abstract algebra - Primitve roots and congruences?

Let $p$ be an odd prime. Show that the congruence $x^4$$\equiv -1\text{ (mod }p\text{)}\ $ has a solution if and only if $p$ is of the form $8k+1$.

Here is what I did

Suppose that $x^4$$\equiv -1\text{ (mod }p\text{)}\ $ and let $y$=$ind_rx$. Then $-x$ is aolso a solution and $ind_r(-x)$ $\equiv ind_r(-1) +ind_r(x)\text{}$ $\equiv (p-1)/2 +y(mod p-1)$ Without loss of generality, we may take $0
\equiv ind_r(p-1)/2(mod p-1) \text{ }\
$. So$4y = (p-1)/2$+$m $$(p-1)$ for some $m$. But $4y<2(p-1)$, so $4y=(p-1)/2$ and so $p=8y+1$ or $4y=3(p-1)/2$. In this case, 3 must divide $y$, so we have $p=8(y/3)+1$. In either case, $p$ is of the form. Conversely, suppose $p=8k+1$ and let $r$ be a primitive root of $p$. Take $x$ $=$ $r^k$. Then $x^4$$ \equiv r^{4k} \text{ }\
\equiv r^{(p-1)/2} \text{ }\
\equiv -1 (mod p) \text{ }\
$. This$x$ is a solution.

We just learned some of this topic. My TA said do not worry about it yet. Is this correct?

Is there another way to do this? If so, please show me.

Answer

There is a simpler way:

If $x^4\equiv-1\pmod p, x^8=(x^4)^2\equiv(-1)^2\pmod p\equiv1$

So,$ord_px$ must divide $8$

But it can not divide $4$ as $x^4\equiv-1\pmod p\implies ord_px=8$

Using this, $8$ divides $\phi(p)=p-1$

Conversely seems to be fine in your approach