lean_workbook_plus_30969

Now we count the number of quadruples modulo $p$ when $p$ is prime. For this we fix any pair $(a,b)$ as long as $(a,b)\ne (0,0)$ . (Since clearly $(a,b)=(0,0)$ cannot lead to a solution of $ad-bc\equiv 1\pmod{p}$ .) This can be done in $p^2-1$ ways. Suppose that $a\ne 0$ (otherwise do the same argument with $b$ ). Pick any $c$ (in $p$ ways). Then $d$ is uniquely determined modulo $p$ as $bc-1$ times the multiplicative inverse of $a$ modulo $p$ . Thus $d$ is uniquely determined in $\{0,1,\ldots,p-1\}$ . In total, $f(p)=p(p^2-1)$ .