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If $f$ is the exponent of a number $a$ ($\mod p$) and $\phi(f)=2$ then, $a$ and its reciprocal are the only numbers, which belong to this exponent.

$\phi$ is the euler-phi function

Nothing written about $p$ (whether it is prime or not)

$a^f\equiv1\mod p$, such that $f$ is minimal called the exponent of $a$

I mean it is clear that $a$ and $a^{-1}$ have the same exponent, but how to bring $\phi(f)$ into play ?

It is written in the first $6$ lines of the paper below (On the Representation of an Integer as the Sum of Consecutive Integers by Mason, Thomas E.) enter image description here

user1161
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  • The paper specifies that $p$ is an odd prime. (Count $5$ lines from the bottom). Probably this is said before. – ajotatxe Sep 09 '16 at 16:56

1 Answers1

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Let $f$ be the exponent of $a$. Let $G_f$ denote the set of all classes modulo $p$ with exponent $f$ and let $H_f$ be the classes with exponent a divisor of $f$ that is $b^f \equiv 1 \pmod{p}$.

Then group of invertible classes modulo $p$ is cyclic (when $p$ is prime) of order $p-1$. And $H_f$ is a subgroup with $f$ elements. The number of elements of maximal order is $\phi(f)$. Thus if $\phi(f)=2$, there are exactly two such elements. Since $a$ and $a^{-1}$ are such elements they are those $2$. (Note the two elements are different, as $f$ is not $2$, since $\phi(f)=2$.)

The claim is in general not true for general $p$, though the condition prime could be weakened. But it seems certain it is intended.

quid
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