Prove that $1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$

Question

I am in search of a correct and shortcut techniques to prove this.

Otherwise I have calculated each remainder according to the power of $20$ to prove this :-

$1 \equiv 1 \bmod 23$
$\Rightarrow 20 \equiv -3\bmod 23$
$\Rightarrow 20^2 \equiv (-3)^2 \equiv 9\bmod 23$
$20^{2n}\equiv (-3)^{2n}\bmod 23$
$20^{2n-1}\equiv (-3)^{2n-1}\bmod 23$.

Therefore

$1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$
For $\sum_{n=1}^{11} 20^{2n-1} \equiv r \bmod 23$
we have $$-3-4+10-2+5-1-9+11+7-6-8 \equiv 0 \bmod 23$$

For $\sum_{n=0}^{10} 20^{2n} \equiv s \bmod 23$
we have $$1+9+12+16+6+8+3+4+13+2-5 \equiv 0\bmod 23$$

Therefore $$\sum_{n=1}^{11} 20^{2n-1}+\sum_{n=0}^{10} 20^{2n} \equiv r+s \bmod 23$$
$$\Rightarrow 1+20+20^2+20^3+....+20^{21} \equiv 0+0 \equiv 0\bmod 23$$

If possible just show any short-cut correct way to prove this. Any help is appreciated.

Answer 1

HINT

Note that by geometric series

$$1+20+20^2+20^3+....+20^{21} \equiv \sum_{k=0}^{21}(-3)^k=\frac{1- (-3)^{22} }{4} \pmod{23}$$

then use FLT for $(-3)^{22} \pmod{23}$.

Answer 2

Bracket expansion and Fermat's little theorem gives $$ (1-20)(1+20+\cdots + 20^{21}) = 1-20^{22}\equiv 1-1 = 0\pmod{23} $$ And since $-19$ multiplied by your sum is congruent to $0$ modulo $23$, and $-19$ is coprime to $23$, that means the sum itself must also be congruent to $0$.

Answer 3

Sum is equal to $\frac{20^{22}-1}{20-1}=\frac{20^{22}-1}{19}$, but $20^{22}\cong 1 \mod 19,23$.

Answer 4

$\sum_{i=0}^{21}{20}^i=\frac{1-{20}^{22}}{1-20}$... But ${20}^{22}\cong 1\pmod{23}$ by Fermat's little theorem...