In a previous blog post, I showed that the last digit of a square number could not be 2, 3, 7, or 8. I will now show that the last digit cycles through the same sequence of 10 digits: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1; as shown in these first 31 square numbers:
0^2: 0 1^2: 1 2^2: 4 3^2: 9 4^2: 16 5^2: 25 6^2: 36 7^2: 49 8^2: 64 9^2: 81 10^2: 100 11^2: 121 12^2: 144 13^2: 169 14^2: 196 15^2: 225 16^2: 256 17^2: 289 18^2: 324 19^2: 361 20^2: 400 21^2: 441 22^2: 484 23^2: 529 24^2: 576 25^2: 625 26^2: 676 27^2: 729 28^2: 784 29^2: 841 30^2: 900
Last time I showed that the reason for only a subset of digits could be the last digit of square numbers was because of the following module equivalence:
$x^2 \text{ mod } 10 = (x \times x) \text{ mod } 10 = ((x \text{ mod } 10) \times (x \text{ mod } 10)) \text{ mod } 10 = (x \text{ mod } 10)^2 \text{ mod } 10$
which says that the last digit of a square number will always be the last digit of the square of a single digit, which limits the possible last digits to the last digit of 0^2, 1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, and 9^2, that is, 0, 1, 4, 9, 6, or 5. From this we can easily see why there is always a cycle among the same 10 digit sequence.
Given that $x^2 \text{ mod } 10 = (x \text{ mod } 10)^2 \text{ mod } 10$, we see that the $(x \text{ mod } 10)$ part is cyclic. This is because it gives the last digit of the number $x$ which cycles through 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore the last digit of the square of these digits must also cycle. Note that this holds for any $n$ in $x^2 \text{ mod } n$ because, given that $x^2 \text{ mod } n = (x \text{ mod } n)^2 \text{ mod } n$, the $(x \text{ mod } n)$ part is cyclic every $n$ numbers.
You'll also notice that the cycle of digits is symmetric (0, 1, 4, 9, 6, 5, 6, 9, 4, 1), that is, goes back through the sequence in reverse order after 5. In a future blog post we'll see why this is.
