Solution: Let these two digits be x and x+ 14 respectively.

Then their squares are x2 and x2+28x+ 196 respectively.

Also, their last two digits are the same, that is, 28x+ 196= 100k k is an integer.

28x-4= 100k

The multiple of 10k+3 and the multiple of 10k+8 are only 28, and the mantissa is 4.

And ∵x and x+ 14 are both double digits.

∴ 10≤x≤85

Substitute x= 13, 23, 33, 43, 53, 63, 73, 83, 18, 28, 38, 48, 58, 68, 78 respectively.

X that makes k an integer is 43,18,68.

∴ Such two digits are (18,32), (43,57), (68,82).