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My attempt was:

By Fermat's little theorem:
$$2^{22} \equiv 1 \pmod{23}$$ $$(2^{11})^2 \equiv 1 \pmod{23}$$

I checked with my calculator the remainder is actually $1$. However, I wonder if I can take the square root on both sides of congruence. Any idea?

Thanks,

roxrook
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3 Answers3

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$\left( \frac{2}{23}\right) = 1 \Rightarrow 2^{11} \equiv 1 \left( \bmod 23 \right)$

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Hint $\rm\ \bmod\ 23\!:\ \ 2 \equiv 5^{\large2}\, \Rightarrow\, 2^{\large11} \equiv 5^{\large 22} \equiv 1\ $ by Fermat's little Theorem.

See Euler's Criterion and quadratic reciprocity to understand what happens generally, and see the Remark here for the analog with higher power residues.

Regarding square-roots, $\rm\ x^2 = a^2\ \iff\ (x-a)\ (x+a) = 0\ \iff\ x = \pm\: a\ \ $ holds true in any integral domain, i.e. it's true in any ring without zero-divisors. More concretely, in $\rm\ \mathbb Z/p\:,\: $ we have prime $\rm\ p\ |\ (x-a)\ (x+a)\ \Rightarrow\ p\ |\ x-a\ $ or $\rm\ p\ |\ x+a\:,\: $ so $\rm\ x \equiv \pm\: a\ \ (mod\ p)\:.$

Bill Dubuque
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    @Siva: If the OP knew such advanced techniques he wouldn't be asking this question. – Bill Dubuque Feb 28 '11 at 06:30
  • @Siv: I think Bill was trying to avoid using the Legendre symbol, and stick to stuff at the level of Fermat's Little Theorem which the OP knows. Presumably if the OP could already read and understand your notation, he may not have asked the question. :-) – ShreevatsaR Feb 28 '11 at 06:30
  • @Shreevatsa: Neither this nor Siva's answers answer the question as asked. The question was "Can I take square roots?" Chan already knew that the remainder was 1. Of course, the title does not exactly match the body. – Aryabhata Feb 28 '11 at 06:36
  • @Bill: I thought you wanted to prove that $2$ was a quadratic residue and hence I thought you wrote $2=5^2$ and hence my previous comment. –  Feb 28 '11 at 06:36
  • @Moron: If you think about it more deeply you will see that it does answer his question generally. – Bill Dubuque Feb 28 '11 at 06:38
  • @Siv: I noted $\rm\ 2\equiv 5^2\ $ which proves quite simply that $2:$ is a square $\rm\ (mod\ 23)$. So I don't understand your comment. – Bill Dubuque Feb 28 '11 at 06:41
  • @Moron: Well, it answers the question in the title. :-) This is always a problem with interpreting what the question asks… – ShreevatsaR Feb 28 '11 at 06:42
  • @Shree: I agree, but I guess OP agrees with my interpretation... – Aryabhata Feb 28 '11 at 06:45
  • @Moron: Your "answer" is far from answering the titled question. – Bill Dubuque Feb 28 '11 at 06:56
  • @all: I really apologize for the confusing title. The true question that I wanted to ask was "if I can take the square root on both sides of congruence". However, I want name my question as a problem statement because it would be more convenient for searching. Most of the time, I want hints; sometimes I want the answer, not because it was my homework, I want to see how the problem being solved in a different way. – roxrook Feb 28 '11 at 20:13
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Yes you can take the square root, the elements $\{0,1,2, \dots, 22\}$ form a finite field, when the operations are taken modulo $23$.

That only tells you that it is $\pm 1$, though.

Aryabhata
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  • @Moron: But the remainder of a number ranging from 0 -> quotient. How could it be -1? Could you explain this? Thanks. – roxrook Feb 28 '11 at 05:49
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    @Chan: 22 = -1... – Aryabhata Feb 28 '11 at 05:51
  • @Moron: Thanks, I understood that 23|(22 + 1). What I did not understand is how $-1$ is called remainder, since $-1 < 0.$ – roxrook Feb 28 '11 at 06:10
  • @Chan: Modular arithmetic allows for negative numbers. -1 becomes 22, if you want a number between 0 and 22. – Aryabhata Feb 28 '11 at 06:32
  • @Chan: mod 23, you have -1 ≡ 22 ≡ 45 ≡ -24 ≡ -47, etc. Squaring any of them gives 1 modulo 23, so each of them is a square root of 1 (well, the same square root). The other square root is 1 ≡ 24 ≡ 47 ≡ -22 ≡ -45, etc. For convenience, we can write these two square roots as -1 and 1, which, if you want to have remainders in the range 0 to 22, is the same as 22 and 1. – ShreevatsaR Feb 28 '11 at 06:40
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    Saying "you can take square roots" because you're in a finite field is a little dangerous to those who might take it at face value: only half the nonzero elements have square roots, and those that do have two of them. (And yes, this is analogous to $\mathbb{R}$, but there the negative sign is very helpful. – Fixee Feb 28 '11 at 07:47
  • @Fixee: I interpreted the question as, if the the square of a number has a remainder which is itself a square, can we take square roots? That seemed to be a natural question one might have. – Aryabhata Feb 28 '11 at 08:05