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I can think of some really contrived or annoying arguments, but my professor gave a simple explanation for this (and general questions like this, such as when there exists a $q^{th}$ root of unity mod $p$), but I can't remember it.

If anyone could help me out that'd be great!

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    $\begingroup$ Well $1$ is a perfect square in $\mathbb{Z}/8\mathbb{Z}$ so it's also automatically a fourth power. $$3^2 \equiv 1 \pmod {8}$$ $$3^4 \equiv (3^2)^2 \equiv 1^2 \equiv 1 \pmod{8}$$ Also there is the trivial solution $x=1$, since $1^4 \equiv 1 \pmod{8}$ $\endgroup$ – Zubin Mukerjee Feb 21 '18 at 20:33
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    $\begingroup$ Also, $-1$ is always its own inverse in $\mathbb{Z}_n$, so $(-1)^4 = ((-1)^2)^2=1$. So, for you, that would be $x=7$. $\endgroup$ – Randall Feb 21 '18 at 20:35
  • $\begingroup$ Note that there is no solution here to $x^4=1$ which isn't also a solution of $x^2=1$ so there is no element of minimum exponent $4$. $\endgroup$ – Mark Bennet Feb 21 '18 at 20:42
  • $\begingroup$ Hmm how can you tell in general if there exists some $x$ such that $x^m = 1$ in $\mathbb{Z}/n\mathbb{Z}$ given $m$ and $n$? $\endgroup$ – CRX111011 Feb 21 '18 at 20:50
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Take any odd integer $x$. Then $$ 8 \mid x^2-1=(x+1)(x-1) $$ (both factors are even, and one of them divisible by $4$). In particular, $8\mid (x^2)^2-1$.

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    $\begingroup$ Fun fact: for any odd numbers $a,b$ $8\mid a^2-b^2.$ $\endgroup$ – Chickenmancer Feb 21 '18 at 20:50
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Why don't you try a bit of modular arithmetic to get a neat general answer (perhaps your professor did this) ? For any integer $n={p_1}^{r_1}...{p_s}^{r_s}$ (prime factorization), the Chinese Remainder theorem gives an isomorphism of rings $\mathbf Z /n \cong \mathbf Z /{p_1}^{r_1} \times ...\times \mathbf Z /{p_s}^{r_s}$, which gives in turn an isomorphism $(\mathbf Z /n)^* \cong (\mathbf Z /{p_1}^{r_1})^* \times ...\times (\mathbf Z /{p_s}^{r_s})^*$ between the multiplicative groups of invertible elements. For a given prime $p$, it is known that $(\mathbf Z /{p}^{r})^*$ is cyclic of order $p({p}^{r}-1)$ if $p$ is odd, $(\mathbf Z /{2}^{r})^* \cong C_2 \times C_{r-2}$ if $p=2$, where the cyclic groups $C_2, C_{r-2}$ are generated resp. by the classes of $1$ and $5$. Now, writing any class of $\mathbf Z /n$ under the form $x=(x_1,...,x_s)$, we have $x^q=1$ iff $x_j^q=1$ for $j=1,...,r$. Lagrange theorem tells you that if $p$ is odd, $(\mathbf Z /{p}^{r})^*$ contains a (unique) subgroup of order $q$ iff $q \mid p({p}^{r}-1)$. If $p=2$, the subgroups of $(\mathbf Z /{2}^{r})^*$ are obtained by composing those of $C_2$ and $C_{r-2}$. In your case here, $x^4=1, p=2, r=3, q=4$, so $(\mathbf Z /8)^* \cong C_2 \times C_2$ and the only possibilities are $<x>= <2>$, or $<5>$, or $<2> \times <5>$ .

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