Show that $\frac{(3^{77}-1)}{2}$ is odd and composite

The question given to me is:

Show that $\large\frac{(3^{77}-1)}{2}$ is odd and composite.

We can show that $\forall n\in\mathbb{N}$:

$$3^{n}\equiv\left\{ \begin{array}{l l} 1 & \quad \text{if $n\equiv0\pmod{2}$ }\\ 3 & \quad \text{if $n\equiv1\pmod{2}$}\\ \end{array} \right\} \pmod{4}$$

Therefore, we can show that $3^{77}\equiv3\pmod{4}$. Thus, we can determine that $(3^{77}-1)\equiv2\pmod{4}$. Thus, we can show that $\frac{(3^{77}-1)}{2}$ is odd as:

$$\frac{(3^{77}-1)}{2}\equiv\pm1\pmod{4}$$

However, I am unsure how to show that this number is composite. The book I am reading simply states two of the factors, $\frac{(3^{11}-1)}{2}$ and $\frac{(3^{7}-1)}{2}$, but I do not know how the authors discovered these factors.

I'd appreciate any help pointing me in the right direction, thanks.

Another way to see this is by writing the number in base 3:

$$3^{77}=1\underbrace{00\dots00}_{77}\ _3$$

Here the index $3$ denotes base 3, and $77$ is the number of digits. Subtracting one, we get:

$$3^{77}-1=\underbrace{22\dots22}_{77}\ _3$$

Therefore, dividing this by two,

$$\frac{3^{77}-1}{2}=\underbrace{11\dots11}_{77}\ _3$$

From this we can directly read that the number is odd, since it is the sum of 77 odd numbers, and composite, since $$\underbrace{11\dots11}_{77}\ _3=1111111_3\cdot\underbrace{10000001000000\dots100000010000001}_{71}\ _3$$

(Although, this is basically the same as some of the other answers.)

Hint: $$a^n-1=(a-1)(a^{n-1}+a^{n-2}+...+a+1)\,,\,\,\forall n\in\mathbb{N} \wedge \forall a\in\mathbb{R}$$

Added Of course, we also have in this case, applying the above: $$3^{77}-1=\left(3^7\right)^{11}-1=(3^7-1)\left(\left(3^7\right)^{10}+\left(3^7\right)^9+...+3^7+1\right)\,,\,etc.$$ and something similar can be done with $\,3^{77}=\left(3^{11}\right)^7$

Well following your congruences idea we have that $$ 3^{77} \equiv 3^{11} \equiv 1 \pmod{23} $$ So $$ 3^{77} - 1\equiv 0 \pmod{23} $$ Since $2^{-1} \equiv 12 \pmod{23}$, we have that $$ \dfrac{3^{77}-1}{2} \equiv 0 \pmod{23} $$ Hence $23 \mid \dfrac{3^{77}-1}{2}$ and is therefore composite.

Note that if $m=kn$ for some $k\in\mathbb{N}$, then $$a^m-1=(a^n)^k-1=(a^n-1)(a^{n(k-1)}+a^{n(k-2)}+\cdots+a^n+1)$$ so that $a^n-1$ divides $a^m-1$.

Hint $\, $ The sought factors demonstrating compositeness of your number arise very simply from a compositional factorization $\rm\:g\:\!f = g\circ f\:$ of a polynomial, combined with the Factor Theorem.

$$\begin{eqnarray}\rm z\,\ -\,\ c\ & | &\,\rm\ g\:\!(\,z\,) - g\:\!(\,c\,) \\ \rm f(x)\!-\!f(a) &|&\,\rm\ g\:\!f(x) - g\:\!f(a) \\ \rm x^7\, -\, a^7\, & | &\,\rm (x^{7})^{11} - (a^{7})^{11} \\ 3^7-\,1\ & | &\:\!\ (3^{7})^{11} - 1 \end{eqnarray}$$

Note that if we employ the notation $\rm\:x^f = f(x)\:$ (e.g. as in Galois theory) then it is clearer

$$\begin{eqnarray}\rm z\,\ -\,\ c\,\:\! & | &\rm\ (\,z\,)^g - \:\!(\,c\,)^g \\ \rm x^f\ -\ a^f\, &|&\rm\ (x^f)^g - (a^f)^g \\ \rm x^7\, -\, a^7\, & | &\:\rm (x^{7})^{11}\!\! - (a^{7})^{11} \\ 3^7-\,1\ & | &\:\!\ (3^{7})^{11\!} - 1 \end{eqnarray}$$

In the Galois case the exponential notation highlights further structure, e.g. from my post here

$\quad\quad\begin{align}{} \rm g^{\:\sigma^4-1} \;=\;& \rm g^{\:(1\:+\;\sigma\:+\;\sigma^2\:+\;\:\sigma^3)\:(\sigma-1)} \\\\ \iff\quad\quad\rm \frac{\sigma^4 g}g \;=\;& \rm (g \;\: \sigma\:g \;\:\sigma^2 g \;\:\sigma^3 g)^{\sigma - 1} \;=\; \frac{\phantom{g\;\;\:} \sigma\:g \;\;\: \sigma^2 g \;\;\:\sigma^3 g \;\;\:\sigma^4 g}{g \;\;\:\sigma\:g \;\;\:\sigma^2 g \;\;\:\sigma^3 g\phantom{\;\;\:\sigma^4 g}} \\\\ \end{align}$