Show that $2^{105} + 3^{105}$ is divisible by $7$ I know that $$\frac{(ak \pm 1)^n}{a}$$ gives remainder $a - 1$ is n is odd or $1$ is n is even.
So, I wrote $ 2^{105} + 3^{105}$ as $8^{35} + 27^{35}$ and then as $(7\cdot 1+1)^{35} + (7\cdot 4-1)^{35}$, which on division should give remainder of $6$ for each term and total remainder of 5 (12/7).
But, as per question, it should be divisible by 7, so remainder should be zero not 5. Where did I go wrong?
[note: i don't know binomial theorem or number theory.]

$$\begin{align}
2^3 &\equiv 1 (\mod 7)\\
(2^3)^{35} &\equiv 1^{35} (\mod 7)\\
2^{105} &\equiv 1 (\mod 7)\tag1
\end{align}$$
Again,
$$\begin{align}
3^3 &\equiv -1 (\mod 7)\\
(3^3)^{35} &\equiv (-1)^{35} (\mod 7)\\
3^{105} &\equiv -1 (\mod 7)\tag2
\end{align}$$
Adding (1) and (2) we get,
$$\begin{align}
2^{105}+3^{105} &\equiv1+(-1)\space (\mod7)\\
\text{or,}\quad 2^{105}+3^{105} &\equiv0\space (\mod7)
\end{align}$$
This implies that $\space2^{105}+3^{105}$ is divisible by 7.
