$a^{27} \equiv 1 \pmod{37}$ implies $a^9 \equiv 1 \pmod{37}$ and similar problems

Question

I am struggling with the following problem.Any help will be appreciated.

If the following statement true then please give a proof otherwise give a counterexample.

1. If $a^{27} \equiv 1 \pmod{37}$, then $a^9 \equiv 1 \pmod{37}$

2. $a^{9} \equiv 1 \pmod{37}$, then $a^3 \equiv 1 \pmod{37}$

3. $a^{5} \equiv 1 \pmod{37}$, then $a^3 \equiv 1 \pmod{37}$

Thank you.

Answer 1

The first one is true: if $a^{27}\equiv 1\pmod{37}$, then $a\bmod{37}$ is one of numbers $1,7,9,10,12,16,26,33,34$. For each of them you have $a^{9}\equiv 1\pmod{37}$.

The second one is false: $7^{9}\equiv 1\pmod{37}$, but $7^3\equiv 10\pmod{37}$.

The third one is also false, try to find a counterexample.

Edit: I thought there was $a^6$. With $a^5$ the statement is true. If $a^5\equiv 1\pmod{37}$, then $a\equiv 1\pmod{37}$ (the only case) and then of course $a^3\equiv 1\pmod{37}$.

Answer 2

$(1)\ \ $ By Fermat $\ 1 \equiv a^{36}\equiv a^{27} a^9\,$ so $\,a^{27}\equiv 1\,\Rightarrow\,a^9\equiv 1$

$(3)\ \ $ Similarly $\ a^{36}\equiv 1\equiv a^5\,\Rightarrow\ a = a^{36}/(a^5)^7\equiv 1$

$(2)\ \ $ By Fermat $\,(2^4)^9\equiv 1\,$ but $\,(2^4)^3\equiv 8^4\equiv (-10)^2\equiv -11\not\equiv 1$

Remark $\ $ Generally $\, a^{36}\equiv 1\equiv a^k\,\Rightarrow\, a^{(36,k)}\equiv 1\,$ by $\, (36,k) = 36\,i + j k\ $ by Bezout. These propertes willl become clearer when you study cyclic groups.