I need help solving congruence: a^{p-1}+a^{p-2}+...+a^2+a+1=0 (mod p), p is prime.
I know that I can use fermat little theorem (a^{p-1}=1 (mod p)) and I know that \Z_p does not have zeros.
I have tried with some primes but can't see the similarity.
I need help solving congruence: a^{p-1}+a^{p-2}+...+a^2+a+1=0 (mod p), p is prime.
I know that I can use fermat little theorem (a^{p-1}=1 (mod p)) and I know that \Z_p does not have zeros.
I have tried with some primes but can't see the similarity.
I assume you mean solve for a? But the only solution is a=1:
$$a^{p-1}+a^{p-2}+\cdots+a+1\equiv0\,(\text{mod }p)$$
$$(a^{p-1}+a^{p-2}+\cdots+a+1)(a-1)\equiv0\,(\text{mod }p)$$
$$a^{p}-1\equiv0\,(\text{mod }p)$$
$$a\equiv a^{p}\equiv1\,(\text{mod }p)$$