<s>[INST] 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 .
 [/INST] Let $p^4-20p^2+19 = (p^2-1)(p^2-19)$, and $180 = 6^2\cdot 5$. But $p\equiv \pm 1\pmod{6}$, so $p^2\equiv 1\pmod{6}$, thus $p^2-1 \equiv 1-1 = 0 \pmod{6}$ and $p^2-19 \equiv 1-19 = -18 \equiv 0 \pmod{6}$, therefore $6^2 \mid p^4-20p^2+19$. On the other hand $p\not \equiv 0\pmod{5}$, so $p^2\equiv \pm 1\pmod{5}$ and $p^4\equiv 1\pmod{5}$, thus $p^4-20p^2 + 19 \equiv 1 + 19 = 20 \equiv 0 \pmod{5}$, therefore $5 \mid p^4-20p^2+19$
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