<s>[INST] 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.
 [/INST] Observe that
\begin{align*}
6^{273}+8^{273} & \equiv (7-1)^{273}+(7+1)^{273} \pmod{7^2}\\
& \equiv \sum_{r=0}^{136}\binom{273}{2r}7^{273-2r}\pmod{7^2}\\
& \equiv \binom{273}{272}7\pmod{7^2}\\
& \equiv 7(273) \pmod{7^2}\\
& \equiv 0 \pmod{7^2}.
\end{align*}
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