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.]
 A: Using Little Fermat, we have:
$$2^{105} +3^{105}\equiv 2^{105\bmod 6} +3^{105\bmod 6}\equiv  2^{3}+ 3^{3}\equiv 1 + 6\equiv 0 \mod 7.$$
A: $$\begin{align}
2^3 &\equiv 1 (\mod 7)\\
(2^3)^{35} &\equiv 1^{35} (\mod 7)\\
2^{105} &\equiv 1 (\mod 7)\tag1
\end{align}$$
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.
A: We don't even need Fermat's Little theorem if we use Proof of $a^n+b^n$ divisible by a+b when n is odd
Now $105=3\cdot5\cdot7$
So, check for $2^3+3^3,2^5+3^5,2^7+3^7$
A: From $2^3=1\ (7)$ we get $2^{105}=1\ (7)$, and from $3^3=-1\ (7)$ we get $3^{105}=-1\ (7)$. It follows that $2^{105}+3^{105}=0\ (7)$.
A: You can write $2^{105} = 2^{6 \cdot 17}\cdot 8$ and $3^{105} = 3^{6 \cdot 17} \cdot 3^3$ and you can use Fermat's little theorem.
$2^{6 \cdot 17} \equiv 1 \pmod 7$, $8 \equiv 1 \pmod 7$ therefore $2^{2015} \equiv 1 \pmod 7$ while $3^{6 \cdot 17} \equiv 1 \pmod 7$ and $3^3 \equiv 6 \pmod 7$ therefore $3^{105} \equiv 6 \pmod 7$. $2^{105} + 3^{105}\equiv 1 + 6 \equiv 7 \equiv 0 \pmod7.$ We demonstrated that the number is divisible for $7.$
