Calculate possible values of $a^4$ mod $120$. Calculate possible values of $a^4$ mod $120$.
I don't know how to solve this, what I did so far:
$120=2^3\cdot3\cdot5$
$a^4 \equiv 0,1 \pmod {\!8}$
$a^4 \equiv 0,1 \pmod {\!3}$
$a^4 \equiv 0,1 \pmod {\!5}$
I could use the CRT to get there's a unique solution mod $30$, but that gets me nowhere, right?
E: I just realized I can calculate $a^4 \bmod 8$ instead of $\!\bmod 2$ and the factors are still $\perp$.
 A: After some heuristics with Excel I saw this:
$$(30b+a)^4 \equiv a^4 \text{ mod 120}$$
This can be verified by expanding the polynomial and noting that $30^4, 4\cdot 30^3, 6\cdot 30^2,$ and $4\cdot 30 \equiv 0 \text{ mod 120}$.
Any integer can be represented as $30b + a$ where $0 \le a \le 29$.
Hence, you can calculate $0^4 \text{ mod 120}$ up to $29^4 \text{ mod 120}$ and catch all of the values:  $0,1,16,25,40,81,96,105$.
A: Property : If $(a,n)=1$ then $a^{\phi(n)}=1\ mod\ n$.
Take $b=3,\ 5$ or $8$.
If $(a,120)=1$, in this case we have for each different values of $b$ : $4$ divide $\phi(b)$ so using the property with $n=b$ we find that $a^{4}=1\ mod\ b$ : each $b$ divide $a^{4}-1$ so do $120$.
Thus if $(a,120)=1$ then $a^{4}=1\ mod\ 120$ 
Now if $a$ and $120$ have some factors in common then $a=2^b3^c5^dr$ for some $b,\ c$ and $d$. And $(a,r)=1$.
So $a^{4}=2^{4b}3^{4c}5^{4d}=2^{4b}3^{4c}5^{2d}\ mod\ 120$ using the previous property. And if $b,\ c$ and $d$ are $>0$ (30 divide a) then $a^{4}=0\ mod\ 120$.
I don't know if we can be more specific without computing.
