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I am supposed to use this fact to help prove it.

If $p$ is an odd prime, then at least one of the numbers $2,5,10$ is a quadratic residue mod $p$

I can prove this by saying let $(\frac{10}{p}) = 1$ Then $(\frac{2}{p}) = -1$ $(\frac{5}{p}) = -1$

let $(\frac{10}{p}) = -1$ Then $(\frac{2}{p})$ or $(\frac{5}{p}) = 1$

So to start off we know that p is a odd prime $>11$
$2,5$ or $10$ is a quadratic residue of p

I am assuming the best way to start this is to say

Let $(\frac{2}{p}) = 1$ where $p>11$ Then try to show that $(\frac{3}{p}) = 1$ where $p>11$ is a quadratic residue

Then do the same for the other two.

But i am unsure about how to go about this


marked as duplicate by user26486, Surb, N. F. Taussig, Chappers, user147263 May 7 '15 at 1:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See this question. $\endgroup$ – user26486 May 6 '15 at 15:37
  • $\begingroup$ In particular, see this answer, which explains how you should finish your proof. $\endgroup$ – user26486 May 6 '15 at 15:40

If $2$ and $3$ were both prime residues, that would suffice. However, note instead that $1 = 1^2$ is always a quadratic residue, so there's no need to examine $\binom{\underline{3}}{p}$, which is not always $1$. Note that $4 = 2^2$ and $9 = 3^2$ are similarly always residues.

(I will note that in your given argument you leave out the possibility that all three of $\binom{\underline{2}}{p}, \binom{\underline{5}}{p}, \binom{\underline{10}}{p}$ are equal to $1$. The cleaner argument would be to note that $\binom{\underline{2}}{p} \cdot \binom{\underline{5}}{p}= \binom{\underline{10}}{p}$ implies that not all three values are $-1$.)


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