Why is there a pattern to the last digits of square numbers?

Solution 1:

The answer to this question is a bit less profound than you might hope. To see why, first note that the last digit of the square of any natural number only depends on the number's last digit - any other digits represent powers of 10 and do not make any difference to the last digit of the square.

So the problem amounts to working out the last digit of the squares of single digit numbers (and 10, if we don't consider 0 a natural number). They are:

1. 1
2. 4
3. 9
4. 6
5. 5
6. 6
7. 9
8. 4
9. 1
10. 0

The relative frequencies of these last digits here explain why they take up the proportions of square numbers that you observe.

Solution 2:

These numbers are the squares modulo 10. Notice that the square of the number $10n+k$ is $$ (10n+k)^2 = 10(10n^2+2nk)+k^2, $$ so the last digit of the square is determined by only the last digit of the original number. In particular, we find $$ 0^2=0 \quad 1^2=1 \quad 2^2 = 4 \quad 3^2 = 9 \quad 4^2 = 10+6 \\ 5^2 = 20+5 \quad 6^2 = 30+6 \quad 7^2 = 40+9 \quad 8^2 = 60+4 \quad 9^2 = 80+1, $$ or writing "$\equiv$" to mean that they have the same last digit, $$ 0^2 \equiv 0 \\ 1^2 \equiv 1 \equiv 9^2 \\ 2^2 \equiv 4 \equiv 8^2 \\ 3^2 \equiv 9 \equiv 7^2 \\ 4^2 \equiv 6 \equiv 6^2 \\ 5^2 \equiv 5, $$ so every last digit except $0$ and $5$ is the last digit of two squares out of a block of 10 consecutive numbers, while $0$ and $5$ are the last digit of only one each.