# I need explanation for this solution for the proof. (Perfect square ends with 0,1,4,5,6,9)

Give a proof to the sentence: "The final decimal digit of a perfect square is 0, 1, 4, 5, 6 or 9."

Solution: A integer $n$ can be expressed as $10a+b$, where $a$ and $b$ are positive integers and $b$ is 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Here $a$ is the integer obtained by subtracting the final decimal digit of $n$ and dividing by 10. (so $a=(n-b)/10$) Next note that $(10a + b)^2 = 100a^2+20ab+b^2=10(10a^2+2b)+b^2$ so the final decimal digit of $n^2$ is same as the final decimal digit of $b^2$.

I understand until this point but not below: Furthermore, note that the final decimal digit of $b^2$ is the same as final decimal digit of $(10-b)^2 = 100 - 20ab + b^2$. (how did you get this equation?) consequently we cab reduce our proof to the consideration of fix cases.

final digit of:

1) $n$ is 1 or 9 is 1

2) $n$ is 2 o 8 is 4

3) $n$ is 3 or 7 is 9

4) $n$ is 4 or 6 is 6

5) $n$ is 5 is 5

6) $n$ is 0 is 0

THANKS!

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Well, it should say "note that the final decimal digit of $b^{2}$ is the same as the final decimal digit of $(10-b)^{2} = 100-20b+b^{2}$." Since $100-20b+b^{2}=10(10-2b)+b^{2}$, the only number affecting the units digit of $(10-b)^{2}$ is $b^{2}$. Thus, the units digit of $b^{2}$ is the same as the units digit of $(10-b)^{2}$.

Note that this whole part of the proof is completely unnecessary. They already showed that since any integer can be written in the form $10a+b$ and since $$(10a+b)^{2}=100a^{2}+20ab+b^{2}=10(10a^{2}+2ab)+b^{2}$$ the only part that matters in determining the units digit is $b^{2}$. From here, you could simply check all integer values of $b$ from 0 to 9, but they instead slightly simplified the number of cases needed by showing that $b^{2}$ and $(10-b)^{2}$ have the same units digit.

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How did you get 100-20b+b^{2}=10(10-2b)+b^{2}? – Mayumi Aug 1 '11 at 14:22
Sorry this part: 10(10-2b)+b^{2} – Mayumi Aug 1 '11 at 14:22
Just factor 10 out of 100-20b, so 100-20b=10(10-2b) – Michael Banaszek Aug 1 '11 at 14:24
Much clearer thanks! – Mayumi Aug 1 '11 at 14:52

The point of the final step is to prove that, upon squaring, $\rm\:b\:$ and $\rm\: 10-b\:$ leave the same remainder modulo $10\:,\:$ i.e. $\rm\:(\pm b)^2 = b^2\pmod{10}\:.\:$ Indeed $\rm\:(10-b)^2 =\: b^2 + 10\ (10-2\:b)\:.\:$ Therefore to find all of the squares modulo $10\:,\:$ it suffices to square only the integers between $0$ and $5\:,\:$ i.e.

$$\:\{0,1,2,3,4,5,6,7,8,9\}^2\equiv\{0,\pm1,\pm2,\pm3,\pm4,5\}^2 \equiv \{0,1,4,9,6,5\}\pmod{10}$$

since $\rm\:\ 9\equiv -1,\:\ 8\equiv -2,\:\ 7\equiv -3,\:\ 6\equiv -4\pmod{10}\:.$

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You might say something like: There are ten possibilities for the unit digit in the decimal representation of an integer. Then go through the cases as you have done. Then you can add in all cases the unit digit in the decimal representation of a perfect square is either 0, 1, 4, 5, 6, or 9.

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don't forget 1 :) – Michael Banaszek Aug 1 '11 at 14:21
Thanks. Corrected the list. My goof. Wasn't my first; won't be my last. – Jay Aug 1 '11 at 15:02