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 =/

The question is answered successfully, I would like find the number for which $a^4\equiv1\pmod n$ for all $a$ such that $(a,n)=1$
Using Carmichael Function, If $\lambda(n)=2^m$ where integer $m\ge 1$
$n$ can not have any odd prime factor with exponent $>1$ and for each prime factor $p_i,p_i-1$ must be of the form $2^{2^{r_i}}$ where integer $r_i\ge0$
$$\text{Then if }n=2^s\cdot\prod p_i=2^s\cdot (2^{2^{r_i}}+1)$$
If $s\ge 3, \lambda(n)=$lcm $(2^{2^{r_i}},2^{s-2})$ as $\lambda(2^s)=2^{s-2}$ for $s\ge3$
$\implies\lambda(n)=$lcm $(2^{s-2},2^{2^{r_1}},2^{2^{r_2}},\cdots)=$lcm$(2^{s-2},2^{2^{\text{max}(r_i)}})$
If  $\lambda(n)$ divides $4=2^2,$ $$2^{\text{max}(r_i)}\le 2 \text{ and }s-2\le 2 \implies 0\le max(r_i)\le1 \text{ and } 3\le s\le 4 $$
For the  maximum value of $n,r_1=0,r_2=1$ and $s=4$
So, the maximum value of $n$ will be $3^1\cdot5^1\cdot2^4=240$
$$\implies x^4\equiv1\pmod {240}\implies x^4\equiv1\pmod {60}\equiv1-60=-59$$
