I wanted to know whether the following is true and if so, then how could it be proved:
$$ p^{20} \equiv 1 \pmod {100} $$
$p$ is a prime number other than 2 and 5
In my earlier question I asked $p^{100} \equiv 1\pmod {100} ,p\neq2,5$ but, unable to get above one.
On
As given by the solution of your prior problem, we see from Wolfram Alpha that $\lambda(100) = 20$, hence the statement is true. As for a proof, it would follow similarly from a proof of your prior question as well.
On
Euler's theorem states that if $p$ is coprime with $100$ (in particular, if $p$ is prime and $p \neq 2$ or $5$), then $p^{\varphi(100)} \equiv 1 \mod 100$, where $\varphi(100) = 40$ is Euler's totient function. Thus we have that $p^{40} = (p^{20})^2 \equiv 1 \mod 100$, which implies two cases: $p^{20} \equiv 1 \mod 100$ or $p^{20} \equiv -1 \mod 100$.
Consider the case when $p^{20} \equiv -1 \mod 100$. Let $p^{10} \equiv k \mod 100$, then $k^2 \equiv -1 \mod 100$ and, in particular, $k^2 \equiv -1 \mod 5$. But one may see that this is impossible for any $k$ (one may easy go over five options for $k$ values), thus we have contradiction in this case.
Another way is write:
$$p^{20}-1=(p^{10}-1)(p^{10}+1)=(p^5-1)(p^5+1)(p^{10}+1)=\\ (p-1)(p+1)(p^4+p^3+p^2+p+1)(p^4-p^3+p^2-p+1)(p^{10}+1)$$
Once $p$ is an odd number then
$$2|(p-1) \text{ and }2|(p+1)\to 4|(p^{20}-1)$$
If $p\equiv 1\pmod{5}$ then
$$5|(p-1) \text{ and }5|(p^4+p^3+p^2+p+1)\to 25|(p^{20}-1)$$
If $p\equiv -1\pmod{5}$ then
$$5|(p+1) \text{ and }5|(p^4-p^3+p^2-p+1)\to 25|(p^{20}-1)$$
If $p\equiv \pm2\pmod{5}$
$$25|(p^{10}+1) \to 25|(p^{20}-1)$$
and then
$$100|(p^{20}-1)$$