Show that $a^{2001} \equiv a \pmod {1000}$

Question

Suppose $a$ is a positive integer which is coprime to 10. Show that $$a^{2001} \equiv a \pmod {1000}$$

I know it has something to do with the Fermat-Euler theorem.

$\phi(1000) = 400$ and $a^{400}\equiv 1 \pmod {1000}$

However, I do not know how to proceed from here to show the congruence

If $a^{400}\equiv1$, then what about $a^{2000}$? and then what about $a^{2001}$? — Gerry Myerson, Feb 18 '20 at 11:15
Yes but 1000 is not a prime number, so I'm not sure how Fermat's little theorem can be applied here? — mathsnerd22, Feb 18 '20 at 11:19
Who said anything about Fermat's little theorem? Look – how are $a^{400}$ and $a^{2000}$ related? If you know $a^{400}$ (and, you do), how can you use that relation to get $a^{2000}$? — Gerry Myerson, Feb 18 '20 at 11:21
Ahhhh yes. $a^{2000} ≡ a^{5(400)} ≡ 1$ and $a^{2001} ≡ a^{2000}a ≡ 1a ≡a$
Thank you so much! — mathsnerd22, Feb 18 '20 at 11:26

score 3 · Answer 1 · edited Feb 18 '20 at 11:39

3

$a^{2000}≡a^{5\cdot400}=\left(a^{400}\right)^5≡1 (\mod 1000)$ therefore $^{2001}≡^{2000}≡[1]≡$

edited Feb 18 '20 at 11:39

Kyan Cheung

answered Feb 18 '20 at 11:32

mathsnerd22

2

Tip: you can use \pmod to automatically get proper spacing on the modulo symbol: $a\equiv b\pmod{c}$ – Wojowu Feb 18 '20 at 12:01

score 1 · Answer 2 · answered Feb 18 '20 at 11:51

1

As $1000=2^35^3$

$\lambda(5^32^3)=100,5^3$ will divide $a^n-1$ if $100|n$ and $(a,10)=1$

answered Feb 18 '20 at 11:51

lab bhattacharjee

2 Answers2