Interesting inverses

$-a(a+1) \equiv 1$ (mod $d$ )

$-a^2-a \equiv 1$ (mod $d$ )

$a^2+a \equiv -1$ (mod $d$ )

$a \equiv 1$ (mod $d$ ) only works for certain values of $d$ . If $a^2 \equiv 1$ (mod $d$ ), then $a \equiv -1$ (mod $d$ ) which is true only if $d=1$ . We claim that $a^3 \equiv 1$ (mod $d$ ) for all positive integers $d$ .

$a^3=a^{2+1}=a^2 \cdot a \equiv (-a-1) \cdot a \equiv -a(a+1) \equiv 1$ (mod $d$ )

Indeed, $r=3$ . We also find that:

$a+1 \equiv -a$ (mod $d$ ).

$(a+1)^3 \equiv (-a)^3$ (mod $d$ )

$(a+1)^3 \equiv -a^3$ (mod $d$ )

$(a+1)^3 \equiv -1$ (mod $d$ )

$(a+1)^{3\times 2} \equiv (-1)^2$ (mod $d$ )

$(a+1)^6 \equiv 1$ (mod $d$ )

Indeed, $s=6$ . Therefore, $r \times s=3\times6 =\boxed{18}$ .