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

Notice how the last digit is always 0, 1, 4, 9, 6, or 5? There also seems to be a cyclic pattern of digits repeating themselves but we'll talk about that in another post. In this post, we'll just talk about why certain digits do not occur at the end of square numbers, which is useful in order to check if a number is a square or not.

We can get the last digit of a number by dividing it by 10 and keeping the remainder, also known as the number modulo 10:

0^2 mod 10: 0 1^2 mod 10: 1 2^2 mod 10: 4 3^2 mod 10: 9 4^2 mod 10: 6 5^2 mod 10: 5 6^2 mod 10: 6 7^2 mod 10: 9 8^2 mod 10: 4 9^2 mod 10: 1 10^2 mod 10: 0 11^2 mod 10: 1 12^2 mod 10: 4 13^2 mod 10: 9 14^2 mod 10: 6 15^2 mod 10: 5 16^2 mod 10: 6 17^2 mod 10: 9 18^2 mod 10: 4 19^2 mod 10: 1 20^2 mod 10: 0 21^2 mod 10: 1 22^2 mod 10: 4 23^2 mod 10: 9 24^2 mod 10: 6 25^2 mod 10: 5 26^2 mod 10: 6 27^2 mod 10: 9 28^2 mod 10: 4 29^2 mod 10: 1 30^2 mod 10: 0

It turns out that this sort of exclusion of certain modulo results from square numbers happens with many other numbers apart from 10. Take 12 for example:

0^2 mod 12: 0 1^2 mod 12: 1 2^2 mod 12: 4 3^2 mod 12: 9 4^2 mod 12: 4 5^2 mod 12: 1 6^2 mod 12: 0 7^2 mod 12: 1 8^2 mod 12: 4 9^2 mod 12: 9 10^2 mod 12: 4 11^2 mod 12: 1 12^2 mod 12: 0 13^2 mod 12: 1 14^2 mod 12: 4 15^2 mod 12: 9 16^2 mod 12: 4 17^2 mod 12: 1 18^2 mod 12: 0 19^2 mod 12: 1 20^2 mod 12: 4 21^2 mod 12: 9 22^2 mod 12: 4 23^2 mod 12: 1 24^2 mod 12: 0 25^2 mod 12: 1 26^2 mod 12: 4 27^2 mod 12: 9 28^2 mod 12: 4 29^2 mod 12: 1 30^2 mod 12: 0

With 12 you only get 0, 1, 4, and 9 as digits which is less than with 10. Why does this happen? We can prove that this is the case using modulo algebra. Let's call the number being squared $x$ and the number after the modulo $n$. Then by the distributive law of the modulo operation:

$x^2 \text{ mod } n = (x \times x) \text{ mod } n = ((x \text{ mod } n) \times (x \text{ mod } n)) \text{ mod } n = (x \text{ mod } n)^2 \text{ mod } n$

What the above derivation says is that when dividing a square number, the remainder will be based on the remainder of the root of the square number. In the case of the last digit of a square number, which is equivalent to the square number modulo 10, the last digit will be one of the last digits of the single digits squared:

**0**,

**1**,

**4**,

**9**, 1

**6**, 2

**5**, 3

**6**, 4

**9**, 6

**4**, 8

**1**. This is because $x \text{ mod } 10$ gives you the last digit of the number $x$, which is a single digit, and then this is squared and the last digit of the square is the answer.

In the general case, this is setting a limit on the number of possible remainders (usually $x \text{ mod } n$ has $n$ different possible answers for all $x$). The number of different remainders is first limited to $n$ with the first inner modulo but then only the remainders resulting from these numbers squared can be returned. This means that if one of the square numbers has the same modulo as another square number, then the total number of final remainders has decreased.

In a future post we will investigate why the remainders of square numbers are cyclic.

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