Let $p$ and $q$ be odd primes, and let $N=pq$. Let $t$ be a positive integer such that $a^{2t}\equiv1\pmod{N}$ for all $a\ \epsilon\ (\mathbb Z/N\mathbb Z)^*$, but the congruence $a^t\equiv1\pmod{N}$ does not hold for all $a\ \epsilon\ (\mathbb Z/N\mathbb Z)^*$
I was able to show that thus either $p-1\not|t$ or $q-1\not|t$. I am now trying to show that for $50$% of the $a\ \epsilon\ (\mathbb Z/N\mathbb Z)^*$, $a^t$ is congruent to $1$ mod one the primes $p,q$ and $-1$ the other prime.
EDIT: Using the work done from before I now have.. We consider this in two cases: the first when exactly one $p-1$ or $q-1$ divides $t$, the second when neither divide $t$. In the first case, since $(p-1)|t$, we have that $a^t\equiv1\pmod{p}$. We also have that $a^{2t}\equiv1\pmod{pq}$ which implies that $(a^t+1)(a^t-1)\equiv0\pmod{q}$. And since $a^t\not\equiv1\pmod{pq}$, $a^t\not\equiv1\pmod{q}$. Therefore $a^t\equiv-1\pmod{q}$. So every time either $p-1$ or $q-1$ divides $t$ we have that $a^t$ is congruent to $1$ modulo one prime and $-1$ modulo the other prime. In the other case, it is my understanding that we would have both congruent to $-1$ so that case would give us zero ouputs we want. I still don't quite understand how this accounts for $50$% of the reqults.