Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 1G / Solution

Write an integral binary quadratic form as $[a,b,c]$. For proper equivalence of binary quadratic forms, the allowed changes of variables have determinant one. Here is the needed reduction rather than an appeal to class-number tables. A shear $(x,y)\mapsto (x+ny,y)$ changes $b$ to $b+2an$ without changing $a$ or the discriminant of a binary quadratic form. Choose $n$ so $-a<b\le a$. If the resulting $c<a$, the determinant-one change $(x,y)\mapsto (-y,x)$ replaces $[a,b,c]$ by $[c,-b,a]$, strictly decreasing the positive integer leading coefficient. Repeating must terminate with $|b|\le a\le c$. If $a=c$ and $b<0$, the same rotation makes $b$ positive; the choice $-a<b\le a$ already fixes the other boundary sign. This is a reduced positive definite binary quadratic form.

Since $4ac-b^2=67$, the reduction inequalities give $67\ge 3a^2$, so $a\in \{1,2,3,4\}$. The integer $b$ is odd. For $a=1$, only $b=1$ survives the boundary convention and $c=17$. For $a=2$, $b=\pm 1$ gives $c=68/8$, not an integer. For $a=3$, $b=\pm 1,\pm 3$ gives $68/12$ or $76/12$; for $a=4$, $b=\pm 1,\pm 3$ gives $68/16$ or $76/16$. None is integral. Therefore every form is properly equivalent to $x^2+xy+17y^2$, establishing class number one for discriminant minus sixty-seven.

For infinitude, its determinant-one shears produceEach has discriminant $-67$ and is positive definite because it is a change of variables of the displayed positive form. Their middle coefficients are distinct, so they are infinitely many distinct elements of the set.