How do I show that:if$p$ is prime $>5$ then $p^4-20p^2+19$ is always divisible by $180$.? Is there someone who can show me How do i show that :If $p$ is a prime number greater than $5$ then :  $$p^4-20p^2+19$$ is always divisible by $180$.
Note : i think should   factor  $p^4-20p^2+19=$ as:$p^4-20p^2+19 = (p^2-1)(p^2-19)$  but how i do continue this Idea ?
Thank you for any help .

$180=5\cdot9\cdot4$
For any integer $p, p^4-20p^2+19\equiv p^4-2p^2+1\pmod9$
Now $p^4-2p^2+1=(p^2-1)^2$
For $(p,3)=1,p\equiv\pm1\pmod3\implies p^2\equiv1\pmod3\iff p^2-1\equiv0$
and for any integer $p, p^4-20p^2+19\equiv p^4-1\pmod{20}$
$\implies p^4\equiv1\pmod5$ by Fermat's Little Theorem for $(5,p)=1$
Now if $(p,2)=1$ $p\pm 1$ are even,$\implies 4|(p^2-1)$ which divides $p^4-1$ 
So we don't need $p(>5)$ to be prime, $(p,2\cdot3\cdot5)=1$ is sufficient
In fact Carmichael function $\lambda(2^n)=2^{n-2}$ for integer $n\ge3$
So, $\lambda(16)=4\implies p^4\equiv1\pmod{16}$
$\implies p^4\equiv1\pmod{\text{lcm}(9,5,16)}$
