How to explain these features of the squares from 01-99?

Question

The last two digits of the square of numbers from 1-99 have some interesting features as observed below. Do we have simple explanations for them? Or are there any other patterns we find?

Let the square be $(10X+Y)^2$ (X from 0-9 and Y from 1-9)

[Case X=0 and 5]The last two digits of the squares of $({Y})^2$ and $({50+Y})^2$ are the same (example: $01^2= 01,51^2=2601$): List 1 \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \ \ 01 & 04&09&16&25&36&49&64&81 \\\hline\hline \end{array}

[Case X=4 and 9]The last two digits of the squares of $({40+Y})^2$ and $({90+Y})^2$ are the same((example: $41^2= 1681,91^2=8281$): List 1' \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \ \ 81 & 64&49&36&25&16&09&04&01 \\\hline\hline \end{array} Note List 1' is List1 in reverse order.

[Case X=1 and 6]The last two digits of the squares of $({10+Y})^2$ and $({60+Y})^2$ are the same(example: $11^2= 121,61^2=3721$): List 2 \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \ \ 21& 44&69&96&25&56&89&24&61 \\\hline\hline \end{array}

[Case X=3 and X=8]The last two digits of the squares of $({30+Y})^2$ and $({80+Y})^2$ are the same(example: $31^2= 961,81^2=6561$): List 2' \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \ \ 61& 24&89&56&25&96&69&44&21 \\\hline\hline \end{array} Note List 2' is List 2 in reverse order.

[Case X=2 and 7]The last two digits of the squares of $({20+Y})^2$ and $({70+Y})^2$ are the same(example: $21^2= 441,71^2=5041$): List 3 \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \ \ 41 & 84&29&76&25&76&29&84&41 \\\hline\hline \end{array}

Note List 3 is symmetric about the center value of 25 (example: $21^2= 441,29^2=841$).

Summary:

1. When the tenth digits differ by 5, last two digits of squares of two digit numbers are the same.

2. When the tenth digits sum to 4,9, or 14, the lists of last two digits of squares of two digit numbers are in reverse order.

3. The list of last two digits of squares for $({20+Y})^2$ and $({70+Y})^2$ is symmetric about its enter of 25.

Answer 1

First, to make your observations a little more rigorous, let's define the last two digits of any one-digit number $D$ as $0$ and $D$. For example, the last two digits of the number $4$ are $04.$ This justifies writing $01$ for the last two digits of $(10\times 0 + 1)^2.$

Now observe that under this definition, if you have a non-negative integer $Q$ and you add any multiple of $100$ to $Q$ (or subtract any multiple of $100$), as long as the result is non-negative it will have the same last two digits as $Q$. (I stipulate "non-negative" in order to avoid questions such as what the last two digits of $33 - 100 = -67$ are.)

Notice that for any integer $N,$ $$ (50 + N)^2 = 2500 + 100N + N^2 = (25+N)\times100 + N^2. $$ The number $(25+N)\times100 + N^2$ is definitely not negative, because it's the square of an integer and the square of an integer cannot be negative. But $(25+N)\times100$ is a multiple of $100$, so $(25+N)\times100 + N^2$ has the same last two digits as $N^2.$

Now observe that $$10(X+5) + Y = 50 + (10X + Y).$$ Consider what happens if we use $10X + Y$ as $N$ in the previous paragraph. Whatever the last two digits are when we square $10X + Y,$ increasing $X$ by $5$ will result in the same last two digits. That's why $X=5$ gives the same results as $X=0,$ $X=6$ gives the same results as $X=1,$ etc.

Next, notice that for any integer $M,$ $$ (50 - M)^2 = 2500 - 100M + M^2 = (25-M)\times100 + M^2. $$ Again, we have a non-negative number (because it is the square of an integer); and since $(25-M)\times100$ is a multiple of $100$, it follows that $(25-M)\times100 + M^2$ has the same last two digits as $M^2.$ This is true even if $(25-M)\times100$ is negative; as long as we begin and end with a non-negative integer, we can subtract a multiple of $100$ without changing the last two digits of the result.

Now observe that $$ 10(4-X) + (10 - Y) = 40 - 10X + 10 - Y = 50 - (10X + Y). $$ This tells us that $(10(4-X) + (10 - Y))^2$ has the same last two digits as $(10X+Y)^2.$

So let's take any digit $X$ from $0$ to $4$ and look at the sequence we get by taking the last two digits of $(10X+Y)^2$ for $Y = 1, 2, 3, \ldots, 9.$ We can replace $X$ by $4-X$ and replace $Y$ by $10-Y$ and get the exact same sequence of two-digit combinations; but $10 - Y$ counts down from $9$ when $Y$ counts up from $1$; to make $10 - Y$ count from $1$ up to $9$ we need $Y$ to count downward from $9$, and so we will go through the sequence "backwards."

That explains why the the list of last two digits for $X=4$ is the reverse of the list for $X=0$ and the list for $X=3$ is the reverse of the list for $X=1$. It also says the list for $X=2$ is the reverse of itself, because $4 - 2 = 2$; therefore the list of $X=2$ must be symmetric around its center entry.

And of course it is still true that adding $5$ to $X$ doesn't change the results of the last two digits, so the list for $X=9$ is the same as for $X=4$ and is the reverse of the list for $X=0$, etc.

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The part of this answer above this line is the elementary explanation using the idea of multiples of an integer but not referring to modular arithmetic. If you understand arithmetic modulo $b$ where $b$ is an integer greater than $1$, we can shorten the proofs:

\begin{align} (50 \pm N)^2 = 2500 \pm 100N + N^2 &\equiv N^2 \pmod{100},\\ (10(X+5)+Y)^2 = (50+(10X+Y))^2 & \equiv (10X+Y)^2 \pmod{100}, \\ (10(4-X)+(10-Y))^2 = (50-(10X+Y))^2 & \equiv (10X+Y)^2 \pmod{100}. \end{align}

Answer 2

$(10a +i)^2 = 100a^2 + 20ai + i^2$ while $(10(a+ 5) + i)^2 = 100(a^2 + 10a + 25) + 20(a+5)i + i^2 = 100(a^2 + 10a +25 + i) + 20ai + i^2$.