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