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Apparently I already proved this but I am having trouble the second time around. It is important to use only basic properties of congruences and orders of elements and Euler's generalization of Fermat's theorem (as that is the section this set of exercises is based on).

It is clear that $a^{2^{n+1}} \equiv 1 \pmod p$ here by squaring both sides.

I know that $a^{p-1} \equiv 1 \pmod p$ here since $a$ is not divisible by $p$ .

It seems obvious to me that the only solutions modulo $p$ to $x^{2} \equiv 1 \pmod p$ are $x = 1,p-1$.

It seems this would help because it would mean that if $2^{n+1}$ was not the order of $a$ modulo $p$, then there must be an $x < 2^{n}$ such that $a^{x} \equiv -1 \pmod p$. I wasn't sure if there was some argument by descent here either.

I also figured that by the division algorithm we can do a similar argument for $a^{x} \equiv -1$ instead of $a^{x} \equiv 1$ where if $x < 2^{n}$ is the order then if the quotient in the division algorithm equation is even, $x | 2^{n} \implies x$ is a power of two $\implies p|2 \implies p=2$ and then check that the $p=2$ case is also satisfied. But then the argument obviously doesn't work if the quotient in that equation is odd because then we can't move a $2$ out to the exponent to square the entire expression.

If we assume for some $x < 2^{n+1}$ we have that $a^{x} \equiv 1 \pmod p$ where $x$ is the order of $a$ modulo $p$ then we must have that $x | 2^{n+1} \implies x$ is a power of $2 \implies x = 2^{k}$ Then $a^{2^{k}} \equiv a^{2^{n+1}} \pmod p \implies p|a^{2^{k}}(a^{d}-1)$ where $d_{2^{n+1}} < 2^{n+1}$. Since $p|a^{2{k}}$ results in the contradiction $p|1$, we must have $p|a^{d}-1$ so that $a^{d} \equiv 1 \pmod p $ if $d < x$, a contradiction too However, we don't necessarily have that $d<x$.

There must be a simple way but I really can't see it.

Derek Luna
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    Since $a^{(2^{n+1})}\equiv 1\pmod p$ we see that the order of $a$ divides $2^{n+1}$. Hence it is of the form $2^b$ for some $b≤n+1$. If it were $<n+1$ it would divide $2^n$ . – lulu Nov 15 '20 at 23:35
  • I'm not seeing how this proves it. Can you elaborate in an answer, and I can accept it if I agree. – Derek Luna Nov 15 '20 at 23:37
  • It's a complete answer as it stands. Well, I should conclude by remarking that if the order was a divisor of $2^n$ then of course we'd have $a^{(2^n)}\equiv 1 \pmod p$ which we don't. – lulu Nov 15 '20 at 23:39
  • Can you point to whatever bits are unclear? – lulu Nov 15 '20 at 23:43
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    Special case of the Theorem in the dupe (which sheds light on the general matter- cf. the invoked Order Test. – Bill Dubuque Nov 16 '20 at 00:21

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Note that if $p=2$ then the claim is not true. In that case, we'd have $-1\equiv 1 \pmod 2$ so the assumption wouldn't say much. Indeed, in that case, if $a$ is non-zero, then the order of $a$ is $1$. So assume that $p$ is odd.

Squaring shows that $a^{(2^{n+1})}\equiv 1\pmod p$ from which we see that the order of $a$ divides $2^{n+1}$. Hence it is of the form $2^b$ for some $b≤n+1$. If it were $<n+1$ it would divide $2^n$, but if it were to divide $2^n$ we'd have $a^{(2^n)}\equiv 1 \pmod p$ contrary to assumption.

lulu
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  • Just general confusion, more clear to me with $-1 \equiv a^{2^{n}} \equiv a^{2^{b}d} \equiv ({a^{2^{b}}})^{d}\equiv 1^{d} \equiv 1 \pmod p$ so it must be that $b =n+1$. Thanks. – Derek Luna Nov 15 '20 at 23:46
  • Ah, I see. Yes, I was assuming that $p$ was odd. If $p=2$ then we have the annoyance that $-1\equiv 1 \pmod 2$. I will edit for clarity. – lulu Nov 15 '20 at 23:52
  • old comment* However lulu, I think this was part of my confusion all along, as this implies that either the order is in fact $2^{n+1}$, or that $p=2$ as $p$ is not necessarily an odd prime.
    • – Derek Luna Nov 15 '20 at 23:52
  • The way I see it now is, if $a^{2^{n}} \equiv -1 \pmod 2$, then $a^{2^{n}} \equiv 1 \pmod 2 $ so that $2^{n+1}$ is in fact actually never the order. Very nuanced issue I wish the author just said odd prime from the start. +$1$ – Derek Luna Nov 15 '20 at 23:59
  • Sadly, it's not unusual for writers to neglect $2$ as a prime. It's careless and, as you remark, it can be confusing. – lulu Nov 16 '20 at 00:01