# Thread: Let p be an odd prime...

1. ## Let p be an odd prime...

Let p be an odd prime, and suppose a number b has order t>1 mod p. Prove that 1+b+b2+…+bt-1 is congruent to 0 mod p.

2. Originally Posted by NikoBellic
Let p be an odd prime, and suppose a number b has order t>1 mod p. Prove that $1+b+b^2+…+b^{t-1}$ is congruent to 0 mod p.
$p \mid b^t - 1=(b-1)(1 + b + \cdots + b^{t-1})$ but $p \nmid b-1$ because $o(b)=t > 1.$ (do we really need $p$ to be odd?)