Question: I would like to solve this problem using the idea of congruence module m, but I have no idea how to start. Could someone help me?

If $n> 0$ and $n$ is not a multiple of $3$, prove that $a = 3^{2n} + 3^n + 1$ is divisible by $13$.

Answer: 
Here is a solution using the observation in [Cauchy's proof of Basel problem](https://en.wikipedia.org/wiki/Basel_problem#Cauchy%27s_proof).

> 
> **Lemma.** We have
> $$ \prod\_{k=1}^{n} \left( t - \tan^2\left(\frac{k\pi}{2n+1}\right)\right) = \sum\_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k. \tag{\*} $$
> 
> 
> 

Using this, we immediately know that

$$ a = \sum\_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{2} = n(2n+1) $$

and

$$ b = \prod\_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{1} = 2n+1. $$

So the ratio $a/b$ is exactly $n$. In OP's case, $n = 8$ and hence the answer is $8$.

---

*Proof of Lemma.* Write $m = 2n+1$. Then by the de Moivre's formula,

\begin{align\*}
\frac{\cos(mx) + i\sin(mx)}{\cos^m x} = \left( \frac{\cos x + i\sin x}{\cos x} \right)^m = (1 + i\tan x)^m = \sum\_{k=0}^{m} \binom{m}{k} (i\tan x)^k.
\end{align\*}

Taking the imaginary parts only and plugging $m = 2n+1$ back,

\begin{align\*}
\frac{\sin((2n+1)x)}{\cos^{2n+1} x} = \sum\_{k=0}^{n} (-1)^k \binom{m}{2k+1} \tan^{2k+1} x,
\end{align\*}

and so,

\begin{align\*}
(-1)^n \frac{\sin((2n+1)x)}{\sin x \cos^{2n} x} = \sum\_{k=0}^{n} (-1)^{n-k} \binom{m}{2k+1} \tan^{2k} x.
\end{align\*}

However, the left-hand side becomes zero if $x\_k = \frac{k\pi}{2n+1}$ for $k = 1, \dots, n$, and also note that $\tan^2(x\_k)$ for $k = 1, \dots, n$ are all different. So, denoting the right-hand side of $\text{(\*)}$ by $P(t)$, i.e., writing

$$ P(t) = \sum\_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k, $$

then this implies that $P(\tan^2(x\_k)) = 0$ for $k = 1, \dots, n$, hence $\tan^2(x\_k)$ for $k = 1, \dots, n$ are $n$ distinct zeros of $P(t)$. Since the leading coefficient of $P(t)$ is $1$, this proves $\text{(\*)}$.
Like [Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$](https://math.stackexchange.com/questions/951522/trig-sum-tan-21-circ-tan-22-circ-cdots-tan2-89-circ-text/952044#952044),

$$\tan(2n+1)x=\dfrac{\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}}{\binom{2n+1}0-\binom{2n+1}2t^2+\cdots+(-1)^n\binom{2n+1}{2n}t^{2n}}$$

where $t=\tan x$

So, the roots of $$\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}=0$$ i.e., $$t^{2n+1}-\binom{2n+1}{2n-1}t^{2n-1}+\cdots+(-1)^n(2n+1)t=0$$ are $t\_r=\tan\dfrac{r\pi}{2n+1}; r=0,\pm1,\pm2,\cdots,\pm n$

As $r=0\implies \tan\dfrac{r\pi}{2n+1}=0$, the roots of $$t^{2n}-\binom{2n+1}{2n-1}t^{2n-2}+\cdots+(-1)^n(2n+1)=0$$ are $t\_r=\tan\dfrac{r\pi}{2n+1}; r=\pm1,\pm2,\cdots,\pm n$

Writing $t\_r^2=p\_r,$

the roots of $$p^n-\binom{2n+1}{2n-1}p^{n-1}+\cdots+(-1)^n(2n+1)=0$$ are $p\_r=\tan^2\dfrac{r\pi}{2n+1}; r=1,2,\cdots, n$

Use [Vieta](https://mathworld.wolfram.com/VietasFormulas.html)'s [formula](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas)