Given that $$ab\equiv 1 \pmod n$$
and $n$ is a pseudo prime base $a$,
Show that $b$ is also a base for pseudo prime n.
I can raise to the power of $n$: $$(ab)^{n}\equiv 1 \pmod n$$
$$a^{n}{b}^{n}\equiv 1 \pmod n$$
and now I to divide them both by $ab$ which is equivalent 1.
Then I get $$ab(a^{n-1}{b}^{n-1})\equiv 1 \pmod n$$
$$a^{n-1}{b}^{n-1}\equiv 1 \pmod n$$
and $a^{n-1}\equiv 1 \pmod n$ by Fermat's Little Theorem. Then $b^{n-1}$ is congruent $1$ modulo $n$ which suits Fermat's Theorem...
though I am not realy sure of my proof...