I am new to the number theory concept of quadratic reciprocity. I worked this example. My result is that the quadratic residues mod 5 are 0,1, and 4. Can someone tell me if I am correct?
Here is my reference
http://www2.math.ou.edu/~kmartin/nti/chap9.pdf
Here is my table
Thanks, Matt
Your table has a $X=6$ where it should say $X=4$ and $\mathbb Z_5^\times = \{1,2,3,4\}$ so the quadradic residues in $\mathbb Z_5^\times$ are $\{1,4\}$. The $0\equiv 5 \pmod5$ drops out because $$\mathbb Z_n^\times := \{ 1\le k\le n-1 : \gcd(k,n) = 1\}$$ So for a prime $p$ $$\mathbb Z_p^\times = \{1\le k\le p-1\}$$ And quadratic residues only make sense in $\mathbb Z_n^\times$, but since your reference looks at $\mathbb Z_p$, you'd include $\{0\}$ by definition. ($0$ and $1$ are always quadratic residues because $0^2 = 0, 1^2 = 1$ in any $\mathbb Z_p$)
A few pages down he states the definition $\Box_p$ to exclude $0$ as a QR, if he had left it out altogether, he wouldn't need that distinction... Also note that $\mathbb Z_p = \mathbb Z / p\mathbb Z$ in your text.