 26 values x = 0, 1, 2,..., 25.]

Solution

End digits    number    square

00 10 10x10 = 100

01 49 49x 49 = 2401

04 48 48x48 = 2304

09 47 47x47 = 2209

16 46 46x 46 = 2116

21 11 11x11 = 121

24 32 32x32 = 1024

25 15 15x15 = 225

29 23 23x23 = 529

36 44 44x44 = 1936

41 21 21x21= 441

44 12 12x12 =144

49 43 43x43 = 1849

56 16 16x16 = 256

61 31 31x31 = 961

64 42 42x42 = 1764

69 13 13x13 =169

76 24 24x24 = 576

81 41 41x41 = 1681

84 22 22x22 = 484

89 17 17x17 =289

96 14 14x14 =196

Therefore,

these perfect squares are ended with given digits.