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.

  • 2
    $\begingroup$ If $N=pq\; $ divides $a^{2t}-1$ then $p$ and $q$ divide $(a^t-1)(a^t+1)$ so $p$ divides $a^t-1$ or $a^t+1$ and same with $q$ $\endgroup$ – J. W. Tanner May 6 '19 at 16:58
  • $\begingroup$ I am confused by this question as written. But what about $a=1$? Then $a^t \equiv 1$ mod $N$ for every positive integer $t$. $\endgroup$ – Mike May 7 '19 at 17:17
  • $\begingroup$ I just edited the problem. I hope that helped to clarify $\endgroup$ – joseph May 7 '19 at 17:30

\begin{align*} a^{2t} & \equiv 1 \pmod{N} & \iff && (a^t-1)(a^t+1) \equiv 0 \pmod{p}\\ & & && (a^t-1)(a^t+1) \equiv 0 \pmod{q} \end{align*} Using the prime property: $p \mid ab \implies p \mid a \text{ or } p \mid b$. We get \begin{align*} a^ t & \equiv \pm 1 \pmod{p}\\ a^ t & \equiv \pm 1 \pmod{q} \end{align*} This gives you 4 possibilities. But one of the possibilities: $a^t \equiv 1 \pmod{p}$ and $a^t \equiv 1 \pmod{q}$ is NOT viable because we are given $a^t \not\equiv 1 \pmod{N}$.

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    $\begingroup$ I think you mean $a^t$ where you wrote $a^N$ $\endgroup$ – J. W. Tanner May 6 '19 at 17:05
  • $\begingroup$ Very helpful thank you. How would I show that this accounts for $50$% of the $a$ in the domain? Or do I need to show this is also true for some $a$ of the case where neither $p-1$ nor $q-1$ divides $t$? $\endgroup$ – joseph May 6 '19 at 17:10
  • $\begingroup$ I am still confused as to how I proceed from here. It seems to me as if we then have that in the case $p-1|t$, then $a^t\equiv1\pmod{p}$. Thus, $a^t\equiv -1\pmod{q}$ since they can't both be congruent to $1$. In the case where $q-1|t$ we have the opposite. But when neither divide $t$, then both congruences are $-1$. If that's true, then how is this supposed to hold for $50$% of the elements in $(\mathbb Z/N\mathbb Z)^*$? $\endgroup$ – joseph May 7 '19 at 15:39
  • $\begingroup$ More needs to be done here, I think. Why not $a_t \equiv_q -1 \equiv_p a^t$ e.g., $N|a^t+1$? $\endgroup$ – Mike May 7 '19 at 17:13
  • $\begingroup$ I am confused as to what you mean by that notation. $\endgroup$ – joseph May 7 '19 at 17:35

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