Order of an element with a non-negative power

Question

Question: If $a$ has order $15$, write down a non-negative power $m$ such that $a^{m}$=$a^{-1}$ and also find all $m$ such that $a^{m}$=$a^{-1}$

Here is what I have that might relate to the question being asked of me
Does there exists $k$ such that order $a^{k}$=$d$$?$
order of $a^{k}$=$\frac{n}{(n,k)}$ if $0\lt k \lt d$
suppose the order is $15$
then $a^{k}$=$\frac{15}{(15,k)}$=$d$ where $\frac{d}{15}$
$15$=$d(15,k)$
$\frac{15}{d}$=$(15,k)$
so $d$=$3$
then $5$=$(15,k)$ so $k$=$5$

Here is the notes that I have but but I can not seem to solve the question being asked of me. Need help

Answer 1

Since $a$ has order $15$, it means that $a$ has $15$ different powers: $$1,a,a^2,a^3,\dots,a^{13},a^{14}$$ and that $a^{15}=1$.

From this, we have $\ a^{15k+d}=\overbrace{a^{15}\cdot \ldots \cdot a^{15}}^{k\text{ times}}\cdot a^d=a^d$ for all integers $k,d$.

In particular, $a^{15k-1}=a^{-1}$, and these are exactly the integer exponents that result in $a^{-1}$.
We get nonnegative exponents for the values $k\ge0$. The first such exponent is $14$, and indeed: $$a\cdot a^{14}=a^{14}\cdot a=a^{15}=1$$ so $a^{14}$ is an inverse of $a$.