Prove that $7|3^{41}-5$ I am trying to prove that $7|3^{41}-5$. 

The way I have been approaching this problem is by trying to factor the exponent of $41$ into a product of smaller exponents that will help me find a number that is divisible by $7$ with a remainder of $5.$
I have not gotten far, but my idea is:
$3^{41} = (3^{2})^{20}\times 3 - 5$
The part that I am getting hung up on is the fact that $41$ is a prime number. In my class, we have done examples where the exponent is factorable, so there is not a lingering 3 being multiplied into the factored exponent. I believe I understand the process of how this works when the exponent is not a prime number, but I cannot seem to understand how to accomplish this proof given these parameters.
We have covered modular arithmetic in this class and I am thinking maybe I should be using that to handle this problem, but I am not sure how to do so.
 A: Fermat's little theorem:  since $7$ is prime, we have $3^6\equiv1\pmod7$.  Thus $3^{42}\equiv(3^6)^7\equiv1$.  Thus we just need to "divide by $3$".  What's the inverse of $3 \pmod7$?  It's $5$, since $5\cdot3\equiv15\equiv1\pmod7$.  Thus we get the result.
A: Without using specific theorems
Observe that:
$$\frac{3^{41}-5}{7}  \in\mathbb Z\iff \frac{3^{41}+2}{7}  \in\mathbb Z $$
$$k=\frac{3^5+2}{7} =35 \in\mathbb Z$$
Then, we have
$$\begin{align}\frac{3^{41}-5}{7} 
&\equiv \frac{3(7k-2)^8+2}{7}\\
&\equiv \frac{3×2^8+2}{7}\\
&\equiv \frac{3×2^7+1}{7}\\
&=55 \in\mathbb Z.\end{align}$$
A: $3^5+2=245\equiv0\pmod7$, so $0\equiv(3^5+2)(3^5-2)=3^{10}-4$,
so $3^{20}-2\equiv3^{20}-16= (3^{10}-4)(3^{10}+4)\equiv0\pmod7$,
so $3^{40}-4=(3^{20}-2)(3^{20}+2)\equiv0\pmod7$,
so $3^{41}-5\equiv3^{41}-12\equiv0\pmod7$.
A: Without Fermat little theorem approach. Having $a^n + 1 = (a+1)(a^{n-1}-...+1)$, one have:
$$ 3^{41} - 5 =  9(3^{39} + 1) - 14=  9(27^{13}+1) - 14 \equiv 0 \mod 7$$
A: As others have mentioned, the usual approach for this sort of problem is to use Fermat's Little Theorem.
However, we can solve it using your idea:
$$3^{41} = (3^{2})^{20}\times 3 - 5$$
Now
$$3^2 = 9 \equiv 2 \pmod 7$$
So
$$3^{20} \equiv 2^{10} = 1024 \pmod 7$$
But
$$1024 = 7\times 146 + 2$$
i.e.,
$$1024 \equiv 2 \pmod 7$$
So
$$3^{40} \equiv 2^2 = 4 \pmod 7$$
And
$$3^{41} \equiv 12 \equiv 5 \pmod 7$$
Therefore
$$3^{41} - 5 \equiv 0 \pmod 7$$
