1. congruence problem

suppose b is congruent to a^67 (mod 91) and (a,91)=1 (gcd) for some a,b element of Z such that 0<=a<91.

1) find k element of Z such that b^k is congruent to a (mod 91)
2) let b=53. Find a

i think 1 & 2 need to be done as separate questions but I have NO clue how to do this.

any help is appreciated.

2. Are we doing all your homework tonight?

1)
$\displaystyle b \equiv a^{67} \mod 91$
$\displaystyle b^k \equiv a^{67k} \mod 91$

So you want to find $\displaystyle k$ such that $\displaystyle 67k \equiv 1 \mod 91$; i.e. you want to find the inverse of 67 in the group $\displaystyle (\mathbb{Z}/91\mathbb{Z})^\times$. To do this, note that for any $\displaystyle x \in (\mathbb{Z}/91\mathbb{Z})^\times$, $\displaystyle x^{\phi(91)}=1$; hence $\displaystyle 67^{\phi(91)}=1$, or $\displaystyle 67(67^{\phi(91)-1})=1$... can you finish now?

3. no, i have no idea how to finish.

4. Originally Posted by Bruno J.
So you want to find $\displaystyle k$ such that $\displaystyle 67k \equiv 1 \mod 91$
Actually, you have $\displaystyle \varphi(91)=72$ (where $\displaystyle \varphi$ is the Euler function) so $\displaystyle a^{72}\equiv1\,(\bmod\,91).$

So you want to find $\displaystyle k$ such that $\displaystyle 67k\equiv1\,(\bmod\,\color{red}72\color{black})$ (not mod 91).