2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$$100|(p^{20}-1)$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.