Some integer does not have a square root in $\mathbb{Z}_n$ if $n \geq 3$.
I am having difficulty showing this. Can someone provide minimal guidance, or a small hint to point me in the right direction?
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2Consider Prime Numbers. – The Demonix _ Hermit Oct 10 '19 at 17:28
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1Have you tried writing down all the squares in, say, $\mathbb{Z}_7$ and seeing what happens? Try it for several other values of $n$. – Ethan Bolker Oct 10 '19 at 17:28
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2Maybe some integer has two square roots? – Angina Seng Oct 10 '19 at 17:28
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alright, Thanks guys :) – Rafael Vergnaud Oct 10 '19 at 17:29
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1Either $n$ is divisible by a prime $p\geq 3$ or $n$ is divisible by $4.$ – Thomas Andrews Oct 10 '19 at 17:41
1 Answers
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The map $S:x \mapsto x^2$ on $\mathbb{Z}_n$ is not injective because $S(x)=S(-x)$. Since $\mathbb{Z}_n$ is finite, $S$ cannot be surjective.
lhf
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