Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$ 
If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.

If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinking I could maybe try to show that there is also a factor of $3$ and $5$ necessarily giving that $a^4-1\equiv0 \mod{60}$.
Other things I was thinking was that as $Ord_n(a) \mid \phi(n)=\phi(60)=16$ that we just need to show that $Ord_n(a) \in \{1,2,4 \}$.
Any hints? I have the exam soon =/

Continue as you started: As $3 \nmid a$, we have $a \equiv \pm 1 \pmod 3$, giving 
$$ a^4 - 1 \equiv 1 - 1 = 0 \pmod 3 $$
which means $3 \mid a^4 + 1$.
For $5$, we have either $a \equiv \pm 1 \pmod 5$, giving
$$ a^4 - 1 \equiv 1 - 1 \equiv 0 \pmod 5 $$ 
or $a \equiv \pm 2 \pmod 5$,
$$ a^4 - 1 \equiv 16 - 1 = 15 \equiv 0 \pmod 5$$
and we are done.
