The last digit of square numbers cycles/repeats (modulo of square numbers)

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.