Exercise 1.5.1

Let $f=(a,b,c)$ a integral binary quadratic form.

Since $a,b,c\in \{0,\pm 1\}$, $b^2\in \{0,1\}$ and $-4ac\in \{0,-4,4\}$, then

If $f$ is a positive definite form, then $\mathrm{\Delta }<0$, thus $\mathrm{\Delta }\in \{-4,-3\}$.

For each of these values, there is some positive definite quadratic form $f=(a,b,c)$ with $a,b,c\in \{0,\pm 1\}$ whose discriminant is the given value :

So the discriminant of all integral positive definite forms $(a,b,c)$ with $a,b,c\in \{0,\pm 1\}$ is $-4$ or $-3$.

If $f$ is an indefinite form, then $\mathrm{\Delta }>0$, thus $\mathrm{\Delta }\in \{1,4,5\}$.

For the value $5$, there is some positive indefinite quadratic form $f=(a,b,c)$ with $a,b,c\in \{0,\pm 1\}$ whose discriminant is this value :

But if $\mathrm{\Delta }=1$ or $\mathrm{\Delta }=4$, then $\mathrm{\Delta }$ is a square, and by Theorem 1.3.1, $f$ is not irreducible.

So the discriminant of all integral irreducible indefinite forms $(a,b,c)$ with $a,b,c\in \{0,\pm 1\}$ is $5$.