Discrete algebra and exponents (See body text)

Question

Let $a,b\in\mathbb{Z}^+$. If $a \equiv b\bmod 49$, and $\gcd(a,49) = 1$. How can I find any positive integer $n > 1$, so that $b^n\equiv a\bmod 49$?

I'm completely stumped by this. I've been trying various solutions, none of which are even worth mentioning because of how they got me nowhere.

In these attemps I tried using Euler's $\phi(n)$ function, but I had no clue on how to proceed. I also wrote the general expression for $b^n$:

$b^n = x\cdot 49 + a,\quad x\in\mathbb{N}$

Any hints are welcome, however, a solution would be very appreciated as well.

Answer 1

We have $$b^n\equiv a\equiv b\pmod{49}\iff49|b(b^{n-1}-1)$$

As $(a,49)=1, a\equiv b\pmod{49}; (b,49)=1\implies49|(b^{n-1}-1)$

Now using Euler's Totient Theorem, $b^{\phi(49)}\equiv1\pmod{49}$ as $(b,49)=1$

So, it sufficient to have $\phi(49)=42|(n-1)$