Prove that $\frac {7^{7^{2024}} +1}{7^{7^{2023}} +1}$ is composite
My steps were to transform $7^{7^{2024}} = 7^{7^{2023}\times 7}$ and substitute $x = 7^{7^{2023}}$ to get $$\frac {x^7+1}{x+1}= x^6-x^5+x^4-x^3+x^2-x+1$$ The problem can probably be solved in a typical NT way, but the lecturer who gave us the problem suggested an algebraic approach, so to try find factors of an expression therefore disproving it being prime.
Recall the following well-known identity :
$$(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$$
Note that, this is not a new finding . Recall IMO 1984, Problem $2$ .
This identity leads to :
$$\small{\begin{align}\frac{x^7+1}{x+1}=(x+1)^6-7x(x^2+x+1)^2\end{align}}$$
Thus, we reduced the problem to $a^2-b^2=(a-b)(a+b)$ .
Inspired by this question and its answer, the following identity can be found there which is verifiable : $$\begin{align*}7^{7^{n+1}} + 1 \;= &\;\;(7^{7^n}+1)\\ &\times\left[(7^{3\cdot7^n} + 3\cdot 7^{2\cdot7^n} + 3\cdot 7^{7^n} + 1)+(7^{(5\cdot7^n+1)/2} + 7^{(3\cdot7^n+1)/2} + 7^{(7^n+1)/2})\right]\\ &\times\left[(7^{3\cdot7^n} + 3\cdot 7^{2\cdot7^n} + 3\cdot 7^{7^n} + 1)-(7^{(5\cdot7^n+1)/2} + 7^{(3\cdot7^n+1)/2} + 7^{(7^n+1)/2})\right],\end{align*}$$
which instantly shows that $\frac{7^{7^{n+1}}+1}{7^{7^n}+1}$ has two factors. One sees that the second factor is the smaller one, and is greater than $1$ for all $n\geq 1$ by comparing appropriate terms of the addend bracket and subtrahend bracket.
Thus, for $n=2023$ the given quantity is composite.
This kind of factorization arises from the more general identity \begin{align} 7^{14k+7} + 1 &= (7^{2k+1}+1)\\ &\times (7^{6k+3} + 3\cdot 7^{4k+2} + 3\cdot 7^{2k+1} + 1) \\ & \times (7^{5k + 3} + 7^{3k+2} + 7^{k+1}). \end{align}
To see how that identity is derived, one needs to note that the identity above belongs to a special class of factorization identities known as Aurifeuillian factorizations. To spot our factorization, one needs to look at the case $b=7$ in the page provided.
Expanding further on these kinds of factorizations requires us to define the cyclotomic polynomial $$ \Phi_n(x) = \prod_{k=1,\gcd(k,n)=1}^n (x-e^{\frac{2\pi i k}{n}}), $$ For example, if $p$ is a prime, then a standard manipulation yields that $$\Phi_p(x) = 1+x+\ldots+x^{p-1}.$$ In your case, what might be more important to observe is that if $n = 2p$ where $p$ is a prime, then $$ \Phi_p(x) = 1-x+x^2-x^3+\ldots + x^{p-1}. $$ Match this to the expression you got when $p=7$. The polynomial $x^6-x^5+x^4-x^3+x^2-x+1$ is nothing but $\Phi_{14}(x)$. If you need help spotting cyclotomic polynomials, at least some cases may be possible to spot, as suggested here.
Aurifeuillian factorizations begin with the observation that you're factorizing a number of the form $\Phi_{n}(k)$ where $n,k$ is an integer. As you can see, the number you're trying to factorize is precisely of that form, where $n = 14$ and $k = 7^{7^{2023}}$.
Once this is done, then one can look at Schinzel's landmark paper.
Schinzel, Andrzej, On primitive prime factors of (a^ n -b^ n), Proc. Camb. Philos. Soc. 58, 555-562 (1962). ZBL0114.02605.
I won't state Schinzel's main result in his language, but instead defer to the following article :
Granville, Andrew; Pleasants, Peter, Aurifeuillian factorization, Math. Comput. 75, No. 253, 497-508 (2006). ZBL1107.11050.
which gives the following statement :
Suppose that $n$ is a composite number not divisible by $8$ and let $N$ be the largest odd factor of $n$ (note that $n=N,2N,4N$ are the only possibilities now). Suppose that $d$ is any squarefree divisor of $N$ (which can be negative if $n=4N$). There exist polynomials $U_{n,d},V_{n,d}$ with integer coefficients such that $$ \Phi_N(x) = U_{N,d}(x)^2 - d(-1)^{(d-1)/2}xV_{N,d}(x)^2, \\ \Phi_{2N}(x) = U_{2N,d}(x)^2 + d(-1)^{(d-1)/2}xV_{2N,d}(x)^2, \\ \Phi_{4N}(x) = U_{4N,d}(x)^2 -2dxV_{4N,d}(x)^2. $$
Let's try this with the illustrative example given by the same paper. Suppose that you take $x = ay^2$ where $a = d(-1)^{(d-1)/2}$ in the first example above. Then, you get :$$ \Phi_N(x) = U_{N,d}(ay^2)^2 - d(-1)^{(d-1)/2}(ay^2)V_{N,d}(ay^2)^2 \\ = (U_{N,d}(ay^2))^2 - (ad(-1)^{(d-1)/2}yV_{N,d}(ay^2))^2 \\ = (U_{N,d}(ay^2) + ad(-1)^{(d-1)/2}yV_{N,d}(ay^2)) \times (U_{N,d}(ay^2) - ad(-1)^{(d-1)/2}yV_{N,d}(ay^2)). $$ From a cursory reading it is not clear if the polynomials can be constructed using the proof or not. Nevertheless, the Wikipedia page on Aurifeuillian factorizations is a decent enough start for any olympiad type problems.
Some degree comparison constraints will tell you that for large enough values of $a,y$, the above factors are both greater than $1$. In other words, $\Phi_N(ay^2)$ is composite for all choices of $y$ and $a = d(-1)^{(d-1)/2}$ where $d$ is any square-free factor of $N$, as long as $n$ is large enough (I am not sure if the factorization is nontrivial for all $n \geq 1$).
Similar results show that $\Phi_{2N}(d(-1)^{\frac{d+1}{2}}y^2)$ and $\Phi_{2N}(2d)$ are also composite for large enough $N$, where $d$ is any square free factor of $N$ and $y$ is arbitrary.
Thus, we obtain a larger class of numbers that can be shown to be composite using the abstract method of Schinzel. This includes the number in question, which is $\Phi_{14}(7^{7^{2023}})$ using the decomposition $$ \Phi_{14}(7^{7^{2023}}) = \Phi_{2 \times 7} \left(7 \times (-1)^{(7+1)/2} \times \left(7^{\frac{7^{2023}-1}{2}}\right)^2\right) $$ and matching this up with the decomposition I gave earlier.
