TITLE: Show that $2^{105} + 3^{105}$ is divisible by $7$
QUESTION [6 upvotes]: 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.]

REPLY [2 votes]: 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 $2^{105}+3^{105}=0\ (7)$.