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I am trying to solve this.

If I am correct, applying Fermat's little theorem here gives:

$\sqrt{5}^{10} \equiv 1 \pmod{11}$

$\sqrt{5}^{10} = 5^5 \equiv 1 \pmod{11}$

$5^{2} \equiv 3 \pmod{11}$
$5^{3} \equiv 5*5^2 \equiv 3*5 \equiv 4 \pmod{11}$
$5^{5} \equiv 5^2*5^3 \equiv 3*4 \equiv 1 \pmod{11}$

but I have no idea how to use the fact that $11 \equiv 3 \pmod{4}$

and also, since $Z_{11}^*$ = {1,2,3,4,5,6,7,8,9,10} is not the $\sqrt(5)$ in $Z_{11}^*$ the same as in $Z$ ?

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    What is meant by $Z_{11}^*$? – 2017-01-12
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    @MarkFischler Cardinality – 2017-01-12
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    I think he means the multiplicative group inside $\mathbb{Z}/11\mathbb{Z}$. – 2017-01-12

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You've thrashed around a bit, but you've managed to close in on a solution.

Since $5^5 \equiv 1 \bmod 11$ as you verified, also $5^6 \equiv 5 \bmod 11$.

So:

$5^3$ is a square root of $5$ modulo $11$.

However this discrete square root of $5$ mod $11$ is not "the same" as a square root in the integers $\mathbb{Z}$. Not sure how you meant that last remark.

This is a particular case where, given a quadratic residue $n$ mod $p$, when prime $p$ is congruent to $3$ mod $4$, the modular square roots of $n$ mod $p$ can be explicitly computed or expressed as:

$$ \pm n^{\frac{p+1}{4}} $$

This may well have been discussed in your text or other study materials. In any case it is mentioned as the first step of the Tonelli-Shanks algorithm.

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    Yes but this does ot use the hint and barely uses Fermat's little theorem. You could easily say $1^2,2^2,3^2,4^2\equiv 16\equiv5$ bingo, but this is clearly not the point of the problem. – 2017-01-12
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    @MarkFischler: I believe the significance of $11 \equiv 3 \bmod 4$ lies in the "easy" case of extracting a modular square root. Namely when prime $p = 4k+3$, then the square roots of $n$ can be given explicitly as $\pm n^{\frac{p+1}{4}}$. This is often discussed in connection with Fermat's Little Thm. and/or taking modular square roots, but I'm guessing about what material prompted this exercise. – 2017-01-12
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    +1 for your insight that the material was probably teaching the $4k+3$ fact along with FLT. – 2017-01-13
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Some supplementary background to hardmath's answer.

  • Note that $\sqrt{5}$ does not exist in $\mathbb{Z}$.

  • The answer is "obviously" $4$. Indeed, $4^2 = 16 \equiv 5 \pmod{11}$. This would be entirely a sufficient answer. For toy questions like that, the answer is usually not very big, so you can just try the first few numbers which are $5 \pmod{11}$ and see if they're square.

  • There is an efficient algorithm for this kind of problem.