# Thread: order t (mod p)

1. ## order t (mod p)

If $\displaystyle a,$ a not equal to 1 has $\displaystyle order t (mod p)$,
show that
$\displaystyle a^(t-1) + a^(t-2) + .......+1=0 (mod p).$

2. $\displaystyle a^t \equiv 1 \pmod{p} \Leftrightarrow a^t - 1 \equiv 0 \pmod{p}$

Can you factorise $\displaystyle a^t - 1$?

3. Originally Posted by SimonM
$\displaystyle a^t \equiv 1 \pmod{p} \Leftrightarrow a^t - 1 \equiv 0 \pmod{p}$

Can you factorise $\displaystyle a^t - 1$?

I don't know.
Do you think that's how it works?

4. Originally Posted by SimonM
$\displaystyle a^t \equiv 1 \pmod{p} \Leftrightarrow a^t - 1 \equiv 0 \pmod{p}$

Can you factorise $\displaystyle a^t - 1$?
Originally Posted by Sally_Math
I don't know.
Do you think that's how it works?
SimonM gave you basically the solution, if you factorize $\displaystyle a^t - 1$ you get $\displaystyle (a-1)(a^{t-1}+a^{t-2}+...+a+1)\equiv 0(\bmod p)$. Since $\displaystyle a-1\not \equiv 0(\bmod p)$ (unless $\displaystyle t=1$, but ignore that case) it means $\displaystyle a^{t-1} + ... + a + 1\equiv 0(\bmod p)$.