If $p>5$ is a prime, then show that at least one of $2,5$ or $10$ is a quadratic residue. Then suppose, $p$ is an odd prime and $b$ is the smallest positive quadratic non residue mod $p$. prove that $b$ is also prime.

I went about it this way, but you can see where my assumptions lie pretty weak in saying that $a,b$ is equal to $2$ and $5$. Here it is...

Let $a$ and $b$ be integers not divisible by prime $p>5$. such that. $$\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)=\left(\frac{ab}{p}\right)$$ Therefore if $\left(\frac{ab}{p}\right) = 1$ then $\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)$ so either both are residues mod $p$ or both are non residues.

If $\left(\frac{ab}{p}\right) = 1$, then $\left(\frac{a}{p}\right)=-\left(\frac{b}{p}\right)$, so exactly one of $(a/p)$ and $(b/p)$ is a quadratic reside. FIN

I feel like i'm onto something here, don't' really want to give up on it. I need to somehow translate, $a=2$ and $b=5$, and $ab=10$. Is there a direct way to show this. From this its obvious that $b$ is a prime number, but also having difficulty showing that due to my first difficulty. Thanks!!

  • 2
    $\begingroup$ $2 \cdot 5 = 10.$ $\endgroup$
    – Will Jagy
    Dec 19, 2013 at 5:17
  • $\begingroup$ I don't understand your confusion. Just let $a=2, b=5$. $\endgroup$
    – Alexander
    Dec 19, 2013 at 5:22
  • 1
    $\begingroup$ without symbols, if $2$ and $5$ are non-residues, then $(2)(5)$ is a quadratic residue. $\endgroup$ Dec 19, 2013 at 5:26

2 Answers 2


Hint: Let $q$ be the smallest positive integer which is a non-residue. If $q$ is not prime, then $q=ab$ for positive integers $a,b\gt 1$. These are both residues, by the definition of $k$. But then $\dots$.


Suppose none of $2$,$5$, or $10$ are quadratic residues. Then the Legendre symbol of each of these mod $p$ is $-1$. However, $10=2\cdot5$ so the Legendre symbol of $10$ is $(-1)\cdot(-1)=(+1)$ contradicting that none are quadratic residues, i.e. at least one is.


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