there are finitely many primes p for which the congruence 8x==1(mod p) has no solutions x, determine the sum of all such p

Question

there are finitely many primes p for which the congruence 8x==1(mod p)  has no solutions x, determine the sum of  all such p

ganeshie8 · Answer

@ikram002p @oldrin.bataku @mukushla @kc_kennylau

ganeshie8 · Answer

@Zarkon

anonymous · Answer

if $8x\equiv1\mod p$ has no solutions $x$ then clearly $8$ cannot be invertible $\mod p$... this is precisely the case for primes $p$ such that $\gcd(8,p)>1$ i.e. $p=2$.

anonymous · Answer

observe $8=2^3$ hence the only prime it'll be relatively non-prime with is $2$... in general this problem can be solved for $nx\equiv1\pmod p$ where by the unique factorization theorem we have $n=p_1p_2\dots$ and the sum of possible $p$ is just $\sum p_i$

ganeshie8 · Answer

so, if the equation was $10x \equiv 1 \mod p$
then the possible $p$'s for which 10 is not invertible are ${2, 5}$ ?
cuz gcd(10, p) will not be 1

nice :) ty it makes good sense :)

ganeshie8 · Answer

wonder if its easy to prove the statement :  $ax\equiv 1 \mod p$ has no solutions when $gcd(a,p) \ne 1$