Discrete Mathematics - Modular arithmethics I've been trying to solve a task for some while now but I'm currently very stuck.

For the integers $a$ and $b$ it applies that $b \equiv a \pmod{91}$ and $\gcd (a, 91) = 1$. Determine a positive number $k> 1$ such that $b^k \equiv a \pmod{91}.$

I know that  $b^{\phi(91)} = 1\pmod{91}$ and  $b^{\phi(91)} = 72$.
After this I'm pretty empty. Would anyone be so kind to guide me or help me in the right direction?

Since we have $a\equiv b\mod 91$ and also $\varphi(91)=72$ added to $(a,91)=(b,91)=1$ giving $b^{72}\equiv 1\mod 91,$ then we can say $b\cdot b^{72}\equiv b\cdot 1\equiv a\equiv b^{73}\mod 91$.
Thus $b^k\equiv a\mod 91$ using $k=\varphi(91)+1$.
