How do I show that:if$p$ is prime $>5$ then $p^4-20p^2+19$ is always divisible by $180$.?

Question

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 .

Answer 1

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 $5 \mid p^4-20p^2+19$

Answer 2

$180=5\cdot9\cdot4$

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, $\lambda(16)=4\implies p^4\equiv1\pmod{16}$

$\implies p^4\equiv1\pmod{\text{lcm}(9,5,16)}$

Answer 3

We see that $p^4=1(mod.5)$(little Fermat's theorem), every odd number $x$ is such that $x^2=1(mod. 3)$. So we have $(p^2-1)(p^2-19)=0(mod. 9)$. Finally, $p^4-20p^2+19=1+3=0(mod. 4)$. Then, $p^4-20p^2+19=0(mod. 180)$.

Answer 4

For any integer $a$ such that $(a,6)=1, a=6b\pm1$ where $b$ is any integer

Now $(6b\pm1)^2=24b^2+24\cdot\dfrac{b(b\pm1)}2+1\equiv1\pmod{24}=1+24c$(say)

$\implies(a^2-1)(a^2-19)=24c(24c-18)=24c\cdot6(4c-3)\equiv0\pmod{144}$

Now if $a^4-20a^2+19\equiv a^4-1\pmod5$

But $a^{5-1}\equiv1\pmod5$ by Fermat's Little Theorem if $(a,5)=1$

$\implies$lcm$(144,5)|(a^4-20a^2+19)$ if $(a,5\cdot6)=1$