Find $6^{273} + 8^{273}\pmod{49}$ 
The number $6^{273} + 8^{273}$ divided by $49$  has a remainder, what is its value?

I used the totient function to compute for modulo 49.
$6^{42}$ and $8^{42}$ are $-1$ and $1$ mod $49$ respectively, $273/49$ is equal to $5$ with a remainder of $21$.
We would then look for the remainder of $-6^{21}$ + $8^{21}$  which I do not know how to solve.
I am aware of other solutions such as factoring odd exponents, but I wanted to know if we can use this kind of approach.

Note that $6\cdot 8\equiv 48 \equiv -1 \bmod 49$, so $8^{-1}\equiv -6 \bmod 49$.
Then $6^{273}+8^{273} \equiv 6^{273}+(-6)^{-273} \equiv 6^{273} -6^{-273}\bmod 49$ (since $-1^{-1} \equiv -1$)
Since $\phi(49)=42$ Euler gives us $6^{42}\equiv 1\bmod 49$. Then note that $6^{273}\not\equiv 0 \bmod 49$ and $6^{273} (6^{273} -6^{-273}) \equiv 6^{546} -1 \equiv (6^{42})^{13} - 1 \equiv 0 \bmod 49 $ and thus $6^{273} -6^{-273} \equiv 6^{273} + 8^{273} \equiv 0 \bmod 49$
